Achromatism in lenses – Ray Optics, Physics – eSaral

Hey, do you want to learn about achromatism in lenses? If yes. Then keep reading

## Achromatism:

We have just that when a white object is placed in front of a lens, then its images of different colors are formed at different positions and are of different sizes. These defects are called ‘longitudinal chromatic aberration’ and ‘lateral chromatic aberration’ respectively. If two or more lenses be different colors are in the same position and of the same size, then the combination is called ‘achromatic combination of lenses, and this property is called ‘achromatism’.

In practice, both types of chromatic aberrations cannot be removed for all colors. We can remove both types of chromatic aberration only for two colors by placing in contact two lenses of appropriate focal lengths and of an appropriate different material. On the other hand, only lateral chromatic aberration can be removed for all colors when two lenses of an appropriate different material. On the other hand, only lateral chromatic aberration can be removed for all colors when two lenses of the same material are placed at a particular distance apart.

## Condition of Achromatism for two thin lenses in contact:

Suppose two thin lenses are placed in contact. Suppose the dispersive powers of the materials of these lenses between violet and red respectively $n_{V}, n_{R}, n_{y}$ and $\mathrm{n}_{\mathrm{V}}^{\prime}, \mathrm{n}_{\mathrm{R}}^{\prime}, \mathrm{n}_{\mathrm{y}}^{\prime}$. If for these rays the focal lengths of the first lens are respectively $f_{v}, f_{R}, f_{y}$ and the focal lengths of the second lens are $f_{V}^{\prime}, f_{R}^{\prime}, f_{y}^{\prime}$, then for the first lens, we have

$\frac{1}{f_{V}}=\left(n_{V}-1\right)\left[\frac{1}{R_{1}}-\frac{1}{R_{2}}\right]$…..(1)

$\frac{1}{f_{R}}=\left(n_{R}-1\right)\left[\frac{1}{R_{1}}-\frac{1}{R_{2}}\right]$…..(2)

Subtracting the second equation from the first, we get

$\frac{1}{f_{V}}-\frac{1}{f_{R}}=\left(n_{V}-n_{R}\right)\left[\frac{1}{R_{1}}-\frac{1}{R_{2}}\right]$

$=\frac{\left(n_{V}-n_{R}\right)}{\left(n_{y}-1\right)}\left(n_{y}-1\right)\left[\frac{1}{R_{1}}-\frac{1}{R_{2}}\right]$

$\frac{1}{f_{v}}-\frac{1}{f_{r}}=\omega \frac{1}{f_{y}}$….(3)

because $\frac{n_{v}-n_{R}}{n_{y}-1}=\omega$

and $\left(n_{y}-1\right)\left(\frac{1}{R_{1}}-\frac{1}{R_{2}}\right)=\frac{1}{f_{y}}$

Similarly, for the second lens, we have

$\frac{1}{f_{v}^{\prime}}-\frac{1}{f_{R}^{\prime}}=\omega \frac{1}{f_{y}^{\prime}}$….(4)

Adding equations (3) and (4), we get

$\left(\frac{1}{f_{V}}+\frac{1}{f_{V}^{\prime}}\right)-\left(\frac{1}{f_{R}}+\frac{1}{f_{R}^{\prime}}\right)=\frac{\omega}{f_{y}}+\frac{\omega^{\prime}}{f_{y}^{\prime}}$……(5)

If the focal lengths of this lens-combination for the violet and the red rays be $\mathrm{F}_{\mathrm{V}}$ and $\mathrm{F}_{\mathrm{R}}$ respectively, then

$\frac{1}{f_{V}}+\frac{1}{f_{V}^{\prime}}=\frac{1}{F_{V}}$

And

$\frac{1}{f_{R}}+\frac{1}{f_{R}^{\prime}}=\frac{1}{F_{R}}$ from eq. (5), we have $\frac{1}{F_{V}}-\frac{1}{F_{R}}=\frac{\omega}{f_{y}}+\frac{\omega}{f_{y}^{\prime}}$

But for the achromatism of the lens combination, the focal length must be the same for all colors of light i.e. $F_{v}=F_{R}$. Hence from the above equation, we have

$\frac{\omega}{f_{y}}+\frac{\omega^{\prime}}{f_{y}^{\prime}}=0$…..(6)

Or

$\frac{\omega}{f_{y}}=-\frac{\omega^{\prime}}{f_{y}^{\prime}}$…..(7)

This is the condition for a lens combination to achromatic. It gives us the following information

1. Both the lenses should be of a different material. If both the lenses are of the same material, then $\omega=\omega^{\prime}$ and then from equation (6). We have

$\frac{1}{f_{y}}+\frac{1}{f_{y}^{\prime}}=0$

or

$\frac{1}{F_{y}}=0$ or $F_{y}=\infty$

that is the combination will then behave like a plane glass-plate.

2. $\omega$ and $\omega^{\prime}$ are positive quantities. Hence, according to eq. (7), $\mathrm{f}_{\mathrm{y}}$ and $\mathrm{f}_{\mathrm{y}}^{\prime}$ should be of opposite signs, i.e., if one lens is convex, the other should be concave.

3. For the combination of behavior like a convergent (convex) lens–system, the power of the convex lens should be greater than that of the concave lens. On other words, the focal length of the convex lens should be smaller than the concave lens. According to eq. (7), we have

$\frac{\mathrm{f}_{\mathrm{y}}}{\mathrm{f}_{\mathrm{y}}^{\prime}}=-\frac{\omega}{\omega^{\prime}}$ If $\mathrm{f}_{\mathrm{y}}$ is less than $\mathrm{f}_{\mathrm{y}}^{\prime}$

then $\omega$ should be less than $\omega^{\prime}$. Hence for a converging lens system, the convex lens should be made of a material of smaller dispersive power.

The dispersive power of crown glass is smaller than that of flint glass. Hence in an achromatic lens-doublet, the convex lens is of crown glass and the concave lens is of flint glass, and they are cemented together by Canada Balsam (a transparent cement). This achromatic combination is used in optical instruments such as microscopes, telescopes, cameras, etc.

In the condition for achromatism

$\frac{\mathrm{f}_{\mathrm{y}}}{\mathrm{f}_{\mathrm{y}}^{\prime}}=-\frac{\omega}{\omega^{\prime}}$

For a better understanding of this chapter, please check the detailed notes of Ray Optics. To watch Free Learning Videos on physics by Saransh Gupta sir Install the eSaral App.

Sign convention for lens – Concave and Convex Lens – eSaral

Hey, do you want to learn about the sign convention for lens? If yes. Then keep reading.

## Sign Convention

1. Whenever and where possible, rays of light are taken to travel from left to right.

2. The transverse distance measured from the optical center and is taken to be positive while those below it negative.

3. Longitudinal distances are measured from the optical center and are taken to be positive if in the direction of light propagation and negative if opposite to it e.g., according to our convention case of a

While using the sign convention, it must be kept in mind that –

(a) To calculate an unknown quantity the known quantities are substituted with a sign in a given formula.

(b) In the result sign must be interpreted as there are number of sign conventions and the same sign has a different meaning in different conventions.

## Rules for image formation

In order to locate the image formed by a lens graphically following rules are adopted –

1. A ray passing through the optical center proceeds undeviated through the lens. (by definition of optical center).

2. A ray passing through the first focus or directed towards it, after refraction from the lens becomes parallel to the principal axis. (by definition of $F_{1}$)

3. A ray passing parallel to the principal axis after refraction through the lens passes or appears to pass through $\mathrm{F}_{2}$ (by definition of $\mathrm{F}_{2}$).

4. Only two rays from the same point of an object are needed for image formation and the point where the rays after refraction through the lens intersect or appear to intersect is the image of the object. If they actually intersect each other the image is real and if they appear to intersect the image is said to be virtual.

So, that’s all from this article. I hope you get the idea about the Sign convention for the lens. If you found this article informative then please share it with your friends. If you have any confusion related to this topic, then you can ask in the comments section down below.

For a better understanding of this chapter, please check the detailed notes of Ray Optics. To watch Free Learning Videos on physics by Saransh Gupta sir Install the eSaral App.

Reflection through Concave Mirror – Ray Optics – eSaral

Hey, do you want to learn about Reflection through Concave Mirror? If yes. Then keep reading.

## Reflection through the concave mirror:

F = Principal focus

P = Pole of mirror

C = Centre of curvature.

PF = Focal length.

When a narrow beam of light traveling parallel to the principal axis is incident on the reflecting surface of the concave mirror, the beam after reflection converges at a point on the principal axis.

## Rules for ray diagrams:

1. When a ray falls in the direction of the center of curvature of the mirror then it reflects back along the same path.
2. A ray, parallel to the principal axis will after reflection, pass through the focus.
3. A ray, passing through the focus is reflected parallel to the principal axis.

### Image formed by the concave mirror:

For a better understanding of this chapter, please check the detailed notes of Ray Optics. To watch Free Learning Videos on physics by Saransh Gupta sir Install the eSaral App.

Difference between Compound Microscope and Astronomical Telescope – eSaral

Hey Do you want to know the difference between Compound Microscope and Astronomical Telescope? If yes. Then you are at the right place.

So, that’s all from this article. I hope you get the idea about the difference between Compound Microscope and Astronomical Telescope. If you found this article informative then please share it with your friends. If you have any confusion related to this topic then you can ask in the comments section down below.

For a better understanding of this chapter, please check the detailed notes of Ray Optics. To watch Free Learning Videos on physics by Saransh Gupta sir Install the eSaral App.

Working of Telescope – Ray Optics, Class 12 – eSaral

A Telescope is an optical instrument used to increase the visual angle of distant large objects such as a star a planet. If you want to learn about the Working of a Telescope then keep reading.

## Telescope

It is an optical instrument used to increase the visual angle of distant large objects such as a star a planet or a cliff etc. An astronomical telescope consists of two converging lenses. The one facing the object is called an objective or field – lens and has a large focal length and aperture, the distance between the two lenses is adjustable.

A telescope is used to see distant objects, in its object is between $\infty$ and 2F of objective and hence image formed by objective is real, inverted, and diminished and is between F and 2F on the other side of it. This image (called intermediate image) acts as an object for the eyepiece and shifting the position of the eye-piece is brought within its focus. So final image I, with respect to the intermediate image, is erect, virtual, enlarged, and at a distance D to $\infty$ from the eye. This in turns implies that the final image with respect to the object is inverted, enlarged, and at a distance D to $\infty$ from the eye.

Magnifying Power (MP)

magnifying Power of a telescope is defined as

$\mathrm{MP}=\frac{\text { Visual angle with instrument }}{\text { Visual angle for unadded eye }}$

$=\frac{\theta}{\theta_{0}}$

But from fig.

$\theta_{0}=\left(y / f_{0}\right)$ and $\theta=\left(y /-u_{e}\right)$

So $\mathrm{MP}=\frac{\theta}{\theta_{0}}=-\left[\frac{\mathrm{f}_{0}}{\mathrm{u}_{\mathrm{e}}}\right]$ with length of tube

$\mathrm{L}=\left(\mathrm{f}_{0}+\mathrm{u}_{\mathrm{e}}\right)$…….(1)

Now there are two possibilities –

### (a) If the final image is at infinity (far point)

This situation is called normal adjustment as in this situation eye is least strained or relaxed. In this situation as for eye – piece v = $\infty$

$\frac{1}{-\infty}-\frac{1}{u_{e}}=\frac{1}{f_{e}}$

i.e. $u_{e}=f_{e}$

So substituting this value of ue in Eqn. (1) we have

$\mathrm{MP}=-\left(\mathrm{f}_{0} / \mathrm{f}_{\mathrm{e}}\right)$

and $L=\left(f_{0}+f_{e}\right)$…..(2)

Usually, the telescope operates in this mode unless stated otherwise. In this mode, as ue is maximum for a given telescope MP is minimum while the length of tube maximum. In this case, the object and final image are at infinity so both total lights entering and leave the telescope are parallel to its axis as shown in fig.

### (b) If the final image is at D (Near point)

In this situation as for eye – piece v = D

$\frac{1}{-D}-\frac{1}{-u_{e}}=\frac{1}{f_{e}}$

i.e., $\frac{1}{-u_{e}}=\frac{1}{f_{e}}\left[1+\frac{f_{e}}{D}\right]$

So substituting this value of ue in Eqn. (1), we have

$\mathrm{MP}=\frac{\mathrm{f}_{0}}{\mathrm{f}_{\mathrm{e}}}\left[1+\frac{\mathrm{f}_{\mathrm{e}}}{\mathrm{D}}\right]$

With

$L=f_{0}+\frac{f_{e} D}{f_{e}+D}$…..(3)

In this situation $\mathrm{u}_{\mathrm{e}}$ is minimum so for a given telescope MP is maximum while the length of tube minimum and the eye is most strained. In the case of a telescope if the object and final image are at infinity so both total lights entering and leave the telescope are parallel to its axis as shown in fig.

Note:

$\frac{\mathrm{f}_{0}}{\mathrm{f}_{\mathrm{e}}}=\frac{\text { Aperture of object }}{\text { Aperture of eye piece }}$

i.e.,

$\mathrm{MP}=\frac{\mathrm{f}_{0}}{\mathrm{f}_{\mathrm{e}}}=\frac{\mathrm{D}}{\mathrm{d}}$

For a better understanding of this chapter, please check the detailed notes of Ray Optics. To watch Free Learning Videos on physics by Saransh Gupta sir Install the eSaral App.

What is a compound microscope – Ray Optics – eSaral

## Compound Microscope

### Construction:

It consists of two convergent lenses of short focal lengths and apertures arranged co-axially lens (of focal length $\mathrm{f}_{\mathrm{O}}$) facing the object is called objective or field lens while the lens (of focal length $\mathrm{f}_{\mathrm{e}}$) facing the eye, eye-piece or ocular. The objective has a smaller aperture and smaller focal length than the eye-piece. the separation between objective and eye-piece can be varied.

### Image Formation:

The object is placed between F and 2F of the objective so the image IM formed by objective (called intermediate image) is inverted, real enlarged and at a distance greater than $\mathrm{f}_{0}$ on the other side of the lens. This image IM acts as an object for the eye-piece and is within its focus. So, eye-piece forms final image I which is erect, Virtual, and enlarged with respect to intermediate image $\mathrm{I}_{\mathrm{M}}$. So, the final image I with respect to the object is Inverted, virtual, enlarged, and at a distance D to from eye on the same side of eye-piece as $\mathrm{I}_{\mathrm{M}}$. This all is shown in fig.

Magnifying power (MP) magnifying Power of an optical instrument is defined as –

$\mathrm{MP}=\frac{\text { Visual angle with instrument }}{\text { Max } \text { Visual angle for unadded eye }}$

$=\frac{\theta}{\theta_{0}}$

If the size of the object is h, and the least distance of distinct vision is D

$\theta_{0}=\frac{\mathrm{h}}{\mathrm{D}}$

$\theta=\frac{\mathrm{h}^{\prime}}{\mathrm{u}_{\mathrm{e}}}$ $\mathrm{MP}=\frac{\theta}{\theta_{0}}=\left[\frac{\mathrm{h}^{\prime}}{\mathrm{u}_{\mathrm{e}}}\right] \mathrm{x}\left[\frac{\mathrm{D}}{\mathrm{h}}\right]=\left[\frac{\mathrm{h}^{\prime}}{\mathrm{h}}\right]\left[\frac{\mathrm{D}}{\mathrm{u}_{\mathrm{e}}}\right]$

But for objective

$\mathrm{m}=\frac{1}{\mathrm{O}}=\frac{\mathrm{v}}{\mathrm{u}}$

i.e., $\quad \frac{\mathrm{h}^{\prime}}{\mathrm{h}}=-\frac{\mathrm{v}}{\mathrm{u}}[$ as $\mathrm{u}$ is – ve $]$

so $\mathrm{MP}=-\frac{\mathrm{v}}{\mathrm{u}}\left[\frac{\mathrm{D}}{\mathrm{u}_{\mathrm{e}}}\right]$

with the length of the tube

$L=v+u_{e}$……(1)

now there are two possibilities ­–

($\mathbf{b}_{1}$) If the final image is at infinity (far point) :

This situation is called normal adjustment as in this situation eye is least strained or relaxed. In this situation as for eye – piece $v=\infty$

$\frac{1}{-D}-\frac{1}{-u_{e}}=\frac{1}{f_{e}}$

i.e., $\frac{1}{\mathrm{u}_{\mathrm{e}}}=\frac{1}{\mathrm{D}}\left[1+\frac{\mathrm{D}}{\mathrm{f}_{\mathrm{e}}}\right]$

Substituting this value of ue in Eqn. (1), we have

$\mathrm{MP}=-\frac{\mathrm{v}}{\mathrm{u}}\left[1+\frac{\mathrm{D}}{\mathrm{f}_{\mathrm{e}}}\right]$

with $L=v+\frac{f_{e} D}{f_{e}+D}$…….(3)

In this situation as $\mathrm{u}_{\mathrm{e}}$ is minimum, MP is maximum and the eye is most strained.

In microscope focal length of objective lens $f_{0}$ is small and object is placed very closed to the objective lens so–

$\mathrm{u} \approx \mathrm{f}_{\circ}$ $\mathrm{L}=\mathrm{v}+\mathrm{u}_{\mathrm{e}}$

$\left[u_{e}<<v\right]$ $\mathrm{L} \approx \mathrm{v}$ $|\mathrm{MP}|=\frac{\mathrm{LD}}{\mathrm{f}_{0} \mathrm{f}_{e}}$

Discussion :

1. The magnifying power of a microscope is negative so it produces final image always inverted.

2. $\mathrm{m}_{\mathrm{e}}=\frac{\mathrm{v}_{\mathrm{e}}}{\mathrm{u}_{\mathrm{e}}}=\frac{\mathrm{D}}{\mathrm{u}_{\mathrm{e}}}=\left[1+\frac{\mathrm{D}}{\mathrm{f}_{\mathrm{e}}}\right]$,

$\mathrm{m}_{0}=\frac{\mathrm{v}_{0}}{\mathrm{u}_{0}}=\frac{\mathrm{v}}{\mathrm{u}}$

$\mathrm{MP}=\mathrm{m}_{0} \times \mathrm{m}_{\mathrm{e}}$

3. $(\mathrm{MP})_{\min }=-\frac{\mathrm{V}}{\mathrm{u}} \frac{\mathrm{D}}{\mathrm{f}_{\mathrm{e}}}$

$(\mathrm{MP})_{\max }=-\frac{\mathrm{v}}{\mathrm{u}}\left[1+\frac{\mathrm{D}}{\mathrm{f}_{\mathrm{e}}}\right]$

4. $|\mathrm{MP}|=-\frac{\mathrm{LD}}{\mathrm{f}_{0} \mathrm{f}_{\mathrm{e}}}$

5. MP does not change appreciably if the objective lens and eye-piece are interchanged

$\left[\mathrm{MP} \sim\left(\mathrm{LD} / \mathrm{f}_{0} \mathrm{f}_{\mathrm{e}}\right)\right]$.

MP is increased by decreasing the focal length of both lenses.

6. If the distance between the eye and eyepiece lens is D then distance between final image in eye-piece lens D’ = (D – d)

$\mathrm{MP}=\frac{\mathrm{LD}^{\prime}}{\mathrm{f}_{0} \mathrm{f}_{\mathrm{e}}}=\frac{\mathrm{L}(\mathrm{D}-\mathrm{d})}{\mathrm{f}_{0} \mathrm{f}_{\mathrm{e}}}$

$=\mathrm{MP}\left[1-\frac{\mathrm{d}}{\mathrm{D}}\right]<\mathrm{MP}$

7. $\mathrm{RP}=\frac{1}{\mathrm{RL}} \alpha \frac{1}{\lambda}$ (Resolving power)

8. In electron microscope $\lambda=\sqrt{(150 / \mathrm{V})} \mathrm{A}$

where : V = Potential difference

So, that’s all from this article. I hope you get the idea about what is a compound microscope. If you liked this explanation then please share it with your friends. If you have any confusion related to this topic then feel free to ask in the comments section down below.

For a better understanding of this chapter, please check the detailed notes of  Chapter Ray Optics. To watch Free Learning Videos on physics by Saransh Gupta sir Install the eSaral App.

What is simple microscope – Construction, Working – eSaral

Hey, do you want to know what is simple microscope? If yes. Then keep reading

## Microscope

It is an optical instrument used to increase the visual angle of neat objects which are too small to be seen by the naked eye.

## Simple Microscope

It is also known as a magnifying glass or simply magnifier and consists of a convergent lens with the object between its focus and optical center and eye close to it. The image formed by it is erect, virtually enlarged and on the same side of the lens between object and infinity. The magnifying power (MP) or angular magnification of a simple microscope (or an optical instrument) is defined as the ratio of visual angle with an instrument to the maximum visual angle for clear vision when the eye is unadded (i.e., when the object is at least distance of distinct vision)

i.e.,

$\mathrm{MP}=\frac{\text { Visual angle with instrument }}{\text { Max. visual angle for unadded eye }}$

$=\frac{\theta}{\theta_{0}}$

If an object of size h is placed at a distance u (<D) from the lens and its image size h’ is formed at a distance V (> D) from the eye

$\theta=\frac{h^{\prime}}{v}=\frac{h}{u}$

with $\theta_{0}=\frac{\mathrm{h}}{\mathrm{D}}$

So $\quad \mathrm{MP}=\frac{\theta}{\theta_{0}}=\frac{\mathrm{h}}{\mathrm{u}} \times \frac{\mathrm{D}}{\mathrm{h}}=\frac{\mathrm{D}}{\mathrm{u}}$….(1)

Now there are two possibilities ­

$\left(a_{1}\right)$ If their image is at infinity [Far point]

In this situation from lens formula –

$\frac{1}{v}-\frac{1}{u}=\frac{1}{f}$

we have $\frac{1}{\infty}-\frac{1}{-\mathrm{u}}=\frac{1}{\mathrm{f}}$

i.e., $\mathrm{u}=\mathrm{f}$

So $M P=\frac{D}{U}=\frac{D}{f}$……(2)

As here u is maximum [as object is to be with in focus], MP is minimum and as in this situation parallel beam of light enters the eye, eye is least strained and is said to be normal, relaxed or unstrained.

$\left(a_{2}\right)$ If the image is at D [Near point]

In this situation as v = D, from lens formula

$\frac{1}{v}-\frac{1}{u}=\frac{1}{f}$

we have $\frac{1}{-\mathrm{D}}-\frac{1}{-\mathrm{u}}=\frac{1}{\mathrm{f}}$

i.e., $\frac{D}{u}=1+\frac{D}{f}$

So $\mathrm{MP}=\frac{\mathrm{D}}{\mathrm{u}}=\left[1+\frac{\mathrm{D}}{\mathrm{f}}\right]$….(3)

As the minimum value of v for clear vision is D, in this situation u is minimum and hence this is the maximum possible MP of a simple microscope and as in this situation final image is closest to the eye, the eye is under maximum strain.

Special POints :

1. A simple magnifier is an essential part of most optical instruments (such as a microscope or telescope) in the form of eyepiece or ocular.

2. The magnifying power (MP) has no unit. It is different from the power of a lens which is expressed in diopter (D) and is equal to the reciprocal of focal length in meter.

3. With the increase in wavelength of light used, the focal length of the magnifier will increase, and hence its MP will decrease.

So, that’s all from this article. I hope you get the idea about what is a simple microscope. If you found this article informative then please share it with your friends. If you have any confusion related to this topic then you can ask in the comments section down below.

For a better understanding of this chapter, please check the detailed notes of Ray Optics. To watch Free Learning Videos on physics by Saransh Gupta sir Install the eSaral App.

Combination of lenses and mirrors – Ray Optics – eSaral

Hey, do you want to learn about the Combination of lenses and mirrors? If yes. Then you are at the right place.

## Combination of Lenses and Mirror:

When several lenses or mirrors are used co-axially, the image formation is considered one after another in steps. The image formed by the lens facing the object serves as an object for the next lens or mirror the image formed by the second lens (or mirror) acts as an object for the third and so on. The total magnification is such situations will be given by

$\mathrm{m}=\frac{\mathrm{I}}{\mathrm{o}}=\frac{\mathrm{I}_{1}}{\mathrm{o}} \times \frac{\mathrm{I}_{2}}{1_{1}} \times \ldots$

i.e.$\quad m=m_{1} \times m_{2} \times \ldots \ldots$

In the case of two thin lenses in contact if the first lens of focal length $\mathrm{f}_{1}$ forms the image $\left.\right|_{1}$ (of an object)at a distance $\mathrm{V}_{1}$ from it.

$\frac{1}{v_{1}}-\frac{1}{u}=\frac{1}{f_{1}}$ ……(1)

now the image $\left.\right|_{1}$ will act as an object for the second lenses and if the second lenses form an image. I at a distance v from it

$\frac{1}{v}-\frac{1}{v_{1}}=\frac{1}{f_{2}}$……(2)

So adding Eqn. (1) and (2) we have

$\frac{1}{v}-\frac{1}{u}=\frac{1}{f_{1}}+\frac{1}{f_{2}}$

or $\frac{1}{v}-\frac{1}{u}=\frac{1}{f}$

with $\quad \frac{1}{f}=\frac{1}{f_{1}}+\frac{1}{f_{2}}$

i.e. the combination behaves as a single lens of equivalent focal length f given by –

$\frac{1}{F}=\frac{1}{f_{1}}+\frac{1}{f_{2}}$

or $P=P_{1}+P_{2}$…..(3)

note : If the two thin lenses are separated by a distance d apart F is given by

$\frac{1}{F}=\frac{1}{f_{1}}+\frac{1}{f_{2}}-\frac{d}{f_{1} f_{2}}$

or $P=P_{1}+P_{2}-P_{1} P_{2} d$

here is it worthy to note that –

Special points:

1. If two thin lenses of equal focal length, but of opposite nature (i.e. one convergent and other divergent) are put in contact, the resultant focal length of the combination be

$\frac{1}{\mathrm{~F}}=\frac{1}{+\mathrm{f}}+\frac{1}{-\mathrm{f}}=0$

i.e.

$F=\infty$ and $P=0$ i.e. the system will behave like a plane. glass plate.

2. If two thin lens of the same nature are put in contact then as

$\frac{1}{\mathrm{~F}}=\frac{1}{\mathrm{f}_{1}}+\frac{1}{\mathrm{f}_{2}}$

$\frac{1}{F}>\frac{1}{f_{1}}$ and $\frac{1}{\mathrm{~F}}>\frac{1}{\mathrm{f}_{2}}$

i.e. $\quad F<f_{1}$ and $F<f_{2}$

i.e. the resultant focal length will be lesser than the smallest individual.

3. If two thin lenses of opposite nature with different focal lengths are put in contact the resultant focal length will be of the same nature as that of the lens of shorter focal length but its magnitude will be more than that of shorter focal length.

4. If a lens of focal length f is divided into two equal parts as in figure (A) each part has a focal length f then as

$\frac{1}{f}=\frac{1}{f^{\prime}}+\frac{1}{f^{\prime}}$ $f^{\prime}=2 f$

i.e. each part have focal length 2f now if these parts are put in contact as in (B) or (C) the resultant focal length of the combination will be

$\frac{1}{F}=\frac{1}{2 f}+\frac{1}{2 f}$ i.e. $F=f($ =initial value $)$

5. If a lens of focal length f is cut in two equal parts as shows in each will have focal length f. Now of these parts are put in contact as shown in the resultant length will be

$\frac{1}{\mathrm{~F}}=\frac{1}{\mathrm{f}}+\frac{1}{\mathrm{f}}$

i.e. $\mathrm{F}=(\mathrm{f} / 2)$

However if the two parts are put in contact as shown in first will behave as the convergent lens of focal length f while the other divergent of same focal length (being thinner near the axis) so in this situation.

$\frac{1}{F}=\frac{1}{+f}+\frac{1}{-f}$

i.e. $-F=\infty$ or $P=0$

So, that’s all from this article. I hope you get the idea about the Combination of lenses and mirrors. If you found this article informative and valuable then please share it with your friends. If you have any confusion related to this topic then you can ask in the comments section down below.

For a better understanding of this chapter, please check the detailed notes of Physics, Ray Optics. To watch Free Learning Videos on physics by Saransh Gupta sir Install the eSaral App.

What is total internal reflection of light – Physics – eSaral

Hey, do you want to know What is a total internal reflection of light? If yes. Then you are at the right place.

## Total internal Reflection

In case of refraction of light, from Snell’s law, we have

$\mu_{1} \sin \mathrm{i}=\mu_{2} \sin \mathrm{r}$……(1)

If the light is passing from denser to the rarer medium through a plane boundary then

$\mu_{1}=\mu_{\mathrm{D}}$

and

$\mu_{2}=\mu_{\mathrm{R}}$

so with $\mu=\left(\mu_{\mathrm{D}} / \mu_{\mathrm{R}}\right)$

$\sin \mathrm{i}=\frac{\mu_{\mathrm{R}}}{\mu_{\mathrm{D}}} \sin \mathrm{r}$

i.e. $\sin \mathrm{i}=\frac{\sin \mathrm{r}}{\mu}$…..(2)

i.e.$\sin \mathrm{i} \alpha \sin \mathrm{r}$

with $(\angle \mathrm{i})<(\angle \mathrm{r})($ as $\mu>1)$

So as the angle of the incident I increase the angle of refraction r will also increase and for a certain value of i $\left(\angle 90^{\circ}\right)$ r will become $90^{\circ}$. The value of angle of incidence for which $r=90^{\circ}$ is called critical angle and is denoted by $\theta_{\mathrm{c}}$. From eq. (2)

$\sin \theta_{C}=\frac{\sin 90}{\mu}$

i.e., $\sin \theta_{\mathrm{C}}=\frac{1}{\mu}$…..(3)

And hence eqn. (2) in terms of critical angle can be written as

$\sin \mathrm{i}=\sin \mathrm{r} \mathrm{X} \sin \theta_{\mathrm{c}}$

$\sin r=\frac{\sin i}{\sin \theta_{c}}$…..(4)

So if $i>\theta_{c} \sin r>1$. This means that r is imaginary (as the value of sin of any angle can never be greater than unit) physically this situation implies that refracted ray does not exist. So, the total light incident of the boundary will be reflected back into the same medium from the boundary. This phenomenon is called total internal reflection.

Special Note:

1. For total internal reflection to take place light from air to water (or glass) and from water to glass total internal reflection can-never take place.

2. When light is passing from denser to rare medium total internal reflection will take place only if the angle of incidence is greater than a certain value called critical angle given by

$\theta_{\mathrm{C}}=\sin ^{-1}\left[\frac{1}{\mu}\right]$

with

$\mu=\frac{\mu_{\mathrm{D}}}{\mu_{\mathrm{R}}}$

3. In the case of total internal reflection as all (i.e. 100%) incident light is reflected back into the same medium there is no loss of intensity while in the case of reflection from mirror or refraction from lenses there is some loss of intensity as all light can never be reflected or refracted. This is why image formed by TIR is much bright than formed by mirror or lenses.

## Critical angle ($\theta_{\mathrm{c}}$)

In case of propagation of light from denser of rate medium through a plane boundary, the critical angle is the angle of incidence for which angle of refraction is $90^{\circ}$ and so from Snell’s law

$\mu \sin i=\mu_{2} \sin r$

$\mu_{\mathrm{D}} \sin \theta_{\mathrm{C}}=\mu_{\mathrm{R}} \sin 90$

$\sin \theta_{\mathrm{C}}=\frac{\mu_{\mathrm{R}}}{\mu_{\mathrm{D}}}=\left[\frac{1}{\mu}\right]$

with $\mu=\frac{\mu_{\mathrm{D}}}{\mu_{\mathrm{R}}}$

or $\theta_{C}=\sin ^{-1}\left(\frac{1}{\mu}\right)$

1. For a given pair of medium critical angle depends on wavelength of light used i.e., greater the wavelength of light lesser will be $\mu$ $\left[\right.$ as a $\left.\mu \alpha \frac{1}{v} \alpha \frac{1}{\lambda}\right]$

and so greater will be the critical angle. This is why the critical angle is maximum for red and minimum for violet rays.

For a given light is depends on the nature of the pair of medium lesser the greater will the critical angle and vice–versa. This is why in the case of

Some illustration of total internal reflection:

1. Shining of air bubble
2. Sparking of diamond
3. Optical – fiber
4. Action of ‘Porro’ prism
5. Duration of suns visibility
6. Mirrage and looming

So, that’s all from this article. I hope you get the idea about What is a total internal reflection of light. If you found this article informative then please share it with your friends. If you have any confusion related to this topic then you can ask in the comments section down below.

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Refraction through a rectangular glass slab – Physics – eSaral

Hey, do you want to learn about Refraction through a rectangular glass slab? If so. Then keep reading.

## Refraction through slab:

The Refractive index & thickness of the glass slab is µ & t respectively. One light ray AB incidents on a slab, Displacement produced, in emergent ray due to refraction. $x=\frac{t \sin (i-r)}{\cos r}=t \sec r \sin (i-r)$

### (a) When an object is in denser medium & observer in rarer medium:

The thickness of a denser medium is t, in which an object is at a point O. Due to refraction, the image may be seen at a point I. Refractive index

$\mu=\frac{\text { Real depth }}{\text { Virtualdepth }}=\frac{\mathrm{AO}}{\mathrm{Al}}=\frac{\mathrm{t}}{\mathrm{Al}}$

Virtual depth $=\frac{t}{\mu}$

Virtual displacement $(\mathrm{OI})=\mathrm{OA}-\mathrm{Al}=\mathrm{t}\left(1-\frac{1}{\mathrm{\mu}}\right)$

### (b) Refraction through a successive slab of different thickness & refractive index.

Virtual depth $(\mathrm{Al})=\frac{\mathrm{t}_{1}}{\mu_{1}}+\frac{\mathrm{t}_{2}}{\mu_{2}}+\frac{\mathrm{t}_{3}}{\mu_{3}}+\ldots$

Virtual displacement (Ol) $=\mathrm{t}_{1}\left(1-\frac{1}{\mu_{1}}\right)+\mathrm{t}_{2}\left(1-\frac{1}{\mu_{2}}\right)+\mathrm{t}_{3}\left(1-\frac{1}{\mu_{3}}\right)+$…………

### (c) When object & observer both are in rarer medium.

Let observer is in air & object is at a point O in air, as shown in the figure. A glass is there in between observer & object. Images forms at point I Refractive index of glass is $\mu$. Virtual displacement $=\mathrm{Ol}=\left(\mathrm{t}-\frac{1}{\mathrm{\mu}}\right)$

### (d) When an object in a rarer medium & Observer in the denser medium.

The Refractive index of water is $\mu .$ Observer is in water, Image may be seen at a point I when an object at a point O is viewed.

$\frac{\text { Re al height }}{\text { Virtual height }}=\frac{1}{\mu}$

Virtual displacement $(\mathrm{O} \mid)=\mathrm{Al}-\mathrm{AO}=(\mu-1) \mathrm{AO}$

So, that’s all from this article. I hope you get the idea about Refraction through a rectangular glass slab. If you found this article informative then please share it with your friends. If you have any confusion related to this topic then feel free to ask in the comments section down below.

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Snell’s law of refraction – Ray Optics, Physics – eSaral

Hey, do you want to learn about Snell’s law of refraction? If yes. Then keep reading.

## Snell’s Law:

For any two media and for the light of a given wavelength, the ratio of the sine of the angle of incidence to the sine of the angle of refraction is a constant.

$\frac{\operatorname{Sin} \mathrm{i}}{\operatorname{Sin} \mathrm{r}}=$ constant
,

Where i = incidence angle
r = refraction angle.

The incident ray, the refracted ray, and the normal at the incident point all lie in the same plane. Frequency (and hence the color) and phase do not change (while wavelength and velocity changes)

## Application of Snell’s Law:

1. When light passes from rarer to denser medium it bends toward the normal. Using Snell’s Law $\mu_{1} \sin \theta_{1}=\mu_{2} \sin \theta_{2}$ $\frac{\sin \theta_{1}}{\sin \theta_{2}}=\frac{\mu_{2}}{\mu_{1}}$ Thus If $\mu_{2}>\mu_{1}$ then $\theta_{2}<\theta_{1}$

2. When light passes from denser to rarer medium it bends away from the normal. From Snell’s law

$\frac{\sin \theta_{1}}{\sin \theta_{2}}=\frac{\mu_{2}}{\mu_{1}}$ Thus If $\mu_{2}<\mu_{1}$ then $\theta_{2}>\theta_{1}$

3. When light propagates through a series of layers of different medium, then according to Snell’s law

$\mu_{1} \sin \theta,=\mu_{2} \sin \theta_{2}=\mu_{3} \sin \theta_{3}=\ldots \ldots \ldots .=$ constant

4. Conditions of no refraction

(a) If light is incident normally on a boundary i.e., $\angle \mathrm{i}=0^{\circ}$ Then from Snell’s law

$\mu_{1} \sin 0=\mu_{2} \sin r$

$\sin r=0$ i.e. $\angle \mathrm{r}=0$

i.e., light passes undeviated from the boundary.

(so boundary will be invisible)

(b) If the refractive indices of two media are equal i.e., if, $\mu_{1}=\mu_{2}=\mu$

Then from Snell’s law

$\mu_{1} \sin \mathrm{i}=\mu \sin r$

$\Rightarrow \angle \mathrm{i}=\angle \mathrm{r}$ i.e.

ray passes undeviated from the boundary with $\angle \mathrm{i}=\angle \mathrm{r} \neq 0$ and boundary will not be visible.

This is also why a transparent solid is invisible in a liquid if $\mu_{s}=\mu_{L}$

5.  Relation between object and image distance :

An object O placed in first medium (refractive index $\mu_{1}$) is viewed from the second medium (refractive index $\mu_{2}$). Then the image distance $\mathrm{d}_{\mathrm{AP}}$ and the object

distance $\mathrm{d}_{\mathrm{AC}}$ are related as

$\mathrm{d}_{\mathrm{AP}}=\left(\frac{\mu_{2}}{\mu_{1}}\right) \mathrm{d}_{\mathrm{AC}}$

(a.)  If $\mu_{2}>\mu_{1}$, i.e., when the object is observed from a denser medium, it appears to be farther away from the interface, i.e. $\mathrm{d}_{\mathrm{AP}}>\mathrm{d}_{\mathrm{AC}}$

(b.) If $\mu_{2}<\mu_{1}$, i.e., when the object is observed from a rarer medium, it appears to be closer to the interface, i.e.

$\mathrm{d}_{\mathrm{AP}}<\mathrm{d}_{\mathrm{AC}}$

Note: The above formula is applicable only for normal view or paraxial ray assumption.

6. Relation between object and image Velocities:

(a.) If an object O moves toward the plane boundary of a denser medium then the image appears to be farther but moves faster to an observer in the denser medium. If $v_{0}=v$ then $v_{1}=\mu v$ Where $v_{0} \& v_{1}$ represents object and image velocities respectively.

(b.) If an object O moves toward the plane boundary of a rarer medium then the image appears to be closer but moves slower to an observer in rarer medium. If $v_{0}=v$ then $v_{1}=v / \mu$

7. Deviation (d):

(a.) A light ray traveling from a denser to a rarer medium at an angle $\alpha<\theta_{\mathrm{C}}$ then deviation. $\delta=\beta-\alpha=\sin ^{-1}(\mu \sin \alpha)-\alpha$

and $\delta_{\max }=\frac{\pi}{2}-\theta_{\mathrm{c}}$

(b.)  If the light is incident at an angle $\alpha>\theta_{C}$, Then the angle of deviation is $\delta=\pi-2 \alpha$ and $\delta_{\max }=\pi-2 \theta_{c}$

(c.) Graphically the relation between d & a can be shown as

So, that’s all from this article. I hope you get the idea about Snell’s law of refraction. If you found this article informative then please share it with your friends. If you have any confusion related to this topic then feel free to ask in the comments section down below.

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Reflection by plane mirror – Ray Optics, Physics – eSaral

Hey, do you want to learn about the Reflection by a plane mirror? If yes. Then keep reading

## Reflection by plane mirror:

1. The image formed by the plane mirror is always erect, of the same size, and at the same distance as the object is.
2. To see the full image in a plane mirror, its length is just half the height of the man and it has to be kept in a specific position.
3. When the plane mirror or any reflecting surface is turned through an angle $\theta$ and if the incident ray remains stationary, then the reflected ray will turn through $2 \theta$.
4. The image of an object formed by a plane mirror is perverted & if the object is real then a virtual image formed.
5. Image due to a plane mirror is as far behind the mirror as is the object in Infront of it.
6. The magnification produced by the plane mirror is 1 i.e. the size of the image is equal to the size of the object.
7. When the two plane mirrors are parallel to each other, the number of images is infinity.
8. When two plane mirrors is held at angle $\theta$ with their reflecting surfaces facing each other and an object is placed between them, images are formed by successive reflections.

First of all, we will calculate

$n=\frac{360}{\theta}$

Then
9. If the angle between the two mirrors is $\theta$, the deviation produced by successive reflections is

$\delta=\delta_{1}+\delta_{2}=2 \pi-2 \theta$

Note : When reflection takes place by a smooth surface, it is called regular reflection, but when reflection takes place by a rough surface, it is called diffused reflection.

So, that’s all from this article. I hope you get the idea about Reflection by a plane mirror. If you found this article informative then please share it with your friends. If you have any confusion related to this topic then feel free to ask in the comments section down below.

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Young double slit experiment class 12 – Physics – eSaral

Hey, do you want to learn about the Young double-slit experiment? If yes. Then keep reading.

## The experiment of young’s Dual-slit

• This experiment shows interference of light
• $s_{1} \& s_{2}$ slit behaves like two coherent sources.
• On-Screen, Bright & Dark portions are alternatively found.
• The bright portion is called Bright fringe & the dark portion is called dark fringe.
• The central fringe is always bright.
• Energy conserves in interference of light.
• It is explained on the basis of hygen principle.
• This experiment verifies the wave nature of light.

1. At a point on screen to find dark or bright fringe, it depends upon the path difference between $\mathrm{s}_{1} \mathrm{P}$ & $\mathrm{s}_{2} \mathrm{P}$ light waves.

2. Two types of path difference between light waves:
(i) Geometrical path difference.
(ii) Optical path difference.

In above experiment, optical difference s1P & s2P has geometrical path difference so, Total path difference = Geometrical difference

$\mathrm{s}_{1} \mathrm{P}-\mathrm{s}_{2} \mathrm{P}=\frac{\mathrm{xd}}{\mathrm{D}}$ (note $-\sin \theta \approx \tan \theta \approx Q)$

• Bright Fringes: If nth bright fringe forms at point P, then for Bright fringe:

$\mathrm{S}_{2} \mathrm{P}-\mathrm{S}_{1} \mathrm{P}=\frac{\mathrm{X}_{\mathrm{n}} \mathrm{d}}{\mathrm{D}}=\mathrm{n} \lambda$

$X_{n}=\frac{n \lambda D}{d}$

So distance between central fringe & nth bright fringe :

$X_{n}=\frac{n \lambda D}{d}$

n = 1 first bright fringe
n = 2 Second bright fringe

• Dark Fringe: If nth dark fringe forms at a point P, then for dark fringe.

$\frac{X_{n} d}{D}=\frac{(2 n-1)}{2} \lambda \mid$

distance between central fringe & nth fringe.

$X_{n}=\frac{(2 n-1) \lambda D}{2 d}$
n = 1 First bright fringe.
n = 2 Second bright fringe.

Distance between dark & bright fringe which are incoming orders is called fringe width. In Interference Fringe width of dark & bright fringes are the same $\beta=\frac{\lambda D}{d}$

• The angular width of the fringe $\alpha \quad D=\beta$ $\alpha=\beta / D=\frac{\lambda}{d}$

## Comments on young’s interference experiment:

1. Energy is conserved in interference. This indicated that energy is redistributed from the destructive interference region to the constructive interference region.

2. If the entire arrangement of young’s double-slit experiment is immersed in water then fringe width decreases $\frac{\beta_{\text {water }}}{\beta_{\text {air }}}=\frac{\lambda_{\text {water }}}{\lambda_{\text {air }}}=\frac{1}{a_{\mathrm{w}}}=\frac{1}{(4 / 3)}$

3. If white light is used in place of monochromatic light in young’s double-slit experiment.

(a) central fringe is white

(b) Colored fringe around the central white fringe

(c) Inner edge of the dark fringe is red. While the outer edge is violet (or blue)

(d) Inner edge of the bright fringe is violet (or blue) and the outer edge is red.

4. If a filter allowing only $\lambda_{\text {red }}$ (of $\lambda_{1}$) is placed in front of slit $\mathrm{s}_{1}$ and filter allowing only $\lambda_{\text {blue }}\left(\right.$ or $\left.\lambda_{1}\right)$ is placed is front of slit $\mathrm{s}_{2}$. Then there is no interference pattern. (refer to point no.3 of conditions)

5. If a thin glass plate or mica sheet is placed in front of one of the slit, then the central fringe shifts towards that slit, the refractive index of glass is $\mu$ and the thickness of sheet t, then the optical path = $\mu \mathrm{t}$ so extra path difference $(\mu-1) t$

If the central fringe now appears at the location of previously formed nth bright fringe then $(\mu-1) t$ $=n \lambda$ if the central fringe appears at the position of previously formed nth dark fringe then $(\mu-1) t$

$=(2 n-1) \frac{\lambda}{2}$

6. If the width of the slit S increased then the degree of spatial coherence decreases. As a result the interference pattern gradually disappears similar occurs if the distance between $\mathrm{s}_{1}$ and $\mathrm{s}_{2}$ is increased.

7. The fringe visibility :

$V=\frac{I_{\max }-I_{\min }}{I_{\max }+I_{\min }}$

$V_{\max }:$ if $I_{1}=I_{2}=I_{0}$

or $\mathrm{I}_{\min }=0$

If widths of slits $\mathrm{s}_{1}$ and $\mathrm{s}_{2}$ are unequal the brightness of the bright fringe and the darkness of the dark fringe decreases.

If $\left.\right|_{1}>>\left.\right|_{2}$ then $\left.\right|_{\max }=\left.\right|_{\min }$

8. When waves from two coherent sources $S_{1}$ are $S_{2}$ interfere in space the shape of the fringe is hyperbolic with foci at $\mathrm{S}_{1}$ and $\mathrm{S}_{2}$.

So, that’s all from this article. I hope you get the idea about the Young double-slit experiment. If you found this article informative then please share it with your friends. If you have any confusion related to this topic then feel free to ask in the comments section down below.

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Interference by Thin Films – Physics – eSaral

Hey, do you want to learn about the interference by thin films? If so. Then you are at the right place.

## Interference by Thin Films

The interference caused by thin films is due to the interference between the waves reflected from the upper and lower surfaces. These two reflected coherent waves are obtained from the same incident wave by division of amplitude.

A film of thickness t and refractive index $\mu$ produces a path difference of $2 \mu \mathrm{cos} \mathrm {r}$ between the two reflected waves and an additional path difference $\frac{\lambda}{2}$

or phase difference of $\pi$ is produced due to the reflection of one wave from a denser medium. Thus, the total path difference in reflected waves is $\left(2 \mu t \cos r+\frac{\lambda}{2}\right)$

For maxima $2 \mu t \quad \cos r+\frac{\lambda}{2}=n \lambda$

or $2 \mu t \cos r=(2 n-1) \frac{\lambda}{2}$

and for minima $2 \mu t \cos r+\frac{\lambda}{2}=(2 n+1) \frac{\lambda}{2}$

or $2 \mu t \cos r=(2 n-1) \frac{\lambda}{2}$

Here r is the angle of refraction. For normal incidence or near-normal incidence $r \approx 0$ ,so that

for maxima $2 \mu t=(2 n \pm 1) \frac{\lambda}{2}$

for minima $2 \mu t=n \lambda$ If a film is very thin $t \approx 0$, then the condition of minima is satisfied and the film appears dark in reflected light.

In transmitted waves, similar interference is observed, but now the additional path difference of $\frac{\lambda}{2}$ is absent. So, in transmitted waves.

$2 \mu t \cos r=n \lambda$ for maxima

and $2 \mu t \cos r=(2 n \pm 1) \frac{\lambda}{2}$ for minima.

## Points to remember

• The fringe width increases with the increase in distance between the source & the screen.
• Fringe width decreases by increasing distance between two slits $s_{1} \& s_{2}$.
• If the experiment is repeated in water instead of air, then $\beta$ decreases.
• When one of the slits of $s_{1} \& s_{2}$ is close then interference does not take place.
• When the two slits are illuminated by two independent sources then interference fringes are not obtained.
• When one of the slits is closed & the width of another is made of the order of $\lambda$, then diffraction fringes are observed
• When the slit is illuminated with different colors then fringes are obtained of the same color but their fringes width is different.
• In young’s double-slit experiment light waves undergo diffraction at both the slits and the diffracted waves superimpose to produce interference.
• If biprism experiment is a liquid instead of air, then the fringe width increases (whereas in young’s double-slit experiment it decreases)
• The wavelength undergoing destructive interference, the color of that wavelength will be absent.
• The wavelength for which the condition of constructive interference is fulfilled that color will be visible maximum consequently the fringes will be colored.

so, that’s all from this article. I hope you get the idea about the Interference by Thin Films. If you found this article informative then please share it with your friends. If you have any confusion related to this topic then feel free to ask in the comments section down below.

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Fresnel biprism experiment – Physics – eSaral

Hey, do you want to learn about the fresnel biprism experiment? If yes. Then keep reading.

## Fresnel’s Biprism Experiment

1. It is an optical device to obtain two coherent sources by the refraction of lights.
2. The angle of biprism is $179^{\circ}$ & refracting angle is $\alpha=1 / 2^{\circ}$.

3. Distance between source & screen D = a + b. Distance between two coherent sources

$=\mathrm{d}=2 \mathrm{a}(\mu-1) \alpha$

Where a = distance between source & Biprism

b = distance between screen & Biprism

$\mu$ = refractive index of the material of the prism.

$\lambda=\frac{d \beta}{D}=\frac{2 a(\mu-1) \alpha \beta}{(a+b)}$

$=\frac{\sqrt{d_{1} d_{2}} \cdot \beta}{(a+b)}$

Note-

$\alpha$ is in radian $\alpha^{0}=\alpha \times \frac{3.14}{180}$ Suppose refracting angle & refractive index is not known then d can be calculated by a convex lens.

One convex lens whose focal length (f) and 4f < D. First convex lens is kept near biprism & $\mathrm{d}_{1}$ is calculated then it is kept near eyepiece & $\mathrm{d}_{2}$ is calculated. $\mathrm{d}=\sqrt{\mathrm{d}_{1} \mathrm{~d}_{2}}$

Application :

With the help of this experiment the wavelength of monochromatic light, the thickness of thin films, and their refractive index & distance between apparent coherent sources can be determined. When Fresnel’s arrangement is immersed in water

(a) Effect on d

$\mathrm{d}_{\text {water }}<\mathrm{d}_{\mathrm{air}}$. Thus when the Fresnel’s biprism experiment is immersed in water, then the separation between the two virtual sources decreases but in young’s double-slit experiment it does not change.

(b) In young’s double-slit experiment $\beta$ decreases and in Fresnel’s biprism experiment $\beta$ increases.

so, that’s all from this article. I hope you get the idea about the fresnel biprism experiment. If you found this article informative then please share it with your friends. If you have any confusion related to this topic then feel free to ask in the comments section down below.

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What are coherent sources of light – Physics – eSaral

Hey, do you want to know what are coherent sources of light? If yes. Then keep reading.

## Coherent source

The two sources of light, whose frequencies are the same and the phase difference between the waves emitted by which remains constant with respect to time are defined as coherent sources. There are two independent concepts of coherence namely

1. Temporal coherence
2. spatial coherence

## Temporal coherence:

In a typical light source (like sodium lamp) a light wave (photon) is produced when an excited atom goes to the ground state and emits light the duration of this $\sim 10^{-9}$ to $10^{-10} \mathrm{sec}$. Thus the electromagnetic (light) wave remains sinusoidal for this much time. This time is known as coherence time it is denoted by $\tau_{c}$.

## Spatial coherence:

Two waves at a different points in space are said to be space coherent if they preserve a constant phase difference over any time t. Note: Laser is a source of monochromatic light waves of a high degree of coherence. The entire wavefront of the laser beam is spatially coherent. Important points: –

1. They are obtained from the same single source.
2. Their state of polarization is the same

Note:

1. Laser light is highly coherent & monochromatic
2. The light emitted by two independent sources (candles, bulbs, etc.) is non-coherent and interference phenomenon cannot be produced by such two sources

(a) Division of wavefront

(b) Division of Amplitude

so, that’s all from this article. I hope you get the idea about What are the coherent sources of light. If you found this article informative then please share it with your friends. If you have any confusion related to this topic then feel free to ask in the comments section down below.

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Einstein equation of the photoelectric effect – Physics – eSaral

Hey, do you want to learn about the Einstein equation of the photoelectric effect? If yes. Then keep reading.

## Einstein’s photoelectric equation

1. Einstein explained the photoelectric effect using the quantum nature of radiation.
2. The emission of a photoelectron is a result of the interaction of one photon with a loosely bound electron in which the photon is completely absorbed by the electron.
3. Some part of incident energy equal to work function is used to remove an electron from metal and remaining is given to electron as its kinetic energy.

$\mathrm{hv}=\mathrm{W}+\mathrm{E}_{\max }=\mathrm{W}+\frac{1}{2} \mathrm{mv}_{\max }^{2}$

$=h v_{0}+\frac{1}{2} m v_{\max }^{2}$

$\frac{1}{2} \mathrm{mv}_{\max }^{2}=\mathrm{h}\left(\mathrm{v}-\mathrm{v}_{0}\right)$

$=\mathrm{hc}\left(\frac{1}{\lambda}-\frac{1}{\lambda_{0}}\right)=\mathrm{eV}_{\mathrm{S}} \ldots(1)$

equation 1 is called Einstein photoelectric equation. maximum velocity of the emitted electron

$v_{\max }=\sqrt{\frac{2 h\left(v-v_{0}\right)}{m}}=\sqrt{\frac{2 h c\left(\lambda-\lambda_{0}\right)}{m \lambda \lambda_{0}}}$ $=\sqrt{\frac{2 \mathrm{eV}_{\mathrm{S}}}{\mathrm{m}}}$

The stopping potential

$V_{S}=\frac{h\left(v-v_{0}\right)}{e}=\frac{h c\left(\lambda_{0}-\lambda\right)}{e \lambda \lambda_{0}}$

4. The Einstein’s photoelectric equation is in accordance with the conservation of energy. Here light energy is converted into electric energy.
5. The equation explains the laws of photoelectric emission.

(a) The increase in intensity increases the number of photons with the same energy hv. So the number of photoelectrons will proportionally increase.

(b) If $v<v_{0}$ then KE will become negative which is not possible so in this condition photoemission is not possible.

(c) If $v>v_{0}$ Then $\mathrm{KE} \propto\left(v-v_{0}\right)$ so maximum kinetic energy or stopping potential increases linearly with frequency of incident radiation.

## Important points

1. In the photoelectric effect, all photoelectrons do not have the same kinetic energy. Their KE ranges from zero to $E_{\max }$ which depends on the frequency of incident radiation and nature of cathode.
2. The photoelectric effect takes place only when photons strikebound electrons because for free electrons energy and momentum conservations do not hold together.
3. Cesium is the best photo-sensitive material.
4. The efficiency of a photoemission

$\eta=\frac{\text { Number of photoelectrons emitted per unit area per unit time }}{\text { Number of photons incident per unit area per unit time }}$

$=\frac{\mathrm{n}_{\mathrm{e}}}{\mathrm{n}_{\mathrm{p}}}$ $\eta=\frac{\text { Intensity of emitted electrons }}{\text { Intensity of incident radiation }}$

$=\frac{I_{e}}{I_{p}}$ Therefore $\eta=\frac{\mathrm{n}_{\mathrm{e}}}{\mathrm{n}_{\mathrm{p}}}=\frac{\mathrm{I}_{\mathrm{e}}}{\mathrm{I}_{\mathrm{p}}}$

so, that’s all from this article. I hope you get the idea about the Einstein equation of the photoelectric effect. If you found this article informative then please share it with your friends. If you have any confusion related to this topic then feel free to ask in the comments section down below.

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What is photoelectric effect class 12 – Modern Physics – eSaral

Hey, do you want to know What is photoelectric effect in class 12? If yes. Then keep reading.

## What is Photoelectric effect?

The phenomenon of emission of electrons from the surface of the metal when the light of suitable frequency falls on it is called the photoelectric effect.

1. The ejected electrons are called photoelectrons
2. The current produced due to emitted electrons is called photocurrent.
3. The photoelectric effect proves the quantum nature of radiation.
4. The classical electromagnetic theory fails to explain photoelectric effect.
5. Einstein explained the photoelectric effect using quantum nature of radiation.
6. Hallwach is credited with discovery of the photoelectric effect.

## Experimental study of photoelectric effect

### Effect of intensity of incident radiation

1. The number of incident photons per second on a metal plate is called intensity of incident radiation.
2. For a fixed incident frequency, the saturation photocurrent is directly proportional to the intensity of incident radiation e.g. when intensity of radiation is doubled at constant frequency saturation photocurrent is also doubled.
3. Saturation current: When all photo electrons produced reach anode the photocurrent becomes maximum and is independent of applied potential difference. This current is called saturation or maximum current.

### Effect of potential

1. When polarity of electrodes is reversed with commutator the current is reduced but does not become zero. This shows that emitted photoelectrons have kinetic energy.
2. The negative potential of the anode at which the photo current becomes zero is called stopping potential $\left(\mathrm{V}_{\mathrm{s}}\right)$. At this potential the electrons with maximum kinetic energy are stopped from reaching the anode.
3. No photo current is produced even on increasing the intensity of incident radiation when the anode is at stopping potential. Thus, stopping potential is independent of the intensity of incident radiation.
4. The stopping potential is a measure of maximum kinetic energy of photo electrons. $E_{\max }$ $=\mathrm{eV}_{\mathrm{s}}$

### Effect of frequency of incident radiation

1. Threshold frequency the minimum frequency of incident radiation that can eject photo electrons from a metal is known as threshold frequency $\left(v_{0}\right)$.
2. At stopping potential if the frequency of incident radiation is increased then current starts flowing again. This can be made zero by increasing stopping potential.
3. Thus, the maximum kinetic energy of photoelectron or stopping potential increases with an increase in the frequency of incident radiation. The maximum kinetic energy of photoelectron increases linearly with increase in the frequency of incident light.

### Effect of material of cathode

The stopping potential, work function, and threshold frequency depend on the nature of material of cathode. Work function : The minimum energy required for emission of electrons from metal is called work function work function $\phi=\mathrm{hv}_{0}$ where $v_{0}$ is threshold frequency.

so, that’s all from this article. I hope you get the idea about What is photoelectric effect in class 12. If you found this article informative then please share it with your friends. If you have any confusion related to this topic then feel free to ask in the comments section down below.

For a better understanding of this chapter, please check the detailed notes of the Dual Nature of Radiation and matter. To watch Free Learning Videos on physics by Saransh Gupta sir Install the eSaral App.

Types of radioactive decay – Alpha Decay, Beta Decay – eSaral

Hey, do you want to learn about the Types of Radioactive decay or Types of radioactive processes? If yes. Then you are at the right place.

### (a) Alpha decay

$_{z} X^{A}$ (parent nucleus) $\longrightarrow_{Z-2} Y^{\mathrm{A}-4}$ (daughter nucleus)

$+{ }_{2} \mathrm{He}^{4}$ (alpha particle)

e. g. ${ }_{92} \mathrm{U}^{238} \longrightarrow{ }_{90} \mathrm{Th}^{234}+{ }_{2} \mathrm{He}^{4}$

1. Alpha particle consists of 2 neutrons, 2 protons and carries a positive charge in magnitude 2 electrons. It is doubly ionized helium nuclei.

2. $\alpha$ emission takes place when the size of the nucleus becomes too large. The decay reduces the size of the nucleus.

3. $\alpha$ emission is explained on basis of the quantum mechanical tunnel effect.

4. The energy released in $\alpha$ decay $\mathrm{Q}=\left(\mathrm{M}_{\mathrm{x}}-\mathrm{M}_{\mathrm{y}}-\mathrm{M}_{\alpha}\right) \mathrm{c}^{2}$

5. The kinetic energy of $\alpha$ particle $\mathrm{E}_{\alpha}=\left(\frac{\mathrm{A}-4}{\mathrm{~A}}\right) \mathrm{Q}$ where A is mass number and Q is disintegration energy

### (b) Beta decay

• Electron emission

$\left(\beta^{-}\right)$ ${ }_{z} X^{A} \longrightarrow_{z+1} Y^{A}$

$+_{-1} \mathrm{e}^{0}\left(\beta^{-}\right.$ particle $)+\bar{v}$

e.g. ${ }_{6} \mathrm{C}^{14} \longrightarrow{ }_{7} \mathrm{~N}^{14}$

$+_{-1} \mathrm{e}^{0}+\bar{v}$ (antineutrino)

1. $\beta^{-}$ Particles are fast-moving electrons carrying a negative charge

2. $\beta^{-}$ Particles are emitted when the nucleus has too many neutrons relative to a number of protons i.e. N/Z ratio is larger than required.

3. The emission of electrons takes place when a neutron is converted to a proton inside the nucleus. This helps in the correction of the N/Z ratio. ${ }_{0} \mathrm{n}^{1} \longrightarrow{ }_{1} \mathrm{p}^{1}+{ }_{-1} \mathrm{e}^{0}+\overline{\mathrm{v}}$

4. The interaction responsible for $\beta$ decay is weak interaction.

• Positron emission

$\mathrm{z} \mathrm{X}^{\mathrm{A}} \longrightarrow \mathrm{z}_{-1} \mathrm{X}^{\mathrm{A}}$

$+_{+1} e^{0}\left(\beta^{+}\right.$ particle $)+v$

$\mathrm{eg} \cdot{ }_{29} \mathrm{Cu}^{64} \longrightarrow{ }_{28} \mathrm{Ni}^{64}$

$+_{+1} \mathrm{e}^{0}+v$ (neutrino)

1. $\beta^{+}$ Particles are positrons with a mass equal to an electron but carry a unit positive charge.

2. $\beta^{+}$ Particles are emitted when the nucleus has too many protons relative to a number of neutrons i.e. N/Z ratio is smaller than required.

3. The emission of positron takes place when a proton is converted to a neutron inside the nucleus. This increases the N/Z ratio.

${ }_{1} \mathrm{p}^{1}={ }_{0} \mathrm{n}^{1}+{ }_{+1} \mathrm{e}^{0}+\boldsymbol{v}$

### (c) Gamma decay

${ }_{Z} \mathrm{X}^{\mathrm{A}^{*}} \longrightarrow{ }_{Z} \mathrm{X}^{\mathrm{A}}+\gamma$

e.g. $\quad{ }_{5} \mathrm{~B}^{12} \longrightarrow{ }_{6} \mathrm{C}^{12^{\star}}+{ }_{-1} \mathrm{e}^{0}+\overline{\mathrm{v}}$

${ }_{6} \mathrm{C}^{12^{*}} \longrightarrow{ }_{6} \mathrm{C}^{12}+\gamma$

1. $\gamma$ rays are electromagnetic radiations that are chargeless and massless

2. $\gamma$ rays are emitted when the nucleus has excess energy

3. $\gamma$ rays are emitted when the nucleus jumps from the excited state to a lower level or ground state. This reduces the energy of the nucleus.

4. $\gamma$ rays are electromagnetic radiations of short wavelength $\left(\sim 10^{-12} \mathrm{~m}\right)$ which travel with speed of light.

### (d) Electron capture

${ }_{\mathrm{Z}} \mathrm{X}^{\mathrm{A}}+{ }_{+1} \mathrm{e}^{0} \longrightarrow+\underset{\mathrm{z}-1}{\mathrm{Y}}^{\mathrm{A}}+\mathrm{v}$

${ }_{4} \mathrm{Be}^{7}+{ }_{-1} \mathrm{e}^{0} \longrightarrow{ }_{3} \mathrm{Li}^{7}+\mathrm{v}$

1. This process takes place when the nucleus has too many protons relative to a number of neutrons. i.e. N/Z ratio is larger than required.

2. This process occurs when a parent nucleus captures one of its own orbital atomic electrons and emits a neutrino.

For a better understanding of this chapter, please check the detailed notes of Radioactivity. To watch Free Learning Videos on physics by Saransh Gupta sir Install the eSaral App.

What is binding energy in physics – Binding energy per nucleon – eSaral

Hey, do you want to learn about What is binding energy in physics? If yes. Then keep reading.

## Binding energy

1. The energy required to break a nucleus into its constituent nucleons and place them at an infinite distance is called binding energy.

2. The energy equivalent to mass defect is called binding energy.

3. This is the energy with which the nucleons are held together.

4. The binding energies $(\sim \mathrm{MeV})$ are very large as compared to molecular binding energies

$(\sim \mathrm{eV})$

Binding energy

$B E=(\Delta m) c^{2}$

$=c^{2}\left[Z m_{p}+(A-Z) m_{n}-M\left(z X^{A}\right)\right]$

the rest mass of protons + rest mass of neutrons = rest mass of nucleus + BE

## Binding energy per nucleon

1. The binding energy per nucleon of a nucleus is the average energy required to extract a nucleon from the nucleus. Binding energy per nucleon

$\overline{\mathrm{B}}=\frac{\text { Total binding energy }}{\text { Total number of nucleons }}$

$=\frac{\mathrm{BE}}{\mathrm{A}}=\frac{\Delta \mathrm{mc}^{2}}{\mathrm{~A}}$

$=\frac{c^{2}}{A}$ x
$\left[Z m_{p}+(A-Z) m_{n}-M\left(z X^{A}\right)\right]$

2. The plot of binding energy per nucleon with mass number A is shown as
3. Binding energy per nucleon gives a measure of the stability of the nucleus. More is the binding energy per nucleon more is the stability of the nucleus.

4. Binding energy per nucleon gives a measure of the stability of the nucleus. More is the binding energy per nucleon more is the stability of the nucleus.

5. Binding energy per nucleon is small for lighter nuclei i.e. ${ }_{1} \mathrm{H}^{1},{ }_{1} \mathrm{H}^{2}$ etc.

6. For A < 28 at A = 4n the curve shows some peaks at

${ }_{2} \mathrm{He}^{4},{ }_{4} \mathrm{Be}^{8},{ }_{6} \mathrm{C}^{16},{ }_{8} \mathrm{O}^{16},{ }_{10} \mathrm{Ne}^{20},{ }_{12} \mathrm{Mg}^{24}$.

7. The binding energy per nucleon is almost constant about 8.5 MeV in range 40 < A < 120. The binding energy per nucleon is maximum about 8.8 MeV for $\mathrm{F} \mathrm{e}^{56}$.

8. The binding energy per nucleon decreases for A > 200 They become less stable and exhibit radioactivity.

9. In fusion lighter nuclei fuse to form heavier nuclei. The process in accompanied by an increase in binding energy per nucleon.

10. In fission a heavy nucleus splits into two lighter nuclei. Here also increase in binding energy per nucleon takes place.

11. The heaviest stable nuclide is ${ }_{83} \mathrm{Bi}^{209}$.

so, that’s all from this article. I hope you get the idea about What is binding energy in physics. If you found this article informative then please share it with your friends. If you have any confusion related to this topic then feel free to ask in the comments section down below.

For a better understanding of this chapter, please check the detailed notes of Chapter Nuclei. To watch Free Learning Videos on physics by Saransh Gupta sir Install the eSaral App.