Half wave rectifier circuit diagram – Definition, Explanation – eSaral
A device that converts alternating current into Direct current is called a rectifier. If you want to learn about the half-wave rectifier and half-wave rectifier circuit diagram then you are at the right place.

Application of diode as a rectifier:

An electronic device that converts alternating current into Direct current is called a rectifier.

Half wave rectifier:

A rectifier, which rectifies only one half of each ac supply cycle is called a half-wave rectifier.

During the first half of the input cycle, the junction diode gets forward bias. The conventional current will flow. The upper end of $R_{L}$ will be positive potential with respect to the lower end during the second half cycle junction diode will get reverse biased and hence no output will be obtained across $R_{L}$.

Input voltage

$\mathrm{V}_{\mathrm{i}}=\mathrm{V}_{\mathrm{m}} \sin \omega \mathrm{t}$

$\mathrm{i}=\mathrm{I}_{\mathrm{m}} \sin \omega \mathrm{t}$

for $0 \leq \omega t \leq \pi$

$\mathrm{i}=0$

for $\pi<\omega t<2 \pi$

$I_{m}=\frac{V_{m}}{R_{f}+R_{L}}$

here $\mathrm{R}_{\mathrm{f}}=$ forward resistance of diode

$R_{L}$ = load resistance

a. dc output current :

$\mathrm{I}_{\mathrm{dc}}=\frac{1}{2 \pi} \int_{0}^{2 \pi} \mathrm{idt}$

$=\frac{1}{2 \pi}\left[\int_{0}^{\pi} \mathrm{I}_{\mathrm{m}} \sin t \mathrm{dt}+\int_{\pi}^{2 \pi} 0 \mathrm{dt}\right]$

$\mathrm{I}_{\mathrm{dc}}=\frac{\mathrm{I}_{\mathrm{m}}}{\pi}=0.318 \mathrm{I}_{\mathrm{m}}$

b. dc output voltage:

$V_{d c}=I_{d c} \times R_{L}$

$=\frac{I_{m}}{\pi} \times R_{L}$

$=\frac{\mathrm{V}_{\mathrm{m}}}{\pi\left[1+\left(\mathrm{R}_{\mathrm{f}} / \mathrm{R}_{\mathrm{L}}\right)\right]}$

$V_{d c}=\frac{V_{m}}{\pi}=0.318 \mathrm{~V}_{\mathrm{m}}$

c. (Root mean square) RMS current:

$I_{r m s}=\left[\frac{1}{2 \pi} \int_{0}^{2 \pi} i^{2} d(t)\right]^{1 / 2}$

$=\frac{I_{m}}{2}$

same

$V_{r m s}=\frac{V_{m}}{2}$

So, that’s all from this blog. I hope you enjoyed this explanation of the half-wave rectifier and half-wave rectifier circuit diagram. If you liked this article then please share it with your friends.

What is Diode in electronics

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What is Diode in electronics – Definition, Important Points – eSaral
A diode is a PN junction device. If you want to know what is diode in electronics. Then keep reading.

Diode

A diode is a PN junction device

Ideal diode

• Conducts with zero resistance when forward biased.
• Offer an infinite resistance when reverse biased.

Practical Circuit For diode

The practical characteristic curve for a diode:

Important terms related to diode:
1. Knee VoltageKnee voltage is defined as the forward voltage at which the current through the junction starts increasing rapidly.For silicon = 0.7 volt for germanium = 0.3 volt

2. Forward resistance or ac resistance:It is defined as the reciprocal of the slope of the forward characteristic curve.forward resistance

$r_{f}=\frac{1}{\text { slope of forwardcharacteristic }}$

$=\frac{1}{\Delta \mathrm{I}_{\mathrm{f}} / \Delta \mathrm{V}_{\mathrm{f}}}$

$=\frac{\Delta \mathrm{V}_{\mathrm{f}}}{\Delta \mathrm{l}_{\mathrm{f}}}$

3. Junction breakdown :When the reverse voltage is increased a point is reached when the junction breaks down with sudden rise in reverse current. This value of the voltage is known as the breakdown voltage. Two types of breakdown occur:a. Zener breakdown:

a. Zener breakdown:
Takes place in junction which are heavily doped so having narrow depletion layers. A very strong electric field appears across the narrow depletion layer which breaks the bond.

b. Avalanche breakdown :

Occur in junctions which are lightly doped. (having wide depletion layer) so at a high electric field, the minority charge carriers, while crossing the junction acquire very high velocities. A chain reaction is established, giving rise to the high current.

4. Diffusion Current :Some electrons and holes have more kinetic energy $\left[\frac{1}{2} \mathrm{mv}^{2}>\mathrm{eV}\right]$ So $\mathrm{e}^{-}$ diffuse from n to p side and hole diffuse from p to n side due to diffusion of the charge carriers a current will flow known as diffusion current.

a. Because of the concentration difference diffusion occurs.

b. Diffusion results in an electric current from p side to the n side

c. When P-N. Jn is in no Bias = diffusion current = drift current

Net charge flow = 0

Net current = 0

5. Drift Current: Due to thermal collisions, the covalent bond is broken. If an electron-hole pair is created in the depletion region, there is a regular flow of electrons towards the n side and of holes towards the p side. Current flow n side to p side called drift current.Drift current and the diffusion current are in the opposite direction

So, that’s it from this blog. I hope you get the idea about what is a diode in electronics. If you found this Explanation helpful then share it with your friends and followers.

Types of Semiconductor

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What is P-N junction – Definition, Explanation, Types – eSaral
P-N junctions are formed by diffusing trivalent impurity to one-half side and pentavalent impurity to another side. If you want to know what is P-N junction then keep reading this article.

P-N Junction

• By merely Joining the two pieces a P-N Junction cannot be formed.
• P-N junctions are formed by diffusing trivalent impurity to one-half side and pentavalent impurity to another side.
• The plane dividing the two zones is known as a junction.
• P-N junction is unohmic
• As P-type semiconductor has a high concentration of holes and N-type semiconductor has a high concentration of free electrons. there is a tendency of holes to diffuse over to the N side and electron to the P side.
• When the hole diffuses from the P to N side then this will neutralize with free-electron similarly when electron diffuse from the N to P side it will neutralize with the hole. So, a depletion layer is formed near the Jn.

Depletion layer:

There is a barrier near a junction that opposes the flow of charge carrier is known as depletion layer width of the depletion layer is in micrometer order.

Potential Barrier:

Potential developed in depletion layer is called P.B.

P-side is at lower potential and N-side is at higher potential.

P.B. for Ge  0.3 volt

P.B. for Si  0.7 volt

Electric field:

Electric field due to P.B.

$E=\frac{V}{d}$

For Ge

$E=\frac{0.3}{10^{-6}}$

$=3 \times 10^{5} \mathrm{~V} / \mathrm{m}$

order

$\mathrm{E} \approx 10^{5} \mathrm{~V} / \mathrm{m}$

the direction of E due to P.B. N to P-side

P-N Junction with forward bias :

When the Positive terminal of a battery is connected to the P side and the negative terminal to the N side. Then PN Junction is in forwarding Bias

Forward bias reduces the potential barrier. More charge carriers diffuse across the junction.

Special Point:
• Potential barrier reduces
• Width of the depletion layer decreases
• P-N junction offers low resistance in forwarding bias.
• Forward current flow in a circuit
• The forward characteristic curve is shown in the figure.
• Forward dynamic resistance $r_{f}=\frac{\Delta V_{f}}{\Delta I_{f}} \cong 100 \Omega$
• Knee or cut in voltageGe  0.3 VSi  0.7 V
• Dependence of forward current on bias voltage $\mathrm{I}=\mathrm{I}_{0}\left[\mathrm{e}^{\frac{\mathrm{qV}}{\mathrm{kT}}}-1\right]$$\mathrm{e}^{\frac{\mathrm{qV}}{\mathrm{kT}}}>>1 I \approx I_{0} e^{\frac{+q V}{k T}} (Approximate exponential growth) I = Forward current \mathrm{I}_{0} = reverse saturation current k = Boltzman constant q = charge of electron V = forward voltage T = temperature P-N junction with reverse bias: When the positive terminal of a battery is connected to the N-side and the negative terminal is connected to the P-side. Holes in the P-region are attracted towards the negative terminal and the electrons in the N-region are attracted towards the positive terminal. Special Point: • The depletion layer increases for reverse biased. • Potential barrier increases • The reverse characteristic curve is shown in figure • Very little current called reverse saturation current flows due to minority carrier flow.For Silicon = 10^{-9} AFor Germanium = 10^{-6} A • In reverse biased condition, junction behaves as a capacitor of few picofarads. • In reverse biased condition, junction behaves like high resistive material between two regions. • In reverse biased P-N diode behaves like an insulator. • Reverse resistance \mathrm{R}_{\mathrm{B}}=\frac{\Delta \mathrm{V}_{\mathrm{B}}}{\Delta \mathrm{I}_{\mathrm{B}}} \cong 10^{6} \Omega$$\frac{R_{B}}{R_{f}}=10^{3}$ : 1 for Geand

$\frac{R_{B}}{R_{f}}=10^{4}$ : 1 for Si
• Dependence of reverse current on bias volt. $\mathrm{I}_{\mathrm{r}}=\mathrm{I}_{0}\left[\mathrm{e}^{-\frac{\mathrm{q} \mathrm{V}}{\mathrm{kT}}}-1\right]$$\mathrm{e}^{-\frac{\mathrm{qV}}{\mathrm{kT}}}<<1$$\mathrm{I} \cong-\mathrm{I}_{0}$

Here $\mathrm{I}_{\mathrm{r}}$ = reverse current

$\mathrm{I}_{0}$ = reverse saturation current

V = applied voltage

q = charge of electron

T = temperature in kelvin
Special Point
• The diffusion current in the p-n junction is greater than the drift current in magnitude if the junction is forward biased.
• A hole diffuses from the p-side to the n-side in a p-n junction. This means that a bond is broken on the n-side and the electron freed from the bond jumps to a broken bond on the p-side to complete it.

What is Fermi energy level in semiconductors

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What is Fermi energy level in semiconductors – Electronics – eSaral

Hey, Do you want to know what is fermi energy level in semiconductors? If yes. Then keep reading.

Fermi Energy Level.

Fermi energy is the maximum kinetic energy of an electron at 0 K these electrons are called Fermi electrons and energy level is known as Fermi energy level.

• It is always found between the conduction band and valance band

• Fermi Level is the energy that corresponds to the center of gravity of the conduction electrons and holes weighted according to their energies.

• In pure germanium semiconductor, the Fermi level is about halfway in the forbidden gap.

• In an n-type semiconductor, the Fermi level lies in the forbidden gap, very close to the conduction band.

• In p-type semiconductor, the Fermi level lies in the forbidden gap, very close to the valence band.

• With rising in temperature the Fermi level moves towards the center of the forbidden gap, for both types of semiconductors.

• An n-type semiconductor is better than a p-type semiconductor as electrons have more mobility than holes.

Mass Action Law

Under thermal equilibrium, the product of concentration $\mathrm{n}_{\mathrm{e}}$ of free electrons and the concentration $\mathbf{n}_{\mathrm{h}}$ of holes is constant & independent of the amount of doping by donor & acceptor impurity.

$\mathrm{n}_{e} \mathrm{n}_{\mathrm{h}}=\mathbf{n}_{\mathrm{i}}^{2}$

where

$\mathrm{n}_{\mathrm{i}}=$ intrinsic concentration

However, the intrinsic concentration is a function of temperature.

• In n-type semiconductors, the number density of electrons is nearly equal to the number density of donor atoms $\mathrm{N}_{\mathrm{d}}$ and is very large as compared to the number density of holes.

$\mathrm{n}_{\mathrm{e}} \approx \mathrm{N}_{\mathrm{D}}$

$N_{D}>>n_{h}$

• In p-type semiconductor, the number density of holes is nearly equal to the number density of acceptor atoms $\mathrm{N}_{\mathrm{a}}$ and is very large as compared to number density of electrons.

$\mathbf{n}_{\mathrm{h}} \approx \mathrm{N}_{\mathrm{A}}$

$\mathrm{N}_{\mathrm{A}}>>\mathrm{n}_{\mathrm{e}}$

So, that’s all from this article. I hope you get the idea about what is fermi energy level in semiconductors. 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 Electronics. To watch Free Learning Videos on physics by Saransh Gupta sir Install the eSaral App.

Types of Semiconductors – Types, Examples – eSaral

Semiconductor in an extremely pure form is known as an intrinsic semiconductor and The impure semiconductor material is called an extrinsic semiconductor. If you want to learn about the Types of Semiconductors then keep reading.

Types of Semiconductors

Pure or intrinsic semiconductor:

A semiconductor in an extremely pure form is known as an intrinsic semiconductor. In an intrinsic semiconductor, the number of free electrons is always equal to the number of holes when an external field is applied across the intrinsic semiconductor the conduction through the semiconductor is by both free electrons and holes.

• Total current $I=I_{e}+I_{h}$

• It is perfectly neutral

• Number of free electron = Number of holes $\left(n_{e}=n_{h}\right)$

Extrinsic semiconductor:

The intrinsic semiconductors are of little importance because of negligible conductivity and moreover, the conductivity has little flexibility.

The electrical conductivity of an intrinsic semiconductor is zero at absolute zero and very small at ordinary (room) temperatures.

The conductivity is considerably increased by adding some impurity element to the pure (intrinsic) semiconductors.

The impure semiconductor material is called an extrinsic semiconductor.

The electrical conductivity of extrinsic semiconductors is called extrinsic conductivity.

Depending on impurity added, the extrinsic semiconductor can be divided into two parts-

N-Type:

An n-type Ge is obtained by adding a small quantity, for $10^{6}$ Ge atoms approximately one impurity atom of a pentavalent impurity

added the ratio of Ge and As atoms is $10^{6}: 1$.

Generally, Arsenic (As) is taken for this purpose.

Each arsenic atom replaces one Ge atom at its crystal lattice site without changing its structure. Four of the five valence electrons of As occupy the same positions in four covalent bonds as earlier occupied by four electrons of replaced Ge atom. But the fifth excess electron remains free. The energy level of this excess electron is only slightly smaller than the lowest energy level of the conduction band. Very small energy of about 0.01 eV (0.05 eV for Si) can detach this electron from the impurity atom.

The thermal energy at room temperature is sufficient to enable this electron to detach itself and become a member of the conduction band and take part in conduction.

Why n-type

Electrons with negative charges help in current conduction, the impure Ge is called n-type. Since impurity atoms donate electrons, the impurity is called donor impurity.

Each $1 \mathrm{~cm}^{3}$ of Ge crystal having $4.52 \times 10^{22}$ Ge atom, will have $4.52 \times 10^{16}$ (one millionth) As atoms and hence as many free electrons. The conductivity of the crystal is increased considerably.

At high temperatures, some covalent bonds are broken. Then more electrons and holes become free.

Free holes in n-type Ge act as minority carriers.

• $\mathrm{n}_{\mathrm{e}}$$>>$$\mathrm{n}_{\mathrm{h}}$

• $\mathrm{I}=\mathrm{I}_{\mathrm{e}}+\mathrm{I}_{\mathrm{h}}$

$\left(\mathrm{I}_{\mathrm{e}}>>\mathrm{I}_{\mathrm{h}}\right)$ $I \approx I_{e}$

In N-Type semiconductor current mainly flow due to free electrons.

• Energy needed to detach fifth electron from impurity for 0.01 eV for Ge, 0.05 eV for Si.

• Electrons are majority carriers due to the addition of pentavalent impurity.

• Holes are minority carriers due to the breaking of covalent bonds.

• N-type semiconductor has an excess of free electrons but it is electrically neutral.

• It is called donor-type impurity because it gives one electron to a crystal.

• The type of conductivity is called negative or N-type conductivity.

P-type:

A p-type Ge is obtained by adding a small quantity for $10^{6}$ Ge atoms approximately one impurity atom of a trivalent impurity added. Ratio of Ge and Al atoms is $10^{6}: 1$.
Generally, Aluminum (Al) or Boron (B) is taken for this purpose.

Each aluminum atom replaces one Ge atom at its crystal lattice site without changing its structure. The three valence electrons of Al occupy the same positions in three covalent bounds as earlier occupied by four electrons of replaced Ge atom.

One covalent bond remains unfilled which shown an electron vacancy.

This electron vacancy is called a hole. The energy of this hole is slightly more than the highest energy level of the valence band (figure).

The electrons from the valence band get easily excited by thermal energy at room temperature to enter the hole. But the electrons filling these holes create new holes in the valence band. These holes in the valence band are filled by more electrons and this continues.

Holes can move freely through a crystal lattice and take part in conduction.

Why p-type

Since holes with positive charges help in current conduction, the impure Ge is called p-type. Since impurity atoms accept electrons (for their holes), the impurity is called acceptor impurity. Each $1 \mathrm{~cm}^{3}$ of Ge crystal having $4.52 \times 10^{22}$ Ge many holes. The conductivity of the crystal is increased considerably.

At high temperatures, some covalent bonds are broken. Then more electrons and holes become free.

Free electrons in p-type Ge act as minority carriers.

• $n_{h}>>n_{e}$

• $\mathrm{I}=\mathrm{I}_{\mathrm{e}}+\mathrm{I}_{\mathrm{h}}$

$\left(\mathrm{I}_{\mathrm{h}}>>\mathrm{I}_{\mathrm{e}}\right)$
$\mathrm{I} \approx \mathrm{I}_{\mathrm{h}}$

In p-type semiconductor current mainly flow due to holes.

• In a p-type semiconductor, majority carriers are positive holes due to the addition of trivalent impurity.

• In a p-type semiconductor, minority carriers are electrons due to the breaking of covalent bonds.

• This impurity is called acceptor type impurity

• In P-type, the valence electrons move from one covalent bond to another bond.

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

Effect of Temperature on semiconductor – Electronics – eSaral

Hey, do you want to learn about the Effect of Temperature on Semiconductor? If yes. Then keep reading.

Effect of Temperature on Semiconductor

At Absolute zero temperature

Covalent Bonds are very strong and there is no free electron in the conduction band. For semiconductors at 0 K, the conduction band is empty while the valence band is full.

Above Absolute zero temperature

When temperature increases some of the covalent bonds are break due to thermal energy supplied.

At room temperature, due to thermal energy, some electrons jump to the conduction band.

In the conduction band electrons are free to move. If a potential difference is applied across semiconductor crystals, these free electrons constitute an electric current.

The variation in the resistivity of a pure semiconductor is mainly due to a change in carrier concentration.

The number of electrons raised from VB to the CB

$n_{e} \propto e^{-E_{o} / k T}$

$\mathbf{n}_{\mathrm{e}}=\mathrm{AT}^{3 / 2} \mathrm{e}^{-\Delta \mathrm{Eg} / 2 \mathrm{kT}}$

$\mathrm{T}=$ Absolute Temp.

$\mathrm{k}=$ Boltzmann Const. $=1.38 \times 10^{-23} \mathrm{~J} / \mathrm{K}$ As temperature (T) increases,

$\mathrm{n}_{\mathrm{e}}$ also increases

i.e. conductivity of the semiconductor increases.

As the temperature of the semiconductor increases, its resistivity decreases.

Hole:

At higher temperatures, some of the electrons gain energy due to thermal agitation and move from the VB to the CB.

The electron leaving the VB to enter CB leave behind an equal number of vacant sites near the top of the VB. These vacant sites are called holes.

These holes represent the vacancy of electrons and behave like a positive charge.

Characteristics of hole:

• The hole carries a positive charge equal to an electronic charge.

• The energy of a hole is high as compared to that of the electron.

• The mobility of the hole is smaller than that of the electron.

• In the external electric field, holes move in a direction opposite of that of the electron.

• The effective mass of hole > Effective mass of the electron.

Electron-hole Recombination:

The completion of a bond may not be necessarily due to an electron from a bond of a neighboring atom. The bond may be completed by a conduction band electron i.e., free electron, and known as electron-hole recombination.

The breaking of bonds or generation of electron-hole pairs and completion of bonds due to recombination is taking place all the time.

At equilibrium, the rate of generation becomes equal to the rate of recombination, giving a fixed number of free electrons and holes.

So, that’s all from this article. I hope you get the idea about the Effect of Temperature on Semiconductor. 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 Electronics. To watch Free Learning Videos on physics by Saransh Gupta sir Install the eSaral App.

Classification of solids in terms of the forbidden energy gap – eSaral

hey, do you want to learn about the Classification of solids in terms of the forbidden energy gap? If yes. Then you are at the right place.

Classification of solids in terms of the forbidden energy gap

The width of the forbidden energy gap between the valence band and the conduction band distinguishes conductors, semiconductors, and insulators from each other.

Difference between Valence, Conduction, and Forbidden Band

Valence band:

This band is never empty. VB may be partially or completely filled with electrons. VB electrons are not capable of gaining energy from the external electric fields. Therefore, the electrons of VB also not contribute to the electric current.

Conduction band:

Electrons are rarely present. CB either empty or partially filled with electrons. CB electrons can gain energy from the external electric field. Electrons in this band contribute to the electric current.

Forbidden energy gap:

Electrons are not found in this band. FEB is completely empty. The minimum energy required for shifting electrons from the valence band to the conduction band is called band gap ($\left(E_{g}\right)$).

So, that’s all from this article. I hope you get the idea about the Classification of the solids according to the Forbidden energy gaps. 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 Electronics. To watch Free Learning Videos on physics by Saransh Gupta sir Install the eSaral App.

Energy band theory in solids – Electronics, Physics – eSaral

Hey, do you want to learn about the Energy band theory in solids? If yes. Then you are at the right place.

Energy Bands in solid

Valence band:

The electrons in the outermost shells are not strongly bonded to their nuclei, these electrons are valence electrons. The band formed by these electrons is called the valence band.

Conduction band:

Valence electrons are loosely attached to the nucleus. Even at ordinary temperature, some of the valence electrons left the valence band. These are called free electrons and also called conduction electrons. The band occupied by these electrons is called the conduction band.

Forbidden energy gap ($\Delta$Eg.):

The separation between the conduction band and valence band is known as the forbidden energy gap.

$\Delta \mathrm{Eg} .=(\mathrm{CB})_{\mathrm{Min}}-(\mathrm{VB})_{\mathrm{Max}}$

• No free electrons are in F.E.G.

• The width of FEG depends upon the nature of the substance

• Width is more than valence electrons are strongly attached with the nucleus

• The width of FEG represent in eV

• If the conduction band is empty, then-current conduction is not possible.

• The Greater is the energy gap, the more tightly the valence electrons are bounded to the nucleus. The energy gap in some semiconductors is as follows-

The energy gap decreases slightly with an increase in temperature.

For a better understanding of this chapter, please check the detailed notes of Electronics Class 12. To watch Free Learning Videos on physics by Saransh Gupta sir Install the eSaral App.

What are the properties of semiconductors – Electronics – eSaral

Hey, do you want to know What are the properties of semiconductors? If yes. Then keep reading

Semiconductor

There are a large number of materials that have resistivities lying between those of an insulator and a conductor (See table). Such materials are known as ‘semiconductors’.

At absolute zero, pure and perfect crystals of the semiconductors are nonconducting, their resistivity approaching to the resistivity of an insulator. They can be made conducting by adding impurities, thermal agitation, and lattice defects etc.

Resistivity of a semiconductor depends upon the temperature, and it decreases with the rise in temperature; consequently, a semiconductor crystal becomes conducting even at room temperature.

At room temperature, their resistivity lies in the range of $10^{2}$ to $10^{9}$ ohm-cm. and is thus intermediate between the resistivity of a good conductor ($10^{-8}$ ohm-cm) and insulator ($10^{14}$ to $10^{22}$ ohm.cm.).

Electrical resistivity of various materials at 20°C in ohm-meter.

At $20^{\circ} \mathrm{C}$ resistivity of semiconductors lying between metals and insulators. But at low temperatures, semiconductor behaves as an insulator, because the resistivity of a semiconductor depends strongly on temperature.

Properties:

• They have a negative temperature coefficient of resistance means the resistance of semiconductors decreases with an increase in temperature and vice-versa.

• Their electrical conductivity is very much affected by even a very minute amount of impurity added to it.
• Covalent Bond

• Crystalline Structure Silicon $\left({ }_{14} \mathrm{Si}^{28}\right)$, germanium $\left({ }_{32} \mathrm{Ge}^{73}\right)$ and tin $\left({ }_{50} \mathrm{Sn}^{119}\right)$ in crystalline form have the same crystal structure and similar electrical properties.

Silicon and germanium are the most popular semiconductors, because of the importance of Ge and Si in present-day electronics.

There are also a number of compound semiconductors, such as metallic oxides and sulphides which are also of great particle importance.

So, that’s all from this article. I hope you get the idea about What are the properties of semiconductors. 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 Electronics. To watch Free Learning Videos on physics by Saransh Gupta sir Install the eSaral App.

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.

For a better understanding of this chapter, please check the detailed notes of Ray Optics class 12 . To watch Free Learning Videos on physics by Saransh Gupta sir Install the eSaral App.

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.

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.

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.

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.