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*Simulator*

**Previous Years AIEEE/JEE Mains Questions**

(1) 442.5 nm (2) 776.8 nm (3) 393.4 nm (4) 885.0 nm

** [AIEEE-2009]**

**Sol.**(1)

$\frac{3 \mathrm{D} \lambda_{1}}{\mathrm{d}}=\frac{4 \mathrm{D} \lambda_{2}}{\mathrm{d}}$

**Direction : Questions number 3 to 5 are based on the following paragraph.**

An initially parallel cylindrical beam travels in a medium of refractive index $\mu(\mathrm{I})=\mu_{0}+\mu_{2} \mathrm{I}$, where $\mu_{0}$ and $\mu_{2}$ are positive constants and I is the intensity of the light beam. The intensity of the beam is decreasing with increasing radius.

(1) planar

(2) convex

(3) concave

(4) convex near the axis and concave near the periphery

**[AIEEE-2010]**

**Sol.**(1)

For a parallel beam wavefronts are plane wave fronts.

(1) maximum on the axis of the beam

(2) minimum on the axis of the beam

(3) the same everywhere in the beam

(4) directly proportional to the intensity I

**[AIEEE-2010]**

**Sol.**(2)

Intensity is maximum at the axis of the cylindrical beam. Hence $\mu$ is maximum at the axis due to

the relation $\mu=\mu_{0}+\mu_{2} \mathrm{I}$

(1) travel as a cylindrical beam

(2) diverge

(3) converge

(4) diverge near the axis and converge near the periphery

**[AIEEE-2010]**

**Sol.**(3)

v is minimum at the axis of beam. Hence beam will converge.

(1) 3 : 2 (2) 2 : 1 (3) $\sqrt{2}: 1$ (4) 4: 1

**[AIEEE-2011]**

**Sol.**(4)

$\mathrm{I}=\mathrm{I}_{\max } \cos ^{2}\left(\frac{\pi}{\lambda} \times \frac{\lambda}{4}\right)$

$\mathrm{I}=\mathrm{I}_{\max } \cos ^{2}\left(\frac{\pi}{4}\right)=\frac{1}{\sqrt{2}} \mathrm{I}_{\max }$

$\frac{\mathrm{I}_{\max }}{2}=\sqrt{2}$

**Statement-1:**On viewing the clear blue portion of the sky through a Calcite Crystal, the intensity of transmitted light varies as the crystal is rotated.

**Statement-1:** The light coming from the sky is polarized due to scattering of sun light by particles in the atmosphere. The scattering is largest for blue light.

(1) Statement-1 is false, statement-2 is true

(2) Statement-1 is true, statement-2 is false

(3) Statement-1 is true, statement-2 true; statement-2 is the correct explanation of statement-1

(4) Statement-1 is true, statement-2 is true; statement -2 is not correct explanation of statement- 1.

** [AIEEE-2011]**

**Sol.**(3)

$\mathrm{I}=\mathrm{I}_{0} \cos ^{2} \theta ;$ Law of Malus

(1) 4 (2) 2 3) 1 (4) 0.5

**[AIEEE-2011]**

**Sol.**(2)

$\mathrm{I}_{1}=4 \mathrm{I}_{0}$

$\mathrm{I}_{2}=2 \mathrm{I}_{0}$

$\frac{\mathrm{I}_{1}}{\mathrm{I}_{2}}=2$

A thin air film is formed by putting the convex surface of a plane-convex lens over a plane glass plate. With monochromatic light, this film gives an interference pattern due to light reflected from the top (convex) surface and the bottom (glass plate) surface of the film

**Statement-1: **

When light reflects from the air-glass plate interface, the reflected wave suffers a phase change of $\pi$.

**Statement-2: **The centre of the interference pattern is dark :-

(1) Statement-1 is true, Statement-2 is true and Statement-2 is not the correct explanation of

Statement-1.

(2) Statement-1 is false, Statement-2 is true

(3) Statement-1 is true, Statement-2 is false

(4) Statement-1 is true, Statement-2 is true and Statement-2 is the correct explanation of statement- 1.

**[AIEEE-2011]**

**Sol.**(1)

**[AIEEE-2012]**

**Sol.**(1)

$\mathrm{I}_{\mathrm{R}}=\mathrm{I}_{1}+\mathrm{I}_{2}+2 \sqrt{\mathrm{I}_{1} \mathrm{I}_{2}} \cos \Delta \phi$

$\mathrm{I}_{\mathrm{R}}=4 \mathrm{I}_{0}+\mathrm{I}_{0}+2 \times 2 \mathrm{I}_{0} \cos \phi$

$\mathrm{I}_{\mathrm{R}}=\mathrm{I}_{0}[5+4 \cos \phi]$

$\left(\mathrm{I}_{\mathrm{R}}\right)_{\max }=\mathrm{I}_{\mathrm{m}}=9 \mathrm{I}_{0}$

$\mathrm{I}_{\mathrm{R}}=\frac{\mathrm{I}_{\mathrm{m}}}{9}[1+4(1+\cos \phi)]$

$\mathrm{I}_{\mathrm{R}}=\frac{\mathrm{I}_{\mathrm{m}}}{9}\left[1+8 \cos ^{2} \frac{\phi}{2}\right]$

**Statement-1 :** Davisson – Germer experiment established the wave nature of electrons.

**Statement-2:** If electrons have wave nature, they can interfere and show diffraction.

(1) Statement-1 is true, Statement-2 is true and Statement-2 is the correct explanation for Statement-1.

(2) Statement-1 is true, Statement-2 is true and Statement-2 is not the correct explanation of

Statement-I

(3) Statement-1 is false, Statement-2 is true.

(4) Statement-1 is true, Statement-2 is false

**[AIEEE – 2012]**

**Sol.**(1)

( 1) $\mathrm{I}_{0}$ $(2) \mathrm{I}_{0} / 2$ (3) $\mathrm{I}_{0} / 4$ (4) $\mathrm{I}_{0} / 8$

** [JEE-Mains 2013]**

**Sol.**(3)

$\mathrm{I}=\left(\frac{\mathrm{I}_{0}}{2}\right) \cos ^{2} 45^{\circ}$

(1) points (2) straight lines (3) semicircles (4) concentric circles

** [JEE-Mains 2013]**

**Sol.**(4)

(1) 1 (2) $\frac{1}{3}$ (3) 3 (4) $\frac{3}{2}$

**[JEE-Main 2014]**

**Sol.**(2)

When polaroid is at Angle $30^{\circ}$ with beam $\mathrm{A},$ it makes $60^{\circ}$ with beam $\mathrm{B}$

by malus law

$\mathrm{I}_{\mathrm{A}} \cos ^{2} 30^{\circ}=\mathrm{I}_{\mathrm{B}} \cos ^{2} 60^{\circ}$

$\Rightarrow \frac{\mathrm{I}_{\mathrm{A}}}{\mathrm{I}_{\mathrm{B}}}=\frac{1}{3}$

(1) $100 \mu \mathrm{m}$

(2) $300 \mu \mathrm{m}$

(3) (3) $1 \mu m$

(4) $30 \mu \mathrm{m}$

**[JEE-Mains 2015]**

**Sol.**(4)

$\frac{\mathrm{d}_{0}}{\mathrm{d}_{1}}=\theta=\frac{1.22 \lambda}{\mathrm{d}}$

$\mathrm{d}_{0}=\frac{1.22 \times 500 \times 10^{-9}}{0.5 \times 10^{-2}} \times 25 \times 10^{-2}$

$\mathrm{d}_{0}=1.22 \times 25 \times 10^{-6}$

$\mathrm{d}_{0} \square 30 \mu \mathrm{m}$

(1) bends downwards

(2) bends upwards

(3) becomes narrower

(4) goes horizontally without any deflection

**[JEE-Mains 2015]**

**Sol.**(2)

(1) $\mathrm{a}=\frac{\lambda^{2}}{\mathrm{L}}$ and $\mathrm{b}_{\mathrm{min}}=\sqrt{4 \lambda \mathrm{L}}$

(2) $\mathrm{a}=\frac{\lambda^{2}}{\mathrm{L}}$ and $\mathrm{b}_{\min }=\left(\frac{2 \lambda^{2}}{\mathrm{L}}\right)$

(3) $\mathrm{a}=\sqrt{\lambda \mathrm{L}}$ and $\mathrm{b}_{\min }=\left(\frac{2 \lambda^{2}}{\mathrm{L}}\right)$

(4) $\mathrm{a}=\sqrt{\lambda \mathrm{L}}$ and $\mathrm{b}_{\min }=\sqrt{4 \lambda \mathrm{L}}$

**[JEE-Mains 2016]**

**Sol.**(4)

Spot size (diameter) $\mathrm{b}=2\left(\frac{\lambda \mathrm{L}}{2 \mathrm{a}}\right)+2 \mathrm{a}$

$\mathrm{a}^{2}+\lambda \mathrm{L}-\mathrm{ab}=0$

For Real roots $\mathrm{b}^{2}-4 \mathrm{L} \lambda \geq 0$

$\mathrm{b}_{\min .}=\sqrt{4 \lambda \mathrm{L}}$

by eq. ( i ) $\quad a=\sqrt{\lambda L}$

(1) 9.75 mm (2) 15.6 mm (3) 1.56 mm (4) 7.8 mm

** [JEE-Mains 2017]**

**Sol.**(4)

For common maxima

$\mathrm{n}_{1} \lambda_{1}=\mathrm{n}_{2} \lambda_{2}$

$\mathrm{n}_{1} \times 650=\mathrm{n}_{2} \times 520$

$\frac{\mathrm{n}_{1}}{\mathrm{n}_{2}}=\frac{4}{5}$

$\frac{\mathrm{n}_{1}}{\mathrm{D}_{2}}=\mathrm{n} \lambda$

$\frac{\mathrm{yd}}{\mathrm{D}}=\mathrm{n} \lambda$

$\mathrm{y}=\frac{4 \times 650 \times 10^{-9} \times 1.5}{0.5 \times 10^{-3}}$

50 cm from the slits. If the observed fringe width is 1 cm, what is slit separation distance? (i.e. distance between the centres of each slit.)

(1) $50 \mu \mathrm{m}$

(2) $75 \mu \mathrm{m}$

(3) $100 \mu \mathrm{m}$

(4) $25 \mu \mathrm{m}$

**[JEE-Mains 2018]**

**Sol.**(4)

In diffraction

d $\sin 30^{\circ}=\lambda$

$\lambda=\frac{\mathrm{d}}{2}$

Young’s fringe width

[d’- separation between two slits]

$$

\beta=\frac{\lambda \times \mathrm{D}}{\mathrm{d}^{\prime}}

$$

$10^{-2}=\frac{\mathrm{d}}{2} \times \frac{50 \times 10^{-2}}{\mathrm{d}}$

$10^{-2}=\frac{10^{-6} \times 50 \times 10^{-2}}{2 \times \mathrm{d}^{\prime}}$

$\mathrm{d}^{\prime}=25 \mu \mathrm{m}$

(1) $30^{\circ}$ ( 2) $45^{\circ}$ (3) $60^{\circ}$ (4) $0^{\circ}$

**[JEE-Mains 2018]**

**Sol.**(2)

Axis of transmission of A & B are parallel.

$\frac{\mathrm{I}}{2} \cos ^{4} \theta=\frac{\mathrm{I}}{8} \Rightarrow \cos ^{4} \theta=\frac{1}{4}$

$\cos \theta=\frac{1}{\sqrt{2}}$

$\theta=45^{\circ}$

Jee main 2014 online kaha h

Thank u so much but in case answer are are not in detailed manner

Thank you