JEE Advanced Wave Optics Previous Year Questions (PYQ) with Detailed Solutions

Q 1 :

A point source $S$ emits unpolarized light uniformly in all directions. At two points A and B, the ratio $r = \frac{I_{A}}{I_{B}}$ of the intensities of light is 2. If a set of two polaroids having $45 °$ angle between their pass-axes is placed just before point B, then the new value of $r$ will be _______. [2024]

(8)

Using $I = I_{0} \cos^{2} θ$

New intensity at B,

$I'_{B} = (\frac{I_{B}}{2}) \cos^{2} 45 ° = \frac{I_{B}}{4}$

New value of $r$ $= \frac{I_{A}}{I_{B}} = \frac{4 I_{A}}{I_{B}} = 4 \times 2 = 8$

Q 2 :

A solid glass sphere of refractive index $n = \sqrt{3}$ and radius $R$ contains a spherical air cavity of radius $\frac{R}{2}$ , as shown in the figure. A very thin glass layer is present at the point O so that the air cavity (refractive index $n = 1$ ) remains inside the glass sphere. An unpolarized, unidirectional and monochromatic light source $S$ emits a light ray from a point inside the glass sphere towards the periphery of the glass sphere. If the light is reflected from the point O and is fully polarized, then the angle of incidence at the inner surface of the glass sphere is $θ$ . The value of $\sin θ$ is _______. [2025]

[IMAGE 1166]

(0.75)

[IMAGE 1167]

Using Brewster's law, $\tan θ = μ$

$\tan α = \sqrt{3}$ $∴$ $α = 60 °$

From Snell's law,

$\sqrt{3} \sin β = 1 \times \sin α$ $∴$ $β = 30 °$

$\frac{R}{2 \sin 30 °} = \frac{x}{\sin 120 °}$

$\frac{R}{\sin 120 °} = \frac{R \sqrt{3}}{2 \sin θ}$ $\Rightarrow \sin θ = \frac{\sqrt{3}}{2} \times \frac{\sqrt{3}}{2}$

$∴$ $\sin θ = \frac{3}{4} = 0.75$

Q 3 :

A plane polarized blue light ray is incident on a prism such that there is no reflection from the surface of the prism. The angle of deviation of the emergent ray is $δ = 60 °$ (see Figure-1). The angle of minimum deviation for red light from the same prism is $δ_{\min} = 30 °$ (see Figure-2). The refractive index of the prism material for blue light is $\sqrt{3}$ . Which of the following statement(s) is(are) correct? [2023]

[IMAGE 1168]

The blue light is polarized in the plane of incidence.
The angle of the prism is $45 °$ .
The refractive index of the material of the prism for red light is $\sqrt{2}$ .
The angle of refraction for blue light in air at the exit plane of the prism is $60 °$ .

1

2

3

4

Select one or more options

(1, 3, 4)

[IMAGE 1169]

Since for blue light ray, there is no reflection from the surface of the prism, so $\tan θ_{B} = μ_{B} = \sqrt{3}$

Angle of incidence, $i = θ_{B} = 60 °$

From Snell's law,

$1 \sin 60 ° = \sqrt{3} \sin r_{1}$ $\Rightarrow 1 \times \frac{\sqrt{3}}{2} = \sqrt{3} \sin r_{1}$

$∴$ $r_{1} = 30 °$ $∵$ $r_{1} + r_{2} = A$

$δ = (i + e) - A$ $∴$ $60 ° = 60 ° + e - A$ $\Rightarrow e = A$

Again, $\sqrt{3} \sin r_{2} = 1 \sin e$

$\Rightarrow \sqrt{3} \sin (A - 30 °) = \sin A$ $\Rightarrow \sqrt{3} \sin 30 ° = \sin A$

$∴$ $\frac{\sqrt{3}}{2} = \sin A$ $∴$ $A = 60 °$

So, $e = 60 °$

For red light,

$μ = \frac{\sin (\frac{A + δ_{\min}}{2})}{\sin (\frac{A}{2})}$ $= \frac{\sin (\frac{60 ° + 30 °}{2})}{\sin (\frac{60 °}{2})}$ $= \frac{\sin 45 °}{\sin 30 °}$

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

Therefore, options (1, 3, 4) are correct.

Q 4 :

The electric field associated with an electromagnetic wave propagating in a dielectric medium is given by $\vec{E} = 30 (2 \hat{x} + \hat{y}) \sin [2 π (5 \times 10^{14} t - \frac{10^{7}}{3} z)] {Vm}^{- 1} .$

Which of the following option(s) is(are) correct? [Given: The speed of light in vacuum, $c = 3 \times 10^{8} {ms}^{- 1}$ ] [2023]

$B_{x} = - 2 \times 10^{- 7} \sin [2 π (5 \times 10^{14} t - \frac{10^{7}}{3} z)] Wb m^{- 2} .$
$B_{y} = 2 \times 10^{- 7} \sin [2 π (5 \times 10^{14} t - \frac{10^{7}}{3} z)] Wb m^{- 2} .$
The wave is polarized in the $x y$ -plane with polarization angle $30 °$ with respect to the $x$ -axis.
The refractive index of the medium is 2.

1

2

3

4

Select one or more options

(1, 4)

Speed of light in the medium, $C_{medium} = \frac{ω}{k} = \frac{5 \times 10^{14}}{\frac{10^{7}}{3}}$

$= 1.5 \times 10^{8} m/s$ $∴$ $μ = \frac{C_{air}}{C_{medium}} = 2$

Also,

$C_{medium} = \frac{E}{B}$ $\Rightarrow B = \frac{E}{C_{m}} = \frac{30 \sqrt{5}}{1.5 \times 10^{8}} = 2 \sqrt{5} \times 10^{- 7}$

[IMAGE 1170]

$B_{d i r e c t i o n}$ $= \vec{V} \times \vec{E} \equiv \hat{k} \times (2 \hat{i} + \hat{j}) = \frac{2 \hat{j} - \hat{i}}{\sqrt{5}}$

[IMAGE 1171]

$∴$ $B_{x} = 2 \times 10^{- 7} (- \hat{i} + 2 \hat{j}) \sin [2 π (5 \times 10^{14} t - \frac{10^{7}}{3} z)]$

Topic Question Set

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Topic Question Set

Q 1 : A point source S emits unpolarized light uniformly in all directions. At two points A and B, the ratio r=IAIB of the intensities of light is 2. If a set of two polaroids having 45° angle between their pass-axes is placed just before point B, then the new value of r will be _______. [2024]

Q 4 : The electric field associated with an electromagnetic wave propagating in a dielectric medium is given by E→=30(2x^+y^)sin[2π(5×1014t-1073z)] Vm-1. Which of the following option(s) is(are) correct? [Given: The speed of light in vacuum, c=3×108 ms-1] [2023]

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