Wave Optics | Physics JEE previous year questions with Solutions

Q 1 :

When a polaroid sheet is rotated between two crossed polaroids, the transmitted light intensity will be maximum for a rotation of [2024]

60°
30°
90°
45°

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(4)

Let $I_{0}$ be intensity of unpolarised light incident on first polaroid.

$I_{1} =$ Intensity of light transmitted from $1^{s t}$ polaroid $= \frac{I_{0}}{2}$

$θ$ be the angle between $1^{s t}$ and $2^{n d}$ polaroid

$ϕ$ be the angle between $2^{n d}$ and $3^{r d}$ polaroid

$θ + ϕ = 90^{o}$ (as $1^{s t}$ and $3^{r d}$ polaroid are crossed)

$ϕ = 90 ° - θ$

$I_{2} =$ Intensity from second polaroid

$I_{2} = I_{1} \cos^{2} θ = \frac{I_{0}}{2} \cos^{2} θ$

$I_{3} =$ Intensity from $3^{r d}$ polaroid

$I_{3} = I_{2} \cos^{2} ϕ = I_{1} \cos^{2} θ \cos^{2} ϕ$

$I_{3} = \frac{I_{0}}{2} \cos^{2} θ \cos^{2} ϕ$

$ϕ = 90 - θ$

$I_{3} = \frac{I_{0}}{2} {[\frac{2 \sin θ \cos θ}{2}]}^{2}$

$I_{3} = \frac{I_{0}}{8} \sin^{2} 2 θ$

$I_{3}$ will be maximum when sin $2 θ = 1$

$2 θ = 90$

Q 2 :

A beam of unpolarised light of intensity $I_{0}$ is passed through a polaroid A and then through another polaroid B which is oriented so that its principal plane makes an angle of $45^{o}$ relative to that of A. The intensity of emergent light is [2024].

$I_{0} / 4$
$I_{0}$
$I_{0} / 2$
$I_{0} / 8$

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(1)

$I_{1} = \frac{I_{0}}{2} and I_{2} = I_{1} \cos^{2} θ$

$= \frac{I_{0}}{2} \cos^{2} 45 ° I_{2} = \frac{I_{0}}{4}$

Q 3 :

When unpolarized light is incident at an angle of 60 on a transparent medium from air, the reflected ray is completely polarized. The angle of refraction in the medium is [2024]

30
60
90
45

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(1)

By Brewster's law

At complete reflection refracted ray and reflected ray are perpendicular.

Q 4 :

In a Young's double slit experiment, three polarizers are kept as shown in the figure. The transmission axes of $P_{1}$ and $P_{2}$ are orthogonal to each other. The polarizer $P_{3}$ covers both the slits with its transmission axis at 45° to those of $P_{1}$ and $P_{2}$ . An unpolarized light of wavelength $λ$ and intensity $I_{0}$ is incident on $P_{1}$ and $P_{2}$ . The intensity at a point after $P_{3}$ where the path difference between the light waves from $S_{1}$ and $S_{2}$ is $λ / 3$ , is [2025]

$\frac{I_{0}}{2}$
$\frac{I_{0}}{4}$
$I_{0}$
$\frac{I_{0}}{3}$

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(1)

$I_{1} = \frac{I_{0}}{2} and I_{2} = I_{1} \cos^{2} θ$

$= \frac{I_{0}}{2} \cos^{2} 45 ° I_{2} = \frac{I_{0}}{4}$

Q 5 :

Two polarisers $P_{1}$ and $P_{2}$ are placed in such a way that the intensity of the transmitted light will be zero. A third polariser $P_{3}$ is inserted in between $P_{1}$ and $P_{2}$ , at the particular angle between $P_{2}$ and $P_{3}$ . The transmitted intensity of the light passing through all three polarisers is maximum. The angle between the polarisers $P_{2}$ and $P_{3}$ is: [2025]

$\frac{π}{4}$
$\frac{π}{6}$
$\frac{π}{8}$
$\frac{π}{3}$

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(1)

Through $P_{2}$ , $I_{1} = I_{0} \sin^{2} (\frac{π}{2} - θ)$

$I_{1} = I_{0} \cos^{2} θ$

Through $P_{3}$ , $I_{net} = (I_{0} \cos^{2} θ) \sin^{2} θ$

$I_{net} = \frac{I_{0}}{4} {[\sin (2 θ)]}^{2} for \max, I_{net} θ = 45 °$

So angle between $P_{2}$ and $P_{3}$ is $\frac{π}{4}$

Q 6 :

Two plane polarised light waves combine at a certain point whose electric field component are $E_{1} = E_{0} \sin ω t$ , $E_{2} = E_{0} \sin (ω t + \frac{π}{3})$ . Find the amplitude of the resultant wave. [2025]

$0.9 E_{0}$
$3.4 E_{0}$
$1.7 E_{0}$
$0.5 E_{0}$

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(3)

$E_{1} = E_{0} \sin (ω t)$

$E_{2} = E_{0} \sin (ω t + \frac{π}{3})$

$E_{R}^{2} = E_{1}^{2} + E_{2}^{2} + 2 E_{1} E_{2} \cos (ϕ)$

$E_{R}^{2} = E_{0}^{2} + E_{0}^{2} + 2 E_{0}^{2} \cos ϕ = 2 E_{0}^{2} (1 + \cos ϕ)$

$E_{R}^{2} = 2 E_{0}^{2} \times 2 \cos^{2} (\frac{π}{6}) = 4 E_{0}^{2} \cdot \frac{3}{4}$

Hence, $E_{R} = E_{0} \sqrt{3} \approx 1.7 E_{0}$

Q 7 :

Given below are two statements: [2023]

Statement I: If the Brewster's angle for the light propagating from air to glass is $θ_{B}$ , then the Brewster's angle for the light propagating from glass to air is $\frac{π}{2} - θ_{B}$ .

Statement II: The Brewster's angle for the light propagating from glass to air is $\tan^{- 1} (μ_{g})$ where $μ_{g}$ is the refractive index of glass.

In the light of the above statements, choose the correct answer from the options given below:

Both Statement I and Statement II are true
Both Statement I and Statement II are false
Statement I is true but Statement II is false
Statement I is false but Statement II is true

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(3)

$Brewster angle μ \Rightarrow \tan i_{p}$

Q 8 :

Two polaroid A and B are placed in such a way that the pass-axis of polaroids are perpendicular to each other. Now, another polaroid C is placed between A and B bisecting the angle between them. If intensity of unpolarised light is $I_{0}$ then intensity of transmitted light after passing through polaroid B will be [2023]

$\frac{I_{0}}{4}$
$\frac{I_{0}}{2}$
$\frac{I_{0}}{8}$
$Z e r o$

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(3)

$I_{A} = \frac{I_{0}}{2} and I_{C} = \frac{I_{0}}{2} \cos^{2} 45 = \frac{I_{0}}{4}$

$I_{B} = I_{C} \cos^{2} 45 = \frac{I_{0}}{8}$

Q 9 :

$' n'$ polarizing sheets are arranged such that each makes an angle $45 °$ with the preceding sheet. An unpolarized light of intensity $I$ is incident into this arrangement. The output intensity is found to be $\frac{I}{64}$ . The value of $n$ will be [2023]

3
6
5
4

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(2)

$After passing through first sheet I_{1} = \frac{l}{2}$

$After passing through second sheet$

$I_{2} = I_{1} \cos^{2} (45 °) = \frac{l}{4}$

$After passing through n^{th} sheet$

$I_{n} = \frac{I}{2^{n}} = \frac{I}{64} \Rightarrow n = 6$

Q 10 :

As shown in figures, three identical polaroids $P_{1}$ , $P_{2}$ and $P_{3}$ are placed one after another. The pass axis of $P_{2}$ and $P_{3}$ are inclined at an angle of $60 °$ and $90 °$ with respect to axis of $P_{1}$ . The source S has an intensity of $256 \frac{W}{m^{2}}$ . The intensity of light at point O is _____ $\frac{W}{m^{2}}$ . [2023]

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(24)

By first polaroid $P_{1}$ intensity will be halved then $P_{2}$ and $P_{3}$ will make intensity $\cos^{2} 60 °$ and $\cos^{2} 30 °$ times respectively.

$Intensity out = \frac{256}{2} \times \frac{1}{4} \times {(\frac{\sqrt{3}}{2})}^{2} = \frac{256 \times 3}{2 \times 4 \times 4} = 24$

Topic Question Set

Q 1 :

When a polaroid sheet is rotated between two crossed polaroids, the transmitted light intensity will be maximum for a rotation of [2024]

Q 2 :

A beam of unpolarised light of intensity $I_{0}$ is passed through a polaroid A and then through another polaroid B which is oriented so that its principal plane makes an angle of $45^{o}$ relative to that of A. The intensity of emergent light is [2024].

Q 3 :

When unpolarized light is incident at an angle of 60 on a transparent medium from air, the reflected ray is completely polarized. The angle of refraction in the medium is [2024]

Q 6 :

Two plane polarised light waves combine at a certain point whose electric field component are $E_{1} = E_{0} \sin ω t$ , $E_{2} = E_{0} \sin (ω t + \frac{π}{3})$ . Find the amplitude of the resultant wave. [2025]

Q 9 :

$' n'$ polarizing sheets are arranged such that each makes an angle $45 °$ with the preceding sheet. An unpolarized light of intensity $I$ is incident into this arrangement. The output intensity is found to be $\frac{I}{64}$ . The value of $n$ will be [2023]

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