Wave Optics Questions — NEET Physics

Q1

In Young's double slit experiment, using monochromatic light of wavelength

\lambda

, the intensity of light at a point on the screen where the path difference is

\lambda

, is

K

units. The intensity of light at a point where the path difference is

\dfrac{\lambda}{3}

will be

A

\dfrac{K}{4}

B K

C

2 K

D

\dfrac{K}{2}

Correct Answer

Option A

Solution

$I=I_0 \cos ^2 \dfrac{\Delta \phi}{2}\left[I_0 \rightarrow\right.$ maximum intensity $]$

\begin{aligned} & I=I_0 \cos ^2 \frac{K \Delta x}{2} \\ & K=I_0 \cos ^2\left(\frac{2 \pi}{\lambda} \times \frac{\lambda}{2}\right) \\ & K=I_0 \\ & K_1=I_0 \cos ^2\left(\frac{2 \pi}{\lambda} \times \frac{\lambda}{3 \times 2}\right)=I_0 \cos ^2\left(\frac{\pi}{3}\right) \\ & K_1=\frac{I_0}{4} \\ & \frac{K}{K_1}=4 \Rightarrow K_1=\frac{K}{4} \end{aligned}

Q2

The intensity of transmitted light when a polaroid sheet, placed between two crossed polaroids at

22.5^{\circ}

from the polarization axis of one of the polaroids, is (

I 0_0

is the intensity of polarised light after passing through the first polaroid):

A

\dfrac{I_0}{8}

B

\dfrac{I_0}{16}

C

\dfrac{I_0}{2}

D

\dfrac{I_0}{4}

Correct Answer

Option A

Solution

\begin{aligned} & I_1=I_0 \cos ^2\left(\frac{45}{2}\right) \\ & I_2=I_1 \cos ^2\left(90-\frac{45}{2}\right) \\ & =I_0 \cos ^2\left(\frac{45}{2}\right) \sin ^2\left(\frac{45}{2}\right) \\ & =\frac{I_0}{4}\left(4 \cos ^2\left(\frac{45}{2}\right) \sin ^2\left(\frac{45}{2}\right)\right) \\ & =\frac{I_0}{4} \sin ^2 45^{\circ}=\frac{I_0}{8} \end{aligned}

Q3

An unpolarized light beam travelling in air is incident on a medium of refractive index 1.73 at Brewster's angle. Then

A Both reflected and transmitted light are perfectly polarized with angles of reflection and refraction close to

60^{\circ}

and

30^{\circ}

, respectively

B Transmitted light is completely polarized with angle of refraction close to

30^{\circ}

C Reflected light is completely polarized and the angle of reflection is close to

60^{\circ}

D Reflected light is partially polarized and the angle of reflection is close to

30^{\circ}

Correct Answer

Option C

Solution

Using Brewster law

\begin{aligned} &\begin{aligned} & \mu=\tan \theta_p \\ & \Rightarrow 1.73=\tan \theta_p \\ & \Rightarrow \sqrt{3}=\tan \theta_p \\ & \Rightarrow \theta_p=60^{\circ} \end{aligned}\\ &\text{ At this polarising angle, reflected light is perfectly polarized and transmitted light is partially polarised. } \end{aligned}

Q4

Two slits in Young's double slit experiment are

1.5 \mathrm{~mm}

apart and the screen is placed at a distance of

1 \mathrm{~m}

from the slits. If the wavelength of light used is

600 \times 10^{-9} \mathrm{~m}

then the fringe separation is

A

4 \times 10^{-5} \mathrm{~m}

B

9 \times 10^{-8} \mathrm{~m}

C

4 \times 10^{-7} \mathrm{~m}

D

4 \times 10^{-4} \mathrm{~m}

Correct Answer

Option D

Solution

To find the fringe separation (also known as fringe width) in Young's double slit experiment, we use the formula:

\beta = \dfrac{\lambda D}{d}

where: $\beta$ is the fringe width (fringe separation). $\lambda$ is the wavelength of the light used.

D

is the distance from the slits to the screen.

d

is the distance between the two slits. Given values are:

d = 1.5 \, \mathrm{mm} = 1.5 \times 10^{-3} \, \mathrm{m}

D = 1 \, \mathrm{m}

\lambda = 600 \times 10^{-9} \, \mathrm{m}

Substituting these values into the formula:

\beta = \dfrac{600 \times 10^{-9} \, \mathrm{m} \times 1 \, \mathrm{m}}{1.5 \times 10^{-3} \, \mathrm{m}}

Calculating the value:

\beta = \dfrac{600 \times 10^{-9}}{1.5 \times 10^{-3}}

\beta = 4 \times 10^{-4} \, \mathrm{m}

Thus, the fringe separation is

4 \times 10^{-4} \, \mathrm{m}

. The correct answer is: Option D

4 \times 10^{-4} \mathrm{~m}

Q5

A beam of unpolarized light of intensity I0 is passed through a polaroid A, then through another polaroid B, oriented at

60^\circ

and finally through another polaroid C, oriented at 45

^\circ

relative to B as shown. The intensity of emergent light is:

A

\dfrac{I_0}{16}

B

\dfrac{I_0}{4}

C

\dfrac{I_0}{2}

D

\dfrac{I_0}{32}

Correct Answer

Option A

Solution

\begin{aligned} & I_1=\frac{I_0}{2} \\ & I_B=\left(I_1\right) \cos ^2 60^{\circ} \\ & =\frac{I_0}{2}\left(\frac{1}{2}\right)^2=\frac{I_0}{8} \\ & I_C=I_B \cos ^2 45^{\circ} \\ & =\frac{I_0}{8} \times\left(\frac{1}{\sqrt{2}}\right)^2=\frac{I_0}{16} \end{aligned}

Q6

If the monochromatic source in Young's double slit experiment is replaced by white light, then

A Interference pattern will disappear

B There will be a central dark fringe surrounded by a few coloured fringes

C There will be a central bright white fringe surrounded by a few coloured fringes

D All bright fringes will be of equal width

Correct Answer

Option C

Solution

In Young’s double-slit experiment, if a monochromatic source is replaced by white light, which contains multiple wavelengths, the patterns observed on the screen will differ significantly.

First, remember that the pattern formed in the double-slit experiment consists of both bright and dark fringes due to constructive and destructive interference, respectively.

The position and intensity of these fringes depend on the wavelength of the light used.

For monochromatic light (light of a single wavelength), the interference pattern is stable, with bright and dark fringes evenly spaced.

Each fringe is uniformly bright or dark.

When using white light, which is a combination of various wavelengths of light, each color, or each wavelength, forms its own interference pattern with slightly different fringe spacing.

This occurs because the separation between fringes

\Delta y

is given by the formula:

\Delta y = \frac{\lambda L}{d}

Where: $\lambda$ is the wavelength of light

L

is the distance from the slits to the screen

d

is the separation between the slits Since different wavelengths have different values for $\lambda$ , each color’s fringes will be at slightly different positions.

The result is that near the center of the pattern, where there is the least path difference, all wavelengths constructively interfere to form a bright white central fringe.

However, moving away from the center, the fringes start to show different colors as the path difference between the light from the two slits increases.

This causes a dispersion of colors with different orders of fringes dominated by different colors.

The constructive and destructive interference patterns of different wavelengths slightly offset one another.

Thus: Option A - Interference pattern will disappear is incorrect because the interference pattern does not disappear but changes due to the dispersion of colors.

Option B - There will be a central dark fringe surrounded by a few coloured fringes is incorrect as the central fringe in the presence of white light is bright, not dark.

Option C - There will be a central bright white fringe surrounded by a few coloured fringes is correct.

This is because all the wavelengths interfere constructively at the center, creating a bright white fringe, succeeded by colored fringes due to the varying interference conditions for each wavelength.

Option D - All bright fringes will be of equal width is incorrect.

The width and spacing of the fringes vary by wavelength, so the interference pattern will not have uniform fringe widths.

Therefore, the correct answer is Option C : There will be a central bright white fringe surrounded by a few coloured fringes.

Q7

An unpolarised light beam strikes a glass surface at Brewster's angle. Then

A The reflected light will be partially polarised.

B The refracted light will be completely polarised.

C Both the reflected and refracted light will be completely polarised.

D The reflected light will be completely polarised but the refracted light will be partially polarised.

Correct Answer

Option D

Solution

According to Brewster's law, reflected rays are completely polarized and refracted rays are partially polarized.

Q8

Which set of colours will come out in air for a situation shown in figure?

A Yellow, Orange and Red

B All

C Orange, Red and Violet

D Blue, Green and Yellow

Correct Answer

Option A

Solution

\sin {i_c} = {1 \over \mu } \propto \lambda

{i_c} \propto \lambda

Yellow, orange, red emerge from air.

Q9

If the screen is moved away from the plane of the slits in a Young's double slit experiment, then the :

A linear separation of the fringes decreases

B angular separation of the fringes increases

C angular separation of the fringes decreases

D linear separation of the fringes increase

Correct Answer

Option D

Solution

In YDSE, Separation between fringes is Related to fringed width

\beta = {{D\lambda } \over d}

If D is increased $\beta$ increases while Angular separation is independent of distance between slits and screen (D) As angular separation

= {\beta \over D} = {\lambda \over d}

Q10

After passing through a polariser a linearly polarised light of intensity I is incident on an analyser making an angle of 30

^\circ

with that of the polariser. The intensity of light emitted from the analyser will be

A

{{2I} \over 3}

B

{I \over 2}

C

{I \over 3}

D

{{3I} \over 4}

Correct Answer

Option D

Solution

According to Malu's Law

{I_{out}} = {I_{in}}\,{\cos ^2}\theta

{I_{out}} = I\,{\cos ^2}(30^\circ )

{I_{out}} = {{3I} \over 4}