Wave Optics JEE Advanced Previous Year Questions With Solutions

Wave optics, also called Physical optics, deals with the study of various phenomena such as polarisation, diffraction, interference and other occurrences where ray approximation of geometric optics cannot be done. Thus, the section of optics that deals with the behaviour of light and its wave characteristics is said to be wave optics.

Download Wave Optics JEE Advanced Previous Year Questions With Solutions PDF

Question 1) In an interference experiment the ratio of the amplitude of coherent waves is (a₁/a₂) = ⅓. The ratio of maximum and minimum intensities of fringes will be:

(A) 2

(B) 18

(C) 4

(D) 9

Answer: (C) 4

Solution:

Given the amplitude ratio of waves is (a₁/a₂) = (3/1)

So,

\(\begin{array}{l}\frac{I_{max}}{I_{min}}= \left [ \frac{a_{2}+a_{1}}{a_{2}-a_{1}} \right ]^{2}\end{array} \)

\(\begin{array}{l}\frac{I_{max}}{I_{min}}= \left [ \frac{\frac{a_{2}}{a_{1}}+1}{\frac{a_{2}}{a_{1}}-1} \right ]^{2}=\left ( \frac{3+1}{3-1} \right )=\left ( \frac{4}{2} \right )^{2}= 4/1 = 4\end{array} \)

Question 2) Two coherent sources produce waves of different intensities which interfere. After interference, the ratio of the maximum intensity to the intensity is 16. The intensity of the waves are in the ratio:

(A) 16: 9

(B) 25: 9

(C) 4: 1

(D) 5: 3

Answer: (B) 25: 9

Solution:

As we know,

\(\begin{array}{l}\frac{I_{max}}{I_{min}}=\left [ \frac{A_{1}+A_{2}}{A_{1}-A_{2}} \right ]^{2}\end{array} \)

I_max/I_min = 16

⇒ A_max/A_min = 4

⇒ (A₁+A₂)/(A₁ – A₂) = 4/1

Using componendo and dividendo

A₁/A₂ = 5/3

⇒ I₁/I₂ = (5/3)²= 25/9

Question 3) On a hot summer night, the refractive index of air is smallest near the ground and increases with a height from the ground. When a light beam is directed horizontally, Huygen’s principle leads us to conclude that as it travels, the light beam:

(A) bends downwards

(B) bends upwards

(C) becomes narrower

(D) goes horizontally without any deflection

Answer: (B) bends upwards

Solution:

Question 4) A thin slice is cut out of a glass cylinder along a plane parallel to its axis. The slice is placed on a flat glass plate as shown in the figure. The observed interference fringes from this combination shall be

(A) Straight

(B) Circular

(C) Equally speed

(D) having fringe spacing which increases as we go outwards

Answer: (A) Straight

Solution:

Locus of equal path difference are lines running parallel to the axis of the cylinder. Hence straight interference fringes will be observed

Question 5) Two coherent monochromatic light beams of intensities I and 4I are superposed. The maximum and minimum possible intensities in the resulting beam are

(A) 5I and I

(B) 5I and 3I

(C) 9I and I

(D) 9I and 3I

Answer: (C) 9I and I

Solution:

Let I₁ = I

I₂ = 4I

I_max = (√I₁ + √I₂)² = (√I + √4I)² = (3√I)² = 9I

and I_min = (√I₁ -√I₂)² = (√I -√4I)² = I

Question 6) In Young’s double slit experiment, light of 500 nm is used to produce an interference pattern. When the distance between the slits is 0.05 mm, the angular width (in degree) of the fringes formed on the distance screen is close to:

(A) 0.17⁰

(B) 0.57⁰

(C) 1.7⁰

(D) 0.07⁰

Answer: (B) 0.57⁰

Solution:

Given,

Wavelength, λ = 500 nm

Distance between the slits, d = 0.05 mm

The angular width of the fringe formed,

θ = λ/d

= (500 x 10^-9)/(0.05 x 10^-3) = 0.01 rad = 0.57⁰

Question 7) In a Young’s double slit experiment, 16 fringes are observed in a certain segment of the screen when light of wavelength 700 nm is used. If the wavelength of light is changed to 400 nm, the number of fringes observed in the same segment of the screen would be:

(A) 24

(B) 30

(C) 18

(D) 28

Answer: (D) 28

Solution:

Let n₁ fringes are visible with light of wavelength λ₁ and n₂ with light of wavelength λ₁. Then

\(\begin{array}{l}\beta =\frac{n_{1}D\lambda _{1}}{d}=\frac{n_{2}D\lambda _{2}}{d}\end{array} \)

⇒ n₂ = (λ₁/λ₂) x n₁

= (700/400) x 16 = 28

Question 8) In Young’s double-slit experiment, the separation between the slits is 0.15 mm. In the experiment, a source of light of wavelength 589 nm is used and the interference pattern is observed on a screen kept 1.5 m away. The separation between the successive bright fringes on the screen is

(A) 6.9 mm

(B) 3.9 mm

(C) 5.9 mm

(D) 4.9 mm

Answer: (C) 5.9 mm

Solution:

Fringe width = (D/d) λ = (1.5/0.15 x 10^-3) x 589 x 10^-9 m

≈ 5.9 mm

Question 9) In a double-slit experiment, when a thin film of thickness t having a refractive index μ is introduced in front of one of the slits, the maximum at the centre of the fringe pattern shifts by one fringe width. The value of t is (≈ is the wavelength of the light used)

(A) 2λ/(µ – 1)

(B) λ/2(µ – 1)

(C) λ/(µ – 1)

(D) λ/(2µ – 1)

Answer: (C) λ/(µ – 1)

Solution:

\(\begin{array}{l}\frac{D(\mu -1)t}{d}=\frac{D\lambda }{d}\end{array} \)

Therefore, t = λ/(µ – 1)

Question 10) In a Young’s double slit experiment, the slits are placed 0.320 mm apart. Light of wavelength λ = 500 nm is incident on the slits. The total number of bright fringes that are observed in the angular range -30⁰ ≤ θ ≤ 30⁰

(A) 640

(B) 320

(C) 321

(D) 641

Answer: (D) 641

Solution:

For ‘n’ number of maximas

dsinθ = nλ

0.32 x 10^-3 sin 30⁰ = n x 500 x 10^-9

n = (0.32 x 10^-3/500 x 10^-9) x (½) = 320

Hence total number of maximas observed in angular range -30⁰≤θ≤30⁰

= 320 + 1 + 320 = 641

Waves and Simple Harmonic Motion Important Questions

Waves and Simple Harmonic Motion Important Questions

Young’s Double Slit Experiment

Superposition of Waves

Power Transmitted by a Wave

Also Read:

Wave Optics JEE Main Previous Year Questions With Solutions

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