JEE Main Previous Year Questions with Solutions on Wave Optics

Optics is broadly divided into Geometrical optics and Physical optics. Physical optics deals with the wave nature of light. The phenomenon of interference, diffraction and polarisation can be explained on the basis of the wave nature of light. The light which is a form of energy can be transferred from one point to another either by particle motion or by wave motion. Different theories were proposed to explain the nature of light.

The earliest theory to explain the nature of light was proposed by Sir Issac Newton. The theory was called the Corpuscular theory. According to this theory, the light was considered as a stream of tiny particles known as corpuscles.

The next theory about light was proposed and developed by Christian Huygens in the year 1679. Huygens proposed that light is in the form of longitudinal waves. He assumed that the entire universe was filled with ether.

Download JEE Main Wave Optics Previous Year Questions with Solutions PDF

JEE Main Previous Year Solved Questions on Wave Optics

Q1: In Young’s double-slit experiment with slit separation 0.1 mm, one observes a bright fringe at angle 1/40 rad by using the light of wavelength λ₁. When the light of wavelength λ₂ is used a bright fringe is seen at the same angle in the same setup. Given that 1 and 2 are in the visible range (380 nm to 740 nm), their values are

(a) 400 nm, 500 nm

(b) 625 nm, 500 nm

(d) 380 nm, 500 nm

Solution

Path difference = dsinθ = d × θ = (0.1mm)(1/40) = 2.5 x 10^-3 mm = 2500 nm

For bright fringes, path difference = nλ

So, 2500 = nλ₁ = mλ₂

n = 4, m = 5

or λ₁ = 2500/4 = 625 nm

λ₂ = 2500/5 = 500 nm

Answer: (b) 625 nm, 500 nm

Q2: In Young’s double-slit experiment, the path difference, at a certain point on the screen, between two interfering waves is (⅛)^th of wavelength. The ratio of the intensity at this point to that at the centre of a bright fringe is close to

(a) 0.80

(b) 0.94

(d) 0.74

Solution

The phase difference between two waves is given as

(Δx) x (2π/λ) = (λ/8) x (2π/λ) = π/4

So, the intensity at this point is

I = I₀cos²(Φ/2)

I = I₀cos²(π/8)

I = I₀ [(1+cos(π/4))/2] = = I₀[(1+(1/√2)/2] = 0.85I₀

Answer: (c) 0.85

Q3: In a double-slit experiment, green light (5303 Å) falls on a double slit having a separation of 19.44 m and a width of 4.05 m. The number of bright fringes between the first and the second diffraction minima is

(a) 10

(b) 04

(d) 09

Solution

λ_g = 5303 Å, d = 19.44 m, a = 4.05 m

For diffraction location of first minima and second minima

y₁ = Dλ/a, y₂ = 2Dλ/a

y₂ – y₁ = (2Dλ/a) – (Dλ/a) = Dλ/a

β = Dλ/d

Number of bright fringes = (y₂ – y₁)/β

= (Dλ/a) x (d/Dλ)

= d/a

= 19.44/4.05 = 5

Answer: (c) 05

Q4: In an interference experiment the ratio of amplitudes of coherent waves is (a₁/a₂) = (⅓). The ratio of maximum and minimum intensities of fringes will be

(a) 4

(b) 9

(d) 18

Solution

(a₁/a₂) = (⅓)

I_max = a₁ + a₂

I_min = a₁ – a₂

(I_max/I_min) = [(a₁ + a₂)²/(a₁ – a₂)²] = [(1 + 1/3)²/(1 – 1/3)²] = (4/2)²= 4

Answer: (a) 4

Q5: Calculate the limit of resolution of a telescope objective having a diameter of 200 cm, if it has to detect light of wavelength 500 nm coming from a star.

(a) 610 × 10^–9radian

(b) 152.5 × 10^–9 radian

(d) 305 × 10^–9 radian

Solution

The limit of resolution,

θΔ= 1.22λ/a = [(1.22 x 500 x10^-9)/(200 x 10^-2)] = 3.05 × 10^–7 radian = 305 × 10^–9 radian

Answer: (d) 305 × 10^–9 radian

Q6: In Young’s double-slit experiment, the ratio of the slit’s width is 4: 1. The ratio of the intensity of maxima to minima, close to the central fringe on the screen, will be

(a) (√3 + 1)⁴:16

(b) 9: 1

(d) 4: 1

Solution

I₁ = 4I₀

I₂ = I₀

I_max= (√I₁+ √I₂)²

= (2√I₀+ √I₀)²= 9I₀

I_min= (√I₁– √I₂)²

= (2√I₀-√I₀)²= I₀

(I_max/I_min) = 9/1

Answer: (b) 9: 1

Q7: In Young’s double-slit experiment, slits are separated by 0.5 mm and the screen is placed 150 cm away. A beam of light consisting of two wavelengths, 650 nm and 520 nm, is used to obtain interference fringes on the screen. The least distance from the common central maximum to the point where the bright fringes due to both the wavelengths coincide are

(a) 1.56 mm

(b) 7.8 mm

(d) 15.6 mm

Solution

Let y be the distance from the central maximum to the point where the bright fringes due to both the wavelengths coincides.

Now, for λ₁, y = m λ₁D/d

For λ₂, y = nλ₂D/d

mλ₁ = nλ₂

(m/n) = λ₂/ λ₁= (520)/(650) =4/5

i.e. with respect to central maximum 4^th bright fringe of λ₁ coincides with 5^th bright fringe of λ₂

Now, y = (4 x 650 x 10^-9 x 1.5)/(0.5 x 10^-3) m

y = 7.8 × 10^–3m or y = 7.8 mm

Answer: (b) 7.8 mm

Q8: A single slit of width b is illuminated by coherent monochromatic light of wavelength λ. If the second and fourth minima in the diffraction pattern at a distance 1 m from the slit are at 3 cm and 6 cm, respectively from the central maximum, what is the width of the central maximum? (i.e. the distance between the first minimum on either side of the central maximum)

(a) 6.0 cm

(b) 1.5 cm

(d) 3.0 cm

Solution

For single slit diffraction, sin θ = nλ/b (n = 1,2,3 —-)

sin θ ≈ θ ≈ tan θ

btan θ = nλ

btan θ₁ = 2λ

b (y₁/d) = 2λ

btan θ₂ = 2λ

b (y₂/d) = 2λ

(y₂ – y₁) b/D = 2λ

(6 – 3 )b/D = 2λ

3b/D = 2λ

λD/b = 3/2 = 1.5 cm

Answer: (b) 1.5 cm

Q9: Unpolarized light of intensity I₀ is incident on the surface of a block of glass at Brewster’s angle. In that case, which one of the following statements is true?

(a) Transmitted light is partially polarized with intensity I₀ /2

(b) Transmitted light is completely polarized with intensity less than I₀ /2

(d) The reflected light is partially polarized with intensity I₀ /2

Solution

At Brewster’s angle, i = tan^–1(μ), the reflected light is completely polarized, whereas refracted light is partially polarized. Thus, the reflected ray will have lesser intensity compared to the refracted ray.

Answer: (c) The reflected light is completely polarized with intensity less than I₀ /2

Q10: A beam of unpolarized light of intensity I₀ 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° relative to that of A. The intensity of the emergent light is

(a) I₀/8

(b) I₀

(d) I₀/4

Solution

The intensity of light after passing polaroid A is

I₁ = I₀/2

Now this light will pass through the second polaroid B whose axis is inclined at an angle of 45° to the axis of polaroid A. So in accordance with Malus law, the intensity of light emerging from polaroid B is

I₂ = I₁cos²45 = (I₀/2)(1/√2)²= I₀/4

Answer: (d) I₀/4