JEE Advanced · 2024 · 26 May 2024 · Paper 2 · Wave Optics

WhatsApp Email

In a Young's double slit experiment, each of the two slits A and B, as shown in the figure, are oscillating about their fixed center and with a mean separation of 0.8 mm . The distance between the slits at time

t

is given by

d = (0.8 + 0.04 sin ω t) mm

, where

ω = 0.08 rad s^{- 1}

. The distance of the screen from the slits is 1 m and the wavelength of the light used to illuminate the slits is

6000 \overset{˚}{A}

. The interference pattern on the screen changes with time, while the central bright fringe (zeroth fringe) remains fixed at point O.

The

8^{th}

bright fringe above the point O oscillates with time between two extreme positions. The separation between these two extreme positions, in micrometer

(μ m)

, is ____.

Answer: 601.5

Solution

The position of the

n^{th}

bright fringe is

y = n \frac{λ D}{d}

.

For the

8^{th}

fringe:

y = 8 \frac{λ D}{d}

Maximum position (when

d

is minimum):

d_{m i n} = 0.8 - 0.04 = 0.76 mm

y_{m a x} = 8 \times \frac{6000 \times 1 0 ^{- 10} \times 1}{0.76 \times 1 0 ^{- 3}}

Minimum position (when

d

is maximum):

d_{m a x} = 0.8 + 0.04 = 0.84 mm

y_{m i n} = 8 \times \frac{6000 \times 1 0 ^{- 10} \times 1}{0.84 \times 1 0 ^{- 3}}

Separation:

y_{m a x} - y_{m i n} = 8 \times 6000 \times 1 0^{- 10} \times 1 \times [\frac{1}{0.76 \times 1 0 ^{- 3}} - \frac{1}{0.84 \times 1 0 ^{- 3}}]

= 8 \times 6 \times 1 0^{- 4} \times \frac{0.08}{0.76 \times 0.84} = 601.5 μm

JEE Advanced · 2024 · 26 May 2024 · Paper 2 · Numerical answer

Practise more: all JEE Advanced PYQs on Wave Optics →

JEE Advanced · 2024 · 26 May 2024 · Paper 2 · Wave Optics

Solution

More JEE Advanced questions on Wave Optics