JEE Advanced 2024 — Paper 2 — PHY-IV solution

JEE Advanced2024Paper 2PHY-IV

PARAGRAPH I

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 $6000$ Å. The interference pattern on the screen changes with time, while the central bright fringe (zeroth fringe) remains fixed at point $O$ .

Q1. The $8^{t h}$ 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 ______.

Your answer

Solution

For the $n^{t h}$ bright fringe, $y_{n} = \frac{nλ D}{d}$ .

With $n = 8$ , $λ = 6000 \times 1 0^{- 10}$ m, $D = 1$ m, and $d = (0.8 + 0.04 sin ω t) \times 1 0^{- 3}$ m:

$y_{8} (t) = \frac{8 \cdot 6 \times 1 0 ^{- 7}}{( 0.8 + 0.04 sin ω t ) \times 1 0 ^{- 3}} .$

$d$ oscillates between $d_{m i n} = 0.76$ mm and $d_{m a x} = 0.84$ mm. The corresponding extremes of $y_{8}$ are

$y_{m a x} = \frac{8 \cdot 6 \times 1 0 ^{- 7}}{0.76 \times 1 0 ^{- 3}}, y_{m i n} = \frac{8 \cdot 6 \times 1 0 ^{- 7}}{0.84 \times 1 0 ^{- 3}} .$

The separation is

$Δ y = 48 \times 1 0^{- 4} (\frac{1}{0.76} - \frac{1}{0.84}) \times 1 0^{- 3} m .$

$\frac{1}{0.76} - \frac{1}{0.84} = \frac{0.84 - 0.76}{0.76 \cdot 0.84} = \frac{0.08}{0.6384} \approx 0.1253.$

$Δ y \approx 48 \times 1 0^{- 4} \cdot 0.1253 \times 1 0^{- 3} \approx 6.015 \times 1 0^{- 4} m = 601.50 μ m .$

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