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 $O$ .

Q2. The maximum speed in $μ$ m/s at which the $8^{t h}$ bright fringe will move is __________.

Your answer

Solution

The position of the $8^{t h}$ bright fringe is $y_{8} = \frac{8 λ D}{d}$ with $d (t) = (0.8 + 0.04 sin ω t) \times 1 0^{- 3}$ m.

Differentiating:

$\frac{d y _{8}}{d t} = - \frac{8 λ D}{d ^{2}} \cdot \frac{dd}{d t} = - \frac{8 λ D}{d ^{2}} \cdot 0.04 ω cos ω t \times 1 0^{- 3} .$

The speed is maximum when $∣ cos ω t ∣ = 1$ and $d$ takes its mean value $0.8 \times 1 0^{- 3}$ m:

$v_{m a x} = \frac{8 λ D}{d ^{2}} \cdot 0.04 ω \times 1 0^{- 3} .$

Substituting $λ = 6 \times 1 0^{- 7}$ m, $D = 1$ m, $d = 8 \times 1 0^{- 4}$ m, $ω = 0.08$ rad/s:

$v_{m a x} = \frac{8 \cdot 6 \times 1 0 ^{- 7}}{64 \times 1 0 ^{- 8}} \cdot 0.04 \cdot 0.08 \times 1 0^{- 3} = 7.5 \cdot 3.2 \times 1 0^{- 6} m/s = 24 \times 1 0^{- 6} m/s .$

$∴ v_{m a x} = 24 μ$ m/s.

← Back to full paper

Practice more PHY-IV

Concept-wise practice with instant solutions on Fundamenthol.

Start practicing →