In Young's double-slit experiment, the distance between slits and...

Given, $D = 2m, d = 0. 25mm = 2. 5 \times {10}^{- 4}m$

Velocity of screen, $\dfrac{dD}{dt} = 5ms^{- 1}$

Speed of first maxima $= \dfrac{d\beta}{dt} = ?$

In Young's double-slit experiment, fringe width $\beta = \dfrac{D\lambda}{d}$

On differentiating with respect to $t$, we get

$\dfrac{d\beta}{dt} = \dfrac{λ}{d} \cdot \dfrac{dD}{dt}$

$= \dfrac{800 \times {10}^{- 9}}{2. 5 \times {10}^{- 4}} \times 5$

$= 16 \times {10}^{- 3}ms^{- 1} = 16mms^{- 1}$