Explanation:
Given, wavelength of light,
\(\lambda = 500\;nm = 5 \times {10^{ - 7}}\;m\)
Distance between slit and screen,
\(D = 1.8{\rm{ }}m\)
Distance between two slits,
\(d = 0.4\;mm = 4 \times {10^{ - 4}}\;m\)
velocity of screen \( = \frac{{dD}}{{dt}} = 4\;m{s^{ - 1}}\)
Speed of first maxima \(=\dfrac{d \beta}{d t}=\) ?
In Young's double slit experiment, fringe width,
\(\beta=\dfrac{D \lambda}{d}\)
Differentiating with respect to \(t\), we get
\(\dfrac{d \beta}{d t}=\dfrac{\lambda}{d} \cdot \dfrac{d D}{d t}=\dfrac{5 \times 10^{-7}}{4 \times 10^{-4}} \times 4\)
\( = 5 \times {10^{ - 3}}\;m{s^{ - 1}} = 5\,mm{s^{ - 1}}\).