In Young's double slit experiment, monochromatic light of wavelength $5000\ \text{\AA}$ is used. The slits are $1.0\ \text{mm}$ apart and the screen is placed at $1.0\ \text{m}$ away from the slits. The distance from the centre of the screen where the intensity becomes half of the maximum intensity for the first time is ______ $\times10^{-6}\ \text{m}$.
Numerical value type. Enter your answer.
Answer: 125
$I = I_0\cos^2\dfrac{\phi}{2} = \dfrac{I_0}{2} \Rightarrow \dfrac{\phi}{2} = \dfrac{\pi}{4} \Rightarrow \phi = \dfrac{\pi}{2}$, i.e. path difference $\dfrac{\lambda}{4}$.
$$\frac{yd}{D} = \frac{\lambda}{4} \Rightarrow y = \frac{\lambda D}{4d} = \frac{5\times10^{-7}\times1}{4\times10^{-3}} = 1.25\times10^{-4}\ \text{m} = 125\times10^{-6}\ \text{m}$$
Solution by Sreeraj P, M.Sc Physics