Question

Consider a plane inclined at an angle ${45}^{\circ }$ with the horizontal has two slits ($S_{1}and{S}_2$ separated by a distance d = $\sqrt{2}$ mm The screen is placed at a distance of D = 10 m A parallel monochromatic light beam of wavelength 5000 $\stackrel{\circ }{A}$ is incident on the slits as shown If the fringe width of interference pattern on the screen is $k\times {10}^{-3}$ meter then find the value of k

Answer 1

Path difference

$=S_{1}N-M{S}_2=d\sin (45°+\theta )-\frac{d}{\sqrt{2}}=d[\frac{\cos \theta }{\sqrt{2}}+\frac{\sin \theta }{\sqrt{2}}]-\frac{d}{\sqrt{2}}$, as $\theta$ is small

$\theta =\frac{\sqrt{2n\lambda }}{d}y=\frac{\sqrt{2n\lambda D}}{d}$

Fringe width$=y_{n+1}-y_n=\frac{\sqrt{2D\lambda }}{d}=5\times {10}^{-3}m$
yn+1−yn=2Dλd=5×10−3mHence k=5.