question_answer

A parallel beam of light\[(\lambda =5000\overset{o}{\mathop{A}}\,)\]is incident at an angle \[\alpha =30{}^\circ \]with the normal to the slit plane in YDSE. Assume that the intensity due to each slit at any point on the screen is \[{{I}_{0}}\]. Point O is equidistant from \[{{S}_{1}}\] and \[{{S}_{2}}\]. The distance between slit is 1 mm, then the intensity at

A) O is \[3\,{{I}_{0}}\]

B) O is zero

C) a point 1 m  below O is \[4\,{{I}_{0}}\]

D) a point on the screen 1 m below O is zero

Correct Answer: C

Solution :

[c] The path difference at O,                \[\Delta x=d\,\sin \,\alpha =d\,\sin \,{{30}^{o}}=\frac{{{10}^{-3}}}{2}m\] Now \[\phi =\frac{2\pi }{\lambda }.\Delta x=\frac{2\pi }{5000\times {{10}^{-10}}}\times \frac{{{10}^{-3}}}{2}=2\pi \times {{10}^{3}}\] So \[I={{I}_{o}}+{{I}_{o}}+2\sqrt{{{I}_{o}}{{I}_{o}}}\cos (2\pi \times {{10}^{3}})=4{{I}_{o}}\] The angular position of P, \[\tan \theta =\frac{1}{\sqrt{3}}\]; or \[\theta ={{30}^{o}}\]. It means path \[dif{{f}^{n}}\] at P is also\[\Delta x\] and hence \[I=4{{I}_{o}}\]