question_answer

The angular width of the central maximum in a single slit diffraction pattern is \[\text{60 }\!\!{}^\circ\!\!\text{ }\]. The width of the slit is \[\text{1}\,\,\,\mu \text{m}\]. The slit is illuminated by monochromatic plane waves. If another slit of same width is made near it. Young's fringes can be observed on a screen placed at a distance 50 cm from the slits. If the observed fringe width is 1 cm, what is slit separation distance? (i.e. distance between the centres of each slit.)               [JEE Main Online 08-04-2018]

A) \[75\mu m\]

B) \[100\mu m\]

C) \[25\mu m\]

D) \[50\mu m\]

Correct Answer: C

Solution :

[c]

\[a\,\,\sin \theta =\lambda \]

\[\sin 30{}^\circ =\frac{\lambda }{a}\]                        \[(a=1\mu m)\]

\[\Rightarrow \lambda =\frac{a}{2}=\frac{1}{2}\mu m\]

\[\beta =\frac{\lambda D}{d}=1\times {{10}^{-2}}\]

\[=\frac{\frac{1}{2}\times {{10}^{-6}}\times 50\times {{10}^{-2}}}{d}=1\times {{10}^{-2}}\]

\[\Rightarrow d=25\mu m\]