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. Youngs 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 :

\[a\,\,\sin \theta =\lambda \] \[\sin 30{}^\circ =\frac{\lambda }{a}\]                    \[(a=1\mu m)\] \[\Rightarrow \lambda =\frac{a}{2}=\frac{1}{2}\mu m\] \[\beta =\frac{\lambda D}{d}=1\times {{10}^{-2}}\] \[=\frac{\frac{1}{2}\times {{10}^{-6}}\times 50\times {{10}^{-2}}}{d}=1\times {{10}^{-2}}\] \[\Rightarrow d=25\mu m\]