### JEE Main On 08 April 2018 Question 20

##### Question: The angular width of the central maximum in a single slit diffraction pattern is $ \text{60 }{}^\circ $ . The width of the slit is $ 1,\mu 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]

#### Options:

A) $ 75\mu m $

B) $ 100\mu m $

C) $ 25\mu m $

D) $ 50\mu m $

## Show Answer

#### Answer:

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 $