Optics Question 1152

Question: If prism angle $ \alpha =1{}^\circ ,\ \mu =1.54, $ distance between screen and prism $ (b)=0.7m, $ distance between prism and source $ a=0.3m,\ \lambda =180\pi \ nm $ then in Fresnal biprism find the value of $ \beta $ (fringe width)

[RPMT 2002]

Options:

A) $ {{10}^{-4}}m $

B) $ {{10}^{-3}}mm $

C) $ {{10}^{-4}}\times \pi m $

D) $ \pi \times {{10}^{-3}}m $

Show Answer

Answer:

Correct Answer: A

Solution:

By using $ \beta =\frac{(a+b)\lambda }{2a(\mu -1)\alpha }=\frac{(0.3+0.7)\times 180\pi \times {{10}^{-9}}}{2\times 0.3(1.54-1)\times ( 1\times \frac{\pi }{180} )} $ = $ {{10}^{-4}}m $



NCERT Chapter Video Solution

Dual Pane