Optics Question 1149
Question: In Fresnel’s biprism $ (\mu =1.5) $ experiment the distance between source and biprism is 0.3 m and that between biprism and screen is 0.7m and angle of prism is $ 1{}^\circ $ . The fringe width with light of wavelength $ 6000{AA} $ will be
[RPMT 2002]
Options:
A) 3 cm
B) 0.011 cm
C) 2 cm
D) 4 cm
Show Answer
Answer:
Correct Answer: B
Solution:
$ \beta =\frac{(a+b)\lambda }{2a(\mu -1)\alpha } $
where a = distance between source and biprism = 0.3 m b = distance between biprism and screen = 0.7 m. a = Angle of prism = 1°, m = 1.5, $ \lambda $ = 6000 x 10^-10 m
Hence, $ \beta =\frac{(0.3+0.7)\times 6\times {{10}^{-7}}}{2\times 0.3(1.5-1)\times (1^{o}\times \frac{\pi }{180})} $ = 1.14 x 10^-4 m = 0.0114 cm.
BETA
