Optics Question 1078
Question: In the Young’s double slit experiment, the spacing between two slits is 0.1 mm. If the screen is kept at a distance of 1.0 m from the slits and the wavelength of light is 5000 Å, then the fringe width is
[MP PMT 1993; RPET 1996]
Options:
A) 1.0 cm
B) 1.5 cm
C) 0.5 cm
D) 2.0 cm
Show Answer
Answer:
Correct Answer: C
Solution:
$ \beta =\frac{\lambda D}{d}=\frac{5000\times {{10}^{-10}}\times 1}{0.1\times {{10}^{-3}}}m=5\times {{10}^{-3}}m=0.5\ cm $ .
BETA
