Optics Question 1087
Question: In Young’s double slit experiment when wavelength used is 6000 Å and the screen is 40 cm from the slits, the fringes are 0.012 cm wide. What is the distance between the slits
[MP PMT 1995; Pb PET 2002]
Options:
A) 0.024 cm
B) 2.4 cm
C) 0.24 cm
D) 0.2 cm
Show Answer
Answer:
Correct Answer: D
Solution:
$ \beta =\frac{\lambda D}{d} $
$ \Rightarrow d=\frac{\lambda D}{\beta } $
$ =\frac{6000\times {{10}^{-10}}\times (40\times {{10}^{-2}})}{0.012\times {{10}^{-2}}}=0.2cm. $
BETA
