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. $



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