Optics Question 467

Question: In Young’s double-slit experiment, the angular width of a fringe formed on a distant screen is $ 1{}^\circ . $ The wavelength of light used is $ 6000\overset{\text{o}}{\mathop{\text{A}}} $ . What is the spacing between the slits?

Options:

A) 344 mm

B) 0.1344 mm

C) 0.0344 mm

D) 0.034 mm

Show Answer

Answer:

Correct Answer: C

Solution:

[c] In Young’s double slit fringe with $ \sin \theta =\theta =y/D $

So, $ \Delta y/D $ and hence angular fringe width $ {\theta _{0}}=\Delta \theta (with\Delta y=\beta ) $ Will be $ {\theta _{0}}=\frac{\beta }{D}=\frac{D\lambda }{d}\times \frac{1}{D}=\frac{\lambda }{d} $

$ \Rightarrow {\theta _{0}}=1^{0}=( \frac{\pi }{180} ) $ rad, and $ \lambda =6\times {{10}^{-7}}m $ Or $ d=\frac{\lambda }{{\theta _{0}}}=\frac{180}{\pi }\times (6\times {{10}^{-7}})=3.44\times {{10}^{-5}}m $ Or $ d=0.0344mm $



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