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 $
BETA
