Optics Question 195
Question: The light of wavelength $ 6328{AA} $ is incident on a slit of width 0.2 mm perpendicularly, the angular width of central maxima will be
[MP PMT 1987; Pb. PMT 2002]
Options:
A) $ {{0.36}^{o}} $
B) $ {{0.18}^{o}} $
C) $ {{0.72}^{o}} $
D) $ {{0.09}^{o}} $
Show Answer
Answer:
Correct Answer: A
Solution:
The angular half width of the central maxima is given by $ \sin \theta =\frac{\lambda }{a} $
$ \Rightarrow \theta =\frac{6328\times {{10}^{-10}}}{0.2\times {{10}^{-3}}} $ rad $ =\frac{6328\times {{10}^{-10}}\times 80}{0.2\times {{10}^{-3}}\times \pi } $ degree = 0.18o
Total width of central maxima $ =2\theta ={{0.36}^{o}} $
