Optics Question 886
Question: The first diffraction minimum due to the single slit diffraction is seen at $ 9=30{}^\circ $ for a light of wavelength 5000 A falling perpendicularly on the slit. The width of the slit is
Options:
A) $ 2.5\times {{10}^{-5}}cm $
B) $ 1.25\times {{10}^{-5}}cm $
C) $ 10\times {{10}^{-5}}cm $
D) $ 5\times {{10}^{-5}}cm $
Show Answer
Answer:
Correct Answer: C
Solution:
[c] For first minimum, $ d\sin \theta =\lambda $
$ \Rightarrow d=\frac{\lambda }{\sin \theta }=\frac{5000\times {{10}^{-8}}cm}{\sin 30^{o}} $
$ \frac{=5000\times {{10}^{-8}}cm}{{\scriptstyle{}^{1}/{} _{2}}}=10\times {{10}^{-5}}cm $
