Optics Question 232
Question: Light of wavelength $ \lambda =5000{AA} $ falls normally on a narrow slit. A screen placed at a distance of 1 m from the slit and perpendicular to the direction of light. The first minima of the diffraction pattern is situated at 5 mm from the centre of central maximum. The width of the slit is
Options:
A) 0.1 mm
B) 1.0 mm
C) 0.5 mm
D) 0.2 mm
Show Answer
Answer:
Correct Answer: A
Solution:
Position of nth minima $ x _{n}=\frac{n\lambda D}{d} $
$ \Rightarrow $ $ 5\times {{10}^{-3}}=\frac{1\times 5000\times {{10}^{-10}}\times 1}{d} $
$ \Rightarrow $ $ d={{10}^{-4}}m=0.1mm $ .
BETA
