Optics Question 212
Question: Light of wavelength $ 589.3nm $ is incident normally on the slit of width $ 0.1mm. $ What will be the angular width of the central diffraction maximum at a distance of $ 1m $ from the slit
[BHU (Med.) 1999]
Options:
A) $ 0.68{}^\circ $
B) $ 1.02{}^\circ $
C) $ 0.34{}^\circ $
D) None of these
Show Answer
Answer:
Correct Answer: A
Solution:
Angular width of central maxima $ =\frac{2\lambda }{d} $
$ =\frac{2\times 589.3\times {{10}^{-9}}}{0.1\times {{10}^{-3}}}rad $
$ =0.0117\times \frac{180}{\pi }=0.68{}^\circ $
BETA
