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 $



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