Question

Light of wavelength 6328 $\dot{A}$ is incident on a slit having a width of 0.2 mm. The width of the central maximum, measured from minimum to minimum of the diffraction pattern on a screen 9.0 metre away will be about:

• A. 0.36 degrees
• B. 0.18 degrees
• C. 0.72 degrees
• D. 0.09 degrees

Answer 1

The correct option is A 0.36 degrees

Angular width of central max = $\theta =\frac{2\lambda }{e}=\frac{2\times 6328\times {10}^{-10}}{0.2\times {10}^{-3}},\theta ={0.36}^{\circ }$