Question

A plane wave of wavelength $6250$ $\stackrel{\circ }{A}$ is incident normally on a slit of width $2\times {10}^{-2}cm$. The width of the principle maximum of diffraction pattern on a screen at a distance of $50cm$ will be:

• A. $312\times {10}^{-3}cm$
• B. $312.5\times {10}^{-4}cm$
• C. $312cm$
• D. $312.5\times {10}^{-5}cm$

Answer 1

Answer: (A)

$312\times {10}^{-3}cm$

For single slit diffraction,

$asin\theta =\lambda$ for first minima

$⟹a\frac{y}{D}=\lambda$
$⟹y=\frac{D\lambda }{a}$

$=\frac{0.5\times 6250\times {10}^{-10}}{2\times {10}^{-4}}$

$=1.56\times {10}^{-3}m$

Thus, the width of the principle maxima=$2y=3.12\times {10}^{-3}m$

$=312\times {10}^{-3}cm$