Question

The Fraunhoffer 'diffraction' pattern of a single slit is formed in the focal plane of a lens of focal length $1m$. The width of slit is $0.3mm$. If third minimum is formed at a distance of $5mm$ from central maximum, then wavelength of light will be

• A. $5000\stackrel{o}{A}$
• B. $2500\stackrel{o}{A}$
• C. $7500\stackrel{o}{A}$
• D. $8500\stackrel{o}{A}$

Answer 1

The correct option is A $5000\stackrel{o}{A}$

a sin $\theta$ = n$\lambda$
$\frac{ax}{f}=3\lambda$
(since $\theta$ is very small so sin $\theta$ $\approx$ tan $\theta$ $\approx$ $\theta$ = x / f )
or $\lambda$ = $\frac{ax}{3f}$ = $\frac{0.3\times {10}^{-3}\times 5\times {10}^{-3}}{3\times 1}$
= $5\times {10}^{-7}$m = 5000 $\stackrel{o}{A}$