Question

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

Answer 1

We have

$a\sin \theta =n\lambda$

Also, $\frac{ax}{f}=3\lambda$

where $a$= width of the slit

$x$ = distance from the central maxima

$f$ = focal length

(Since$\theta$ is very small $\sin \theta \approx \tan \approx \theta \approx \theta =\frac{x}{f}$)

or $\lambda =\frac{ax}{3f}=\frac{0.3\times {10}^{-3}\times 5\times {10}^{-3}}{3\times 1}$

=$5\times {10}^{-7}=5000\dot{A}$