Question

Diffraction pattern of single slit of width $0.5cm$ is formed by a lens of local length $40cm$. Calculate the distance between the first dark and the next bright fringe from the axis wavelength of light is $4890\mathring{A}$.

Answer 1

For a minima,

$a\sin \theta =n\lambda$

$\sin \theta =\frac{x_1}{f}$

$x_1=\frac{f\lambda }{a}$

$=\frac{0.4\times 4890\times {10}^{-10}}{5\times {10}^{-3}}$

$=3.912\times {10}^{-5}m$

For a maxima,

$a\sin \theta =(2n+1)\frac{\lambda }{2}$

$\sin \theta =\frac{x_2}{f}$

$n=1\to x_2=\frac{3f\lambda }{2a}$

$=\frac{3\times 0.4\times 4890\times {10}^{-10}}{2\times 5\times {10}^{-3}}$

$=5.868\times {10}^{-5}m$

thus the distance between the first dark and next bright fringe,

$x=x_2-x_1$

$=5.868\times {10}^{-5}-3.912\times {10}^{-5}$

$\therefore x=1.956\times {10}^{-5}m$