Question

Determine the angular separation between central maximum and first order maximum of the diffraction pattern due to a single slit of width 0.25 mm, when light of wavelength $5890 \AA$ is incident on it normally.

Answer 1

Explanation :

It is given that,

Width of slit, $a= 0.25\ mm=0.25\times 10^{-3}\ m$

Wavelength of the light, $\lambda=5890\ A=5890\times 10^{-10}\ m$

The angular separation between central maximum and first order maximum of the diffraction pattern due to a single slit is :

$sin\theta=\dfrac{\lambda}{a}$

$sin\theta=\dfrac{5890\times 10^{-10}\ m}{0.25\times 10^{-3}\ m}$

$sin\theta=23560\times 10^{-7}$

$sin\theta=0.0023$

$\theta=sin^{-1}(0.0023)$

$\theta=0.13^0$

So, the angular separation between central maximum and first order maximum of the diffraction pattern due to a single slit is $0.13^0$.