Question

The angular width $\Delta \theta$ of the central maximum in the diffraction pattern of a slit illuminated by light of wavelength $500\text{nm}$ is measured. The angular width of the central maximum is found to increase by $25\text{\%}$, when the same slit is illuminated by another light of wavelength ${\lambda }^{\prime }$. Then wavelength ${\lambda }^{\prime }$ will be

• A. $375\text{nm}$
• B. $450\text{nm}$
• C. $575\text{nm}$
• D. $625\text{nm}$

Answer 1

The correct option is D $625\text{nm}$

The angular width of the central maximum is given by,

$\Delta \theta =\frac{2\lambda }{a}$

For second case,
$\Delta {\theta }^{\prime }=\frac{2{\lambda }^{\prime }}{a}$

It is given that,

$\Delta {\theta }^{\prime }=\Delta \theta +\frac{25}{100}\Delta \theta$

$\Rightarrow \Delta {\theta }^{\prime }=1.25\Delta \theta$

$\Rightarrow \frac{2{\lambda }^{\prime }}{a}=1.25\times \frac{2\lambda }{a}$

$\Rightarrow {\lambda }^{\prime }=1.25\times \lambda =1.25\times 500\text{nm}$

$\Rightarrow {\lambda }^{\prime }=625\text{nm}$

Hence, option $\left(D\right)$ is correct.