Angular width of the central maximum in the fraunhofer diffraction for $\lambda$ = 6000 $\stackrel{0}{\mathrm{A}}$ is ${\theta }_0$. When
the same slit is illuminated by another monochromatic light, the angular width decreases by
30%. The wavelength of this light is [NEET (Odisha) 2019]

The angular width of central maximum is given by

$2\theta =\frac{2\lambda }{a}$                                 …(i)

where,  $\lambda$ = wavelength of light used

and a = width of the slit.
Angular width,  $2\theta = {\theta }_0$     (given)

So,  ${\theta }_0=\frac{2{\lambda }_1}{a}$                  ... (ii) [from Eq. (i)l

For another light of wavelength ${\lambda }_2$ (says), the angular width decreases by 30%, so

$2\theta =\left(\frac{100-30}{100}\right){\theta }_0$

$2\theta =\frac{70}{100}{\theta }_0=0.7{\theta }_0$

$\therefore$   Angular width   $0.7{\theta }_0=\frac{2{\lambda }_2}{a}$                      …(iii)

As, slit width is constant, so on dividing Eq. (ii) by Eq. (iii), we get

$\frac{{\theta }_0}{0.7{\theta }_0}=\frac{{\lambda }_1}{{\lambda }_2}$

$\Rightarrow {\lambda }_2={\lambda }_1\times 0.7$

= 6000 x 0.7
= 4200 $\stackrel{0}{\mathrm{A}}$