Physics - Wave Optics Question with Solution | TestHub

The angular width of the central maximum in a single slit Fraunhofer diffraction is given by $\theta =\frac{2\lambda }{a}$, where $\lambda$ is the wavelength of light and $a$ is the width of the slit.

From this relation, the angular width is directly proportional to the wavelength, $\theta \propto \lambda$.

Let the initial wavelength be ${\lambda }_1=6480\text{A˚}$ with angular width ${\theta }_1$. When the angular width decreases by $25\%$, the new angular width becomes ${\theta }_2={\theta }_1-0.25{\theta }_1=0.75{\theta }_1$.

Since $\theta \propto \lambda$, the new wavelength $\lambda$ is given by $\lambda =0.75{\lambda }_1$.

Substituting the given value of ${\lambda }_1$:

$\lambda =0.75\times 6480=\frac{3}{4}\times 6480=4860\text{A˚}$