Physics - Wave Optics Question with Solution | TestHub

For a single slit diffraction experiment, the half angular width of the principal maximum for a wavelength $\lambda$ is given by $\theta =\frac{\lambda }{a}$, where $a$ is the slit width.

For a wavelength $p\lambda$, the half angular width of the principal maximum is given as $q\theta$.

Therefore, $q\theta =\frac{p\lambda }{a}=p\left(\frac{\lambda }{a}\right)=p\theta$, which implies $q=p$.

The angular width of the first secondary maximum is the angular separation between the first minimum and the second minimum.

For wavelength $\lambda$, the angular width of the first secondary maximum is $\frac{2\lambda }{a}-\frac{\lambda }{a}=\frac{\lambda }{a}=\theta$.

The half angular width of the first secondary maximum for wavelength $\lambda$ is $\frac{\theta }{2}$.

Similarly, for wavelength $p\lambda$, the half angular width of the first secondary maximum is $\frac{p\lambda }{2a}=\frac{p\theta }{2}=\frac{q\theta }{2}$.

The ratio of the half angular widths of the first secondary maxima in the first case to the second case is:

$\text{Ratio}=\frac{\frac{\theta }{2}}{\frac{q\theta }{2}}=\frac{1}{q}$

Thus, the ratio is $1:q$.

Answer: $1:q$