In a single slit diffraction experiment, for wavelength '$\lambda$', half angular width of the principal maxima is '$\theta$'. Also for wavelength of light '$p\lambda$', the half angular width of the principal maxima is '$q\theta$'. The ratio of the half angular widths of the first secondary maxima in the first case to second case will be

For a single slit diffraction experiment, the half angular width of the principal maximum for a wavelength $\lambda$ is given by $\theta = \dfrac{\lambda}{a}$, 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, $q\theta = \dfrac{p\lambda}{a} = p\left(\dfrac{\lambda}{a}\right) = p\theta$, 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 $\dfrac{2\lambda}{a} - \dfrac{\lambda}{a} = \dfrac{\lambda}{a} = \theta$.

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

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

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

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

Thus, the ratio is $1 : q$.

Answer: $1 : q$