In a single slit diffraction experiment, for a wavelength of light ' $\lambda$ ', half-angular width of the…

In a single slit diffraction experiment, for a wavelength of light ' $\lambda$ ', half-angular width of the principle maxima is ' $\theta$ '. Also for wavelength of light $\mathrm{p} \lambda$, the half angular width of the principle maxima is $q \theta$. The ratio of the halfangular widths of the first secondary maxima in the first case to second case will be
  1. $\mathrm{p}: 1$
  2. $\mathrm{q}: 1$
  3. $\mathrm{p}: \mathrm{q}$
  4. $\mathrm{q}: \mathrm{p}$

Solution

Let \(d\) and \(d\) ' be the width of the slits in the two cases. \(\begin{aligned} & \therefore & \theta & =\frac{\lambda}{d} \text { and } q \theta=\frac{p \lambda}{d^{\prime}} \\ & \therefore & \frac{d}{d^{\prime}} & =\frac{q}{p} \end{aligned}\)

Asked in: MHT CET 2024 (16 May Shift 2)

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