In a single slit diffraction experiment, for a wavelength of light ' λ ', half-angular width of the principle maxima is ' θ '. Also for wavelength of light p λ , the half angular w

Question

In a single slit diffraction experiment, for a wavelength of light ' λ ', half-angular width of the principle maxima is ' θ '. Also for wavelength of light p λ , the half angular width of the principle maxima is qθ . The ratio of the halfangular widths of the first secondary maxima in the first case to second case will be

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The half-angular width of the principal maximum in a single slit diffraction is given by $θ = \frac{λ}{a}$ , where $a$ is the slit width.

For the second case with wavelength $p λ$ , the half-angular width is given as $qθ$ .

$qθ = \frac{p λ}{a}$

Substituting $θ = \frac{λ}{a}$ , we get $q = p$ .

The angular width of the first secondary maximum is the angular separation between the first minimum and the second minimum, which is $\frac{2 λ}{a} - \frac{λ}{a} = \frac{λ}{a}$ .

The half-angular width of the first secondary maximum is therefore $\frac{λ}{2 a}$ .

Let $α_{1}$ and $α_{2}$ be the half-angular widths of the first secondary maxima in the first and second cases respectively.

$α_{1} = \frac{λ}{2 a}$

$α_{2} = \frac{p λ}{2 a}$

The required ratio is $\frac{α _{1}}{α _{2}} = \frac{\frac{λ}{2 a}}{\frac{p λ}{2 a}} = \frac{1}{p}$

Since $p = q$ , the ratio is $\frac{1}{q}$ , which can be written as $1 : q$ .

Answer: $1 : q$

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