In the diffraction pattern, the first maximum | vedclass

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(A) The condition for the $n^{th}$ secondary maximum in a single-slit diffraction pattern is given by $a \sin 	heta = (n + \frac{1}{2}) \lambda$, whe

Vedclass · Accepted Answer

$3 \left( \frac{\sqrt{2} - 1}{\sqrt{2}} \right) \lambda$

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(A) The condition for the $n^{th}$ secondary maximum in a single-slit diffraction pattern is given by $a \sin 	heta = (n + \frac{1}{2}) \lambda$, where $n = 1, 2, 3, \dots$For the first maximum, $n = 1$, so $a \sin 	heta = \frac{3}{2} \lambda$.Case $1$: $a_1 \sin 30^{\circ} = \frac{3}{2} \lambda \implies a_1 (\frac{1}{2}) = \frac{3}{2} \lambda \implies a_1 = 3 \lambda$.Case $2$: $a_2 \sin 45^{\circ} = \frac{3}{2} \lambda \implies a_2 (\frac{1}{\sqrt{2}}) = \frac{3}{2} \lambda \implies a_2 = \frac{3\sqrt{2}}{2} \lambda = \frac{3}{\sqrt{2}} \lambda$.The change in slit width is $\Delta a = a_1 - a_2 = 3 \lambda - \frac{3}{\sqrt{2}} \lambda = 3 \lambda (1 - \frac{1}{\sqrt{2}}) = 3 \lambda \left( \frac{\sqrt{2} - 1}{\sqrt{2}} ight)$.

In the diffraction pattern, the first maximum is at $30^{\circ}$, when a monochromatic light of wavelength $\lambda$ is incident on a slit of width $a$. For the same wavelength, if the slit width is changed, so that the first maximum is at $45^{\circ}$, by how much is the slit width changed?

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