Question 31: Wave Optics

Question

A slit of width $10 	imes 10^{-7} \mathrm{~m}$ is illuminated by light of wavelength $500 \mathrm{~nm}$. Angular position of the first minimum is

Accepted Answer

To find the angular position of the first minimum in single-slit diffraction, we use the condition:
$ a \sin 	heta = m \lambda $
where:
$a$ is the slit width,
$\lambda$ is the wavelength of the light, and
$m$ is the order of the minimum (for the first minimum, $m = 1$).
Given:
$a = 10 	imes 10^{-7} 	ext{ m} = 10^{-6} 	ext{ m}$
$\lambda = 500 	ext{ nm} = 500 	imes 10^{-9} 	ext{ m} = 5 	imes 10^{-7} 	ext{ m}$
Follow these steps:
Substitute the values into the formula for the first minimum ($m = 1$):
$    10^{-6} \sin 	heta = 5 	imes 10^{-7}    $
Solve for $\sin 	heta$:
$    \sin 	heta = \frac{5 	imes 10^{-7}}{10^{-6}} = 0.5    $
Find the corresponding angle $	heta$:
$    	heta = \sin^{-1}(0.5) = 30^\circ    $
Thus, the angular position of the first minimum is $30^\circ$. This corresponds to Option B.

Answer

$\frac{1}{2}^0$

Answer

30$^\circ$

Answer

1$^\circ$

Answer

60$^\circ$

Wave Optics

A slit of width 10 \times 10^{-7} \mathrm{~m} is illuminated by light of wavelength 500 \mathrm{~nm}. Angular position of the first minimum is

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