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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