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(B) The angular width of the central maximum in a single slit diffraction pattern is given by $	heta = \frac{2\lambda}{a}$.Given: $\lambda = 628 	ex

Vedclass · Accepted Answer

$36$

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(B) The angular width of the central maximum in a single slit diffraction pattern is given by $	heta = \frac{2\lambda}{a}$.Given: $\lambda = 628 	ext{ nm} = 628 	imes 10^{-9} 	ext{ m}$, $a = 0.2 	ext{ mm} = 0.2 	imes 10^{-3} 	ext{ m}$.Substituting the values: $	heta = \frac{2 	imes 628 	imes 10^{-9}}{0.2 	imes 10^{-3}} = \frac{1256 	imes 10^{-9}}{0.2 	imes 10^{-3}} = 6280 	imes 10^{-6} = 6.28 	imes 10^{-3} 	ext{ radians}$.To convert radians to degrees, we multiply by $\frac{180}{\pi}$:$	heta^{\circ} = 6.28 	imes 10^{-3} 	imes \frac{180}{3.14}$.Since $\frac{6.28}{3.14} = 2$, we get $	heta^{\circ} = 2 	imes 10^{-3} 	imes 180 = 360 	imes 10^{-3} = 0.36^{\circ}$.Expressing this as $\alpha 	imes 10^{-2}$, we have $0.36 = 36 	imes 10^{-2}$.Therefore, the value of $\alpha$ is $36$.

In a single slit diffraction pattern, the wavelength of light used is $628 \text{ nm}$ and slit width is $0.2 \text{ mm}$. The angular width of the central maximum is $\alpha \times 10^{-2} \text{ degrees}$. The value of $\alpha$ is . . . . . . . (Take $\pi = 3.14$)

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