A monochromatic light source S of wavelength | vedclass

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Question

(B) The given arrangement is shown in the figure.Angle $	heta = 0.5 	imes 10^{-3} \,radians$.The distance between the source $S$ and its image $S_1$

Vedclass · Accepted Answer

$2.64$

Vedclass · Answer

(B) The given arrangement is shown in the figure.Angle $	heta = 0.5 	imes 10^{-3} \,radians$.The distance between the source $S$ and its image $S_1$ is given by $d = 2 	imes (SO 	imes \sin 	heta)$. Since $	heta$ is very small, $\sin 	heta \approx 	heta$.$d = 2 	imes 20.0 \,cm 	imes 0.5 	imes 10^{-3} = 20 	imes 10^{-3} \,cm = 2 	imes 10^{-4} \,m$.The distance of the sources $S$ and $S_1$ from the screen is $D = a + b = 20.0 \,cm + 100.0 \,cm = 120.0 \,cm = 1.2 \,m$.The wavelength of the light used is $\lambda = 440 \,nm = 440 	imes 10^{-9} \,m$.Both $S$ and $S_1$ act as coherent sources and produce interference fringes on the screen.The width of a fringe is given by $\beta = \frac{\lambda D}{d}$.$\beta = \frac{440 	imes 10^{-9} \,m 	imes 1.2 \,m}{2 	imes 10^{-4} \,m} = \frac{528 	imes 10^{-9}}{2 	imes 10^{-4}} \,m = 264 	imes 10^{-5} \,m = 2.64 	imes 10^{-3} \,m = 2.64 \,mm$.

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