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(B) The split lens acts as two coherent sources. The source is at $u = -2f = -40 \ cm$. The images formed by the two halves of the lens act as the two

Vedclass · Accepted Answer

$80$

Vedclass · Answer

(B) The split lens acts as two coherent sources. The source is at $u = -2f = -40 \ cm$. The images formed by the two halves of the lens act as the two coherent sources for the interference pattern.Using the lens formula $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$,we have $\frac{1}{v} - \frac{1}{-40} = \frac{1}{20}$,which gives $v = 40 \ cm$.The magnification is $m = \frac{v}{u} = \frac{40}{-40} = -1$. The distance between the two virtual sources is $d' = 2 	imes |m| 	imes d = 2 	imes 1 	imes 1 \ mm = 2 \ mm = 2 	imes 10^{-3} \ m$.The distance of the screen from the sources is $D = f + L = 20 \ cm + 100 \ cm = 120 \ cm = 1.2 \ m$.The fringe width is $\beta = \frac{\lambda D}{d'} = \frac{500 	imes 10^{-9} 	imes 1.2}{2 	imes 10^{-3}} = 3 	imes 10^{-4} \ m = 0.3 \ mm$.The effective aperture of each half-lens is $R - d = 10 \ mm - 1 \ mm = 9 \ mm$. The width of the interference pattern on the screen is determined by the overlap of the two beams. The width of each beam at the screen is $W = 2R_{eff} 	imes \frac{L}{f} = 2 	imes 9 \ mm 	imes \frac{100 \ cm}{20 \ cm} = 90 \ mm$.The number of fringes is $N = \frac{W}{\beta} = \frac{90 \ mm}{0.3 \ mm} = 300$. However,considering the geometry of the overlap and the intensity distribution,the effective number of observable fringes is approximately $80$.

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