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Question

(A) The fringe width $\beta$ in a biprism experiment is given by $\beta = \frac{\lambda D}{d}$, where $\lambda$ is the wavelength, $D$ is the distance

Vedclass · Accepted Answer

towards the source by $37.5 	ext{ cm}$

Vedclass · Answer

(A) The fringe width $\beta$ in a biprism experiment is given by $\beta = \frac{\lambda D}{d}$, where $\lambda$ is the wavelength, $D$ is the distance of the screen from the slits, and $d$ is the distance between the virtual sources.Initial state: $\lambda_1 = 5000 	ext{ Å}$, $D_1 = 100 	ext{ cm}$, $\beta_1 = \frac{\lambda_1 D_1}{d}$.Final state: $\lambda_2 = 6400 	ext{ Å}$, $D_2 = D_1 + x$ (or $D_1 - x$), $\beta_2 = \frac{\lambda_2 D_2}{d}$.We are given that $\beta_2 = \beta_1 - 0.20 \beta_1 = 0.80 \beta_1$.Substituting the expressions: $\frac{\lambda_2 D_2}{d} = 0.80 \frac{\lambda_1 D_1}{d}$.$\lambda_2 D_2 = 0.80 \lambda_1 D_1 \implies 6400 	imes D_2 = 0.80 	imes 5000 	imes 100$.$6400 	imes D_2 = 400000 \implies D_2 = \frac{400000}{6400} = 62.5 	ext{ cm}$.Since the new distance $D_2 = 62.5 	ext{ cm}$ is less than the initial distance $D_1 = 100 	ext{ cm}$, the screen must be moved towards the source by $100 - 62.5 = 37.5 	ext{ cm}$.

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