In a Young's double slit experiment,15 fringe | vedclass

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(A) The width of the screen portion is constant for both cases.Let the width of the screen portion be $L$.The fringe width $\beta$ is given by $\beta

Vedclass · Accepted Answer

$750$

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(A) The width of the screen portion is constant for both cases.Let the width of the screen portion be $L$.The fringe width $\beta$ is given by $\beta = \frac{\lambda D}{d}$.The number of fringes $n$ observed in a length $L$ is given by $L = n 	imes \beta = n 	imes \frac{\lambda D}{d}$.For the first case: $L = 15 	imes \frac{500 \; nm 	imes D}{d}$.For the second case: $L = 10 	imes \frac{\lambda 	imes D}{d}$.Equating the two expressions for $L$:$15 	imes 500 = 10 	imes \lambda$.$\lambda = \frac{15 	imes 500}{10} = 15 	imes 50 = 750 \; nm$.

In a Young's double slit experiment,$15$ fringes are observed on a small portion of the screen when light of wavelength $500 \; nm$ is used. Ten fringes are observed on the same section of the screen when another light source of wavelength $\lambda$ is used. Then the value of $\lambda$ is (in $nm$):

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