In a biprism experiment,the 4^{text{th}} dark | vedclass

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(D) Concept: In a biprism experiment,the position of the $n^{	ext{th}}$ dark fringe from the central axis is given by the formula: $y_n = (2n - 1) \f

Vedclass · Accepted Answer

$\frac{d^2}{7 D}$

Vedclass · Answer

(D) Concept: In a biprism experiment,the position of the $n^{	ext{th}}$ dark fringe from the central axis is given by the formula: $y_n = (2n - 1) \frac{\lambda D}{2d}$.Given that the $4^{	ext{th}}$ dark band is formed opposite to one of the slits,the distance of the slit from the central axis is $y = \frac{d}{2}$.For $n = 4$,the position is $y_4 = (2(4) - 1) \frac{\lambda D}{2d} = \frac{7 \lambda D}{2d}$.Equating the two expressions for $y_4$: $\frac{7 \lambda D}{2d} = \frac{d}{2}$.Solving for $\lambda$: $7 \lambda D = d^2 \implies \lambda = \frac{d^2}{7D}$.Thus,option $(D)$ is correct.

In a biprism experiment,the $4^{\text{th}}$ dark band is formed opposite to one of the slits. The wavelength of light used is ($D=$ distance between source and screen,$d=$ distance between the slits).

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