(D) The shift in the central fringe due to the introduction of a thin sheet is given by the formula: $\Delta x = \frac{(\mu - 1) t D}{d}$.Given that t

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(D) The shift in the central fringe due to the introduction of a thin sheet is given by the formula: $\Delta x = \frac{(\mu - 1) t D}{d}$.Given that the central fringe shifts to the position of the $n^{th}$ bright fringe,the shift is equal to $n \beta$,where $\beta = \frac{\lambda D}{d}$ is the fringe width.Equating the two: $\frac{(\mu - 1) t D}{d} = n \frac{\lambda D}{d}$.Simplifying,we get: $\lambda = \frac{(\mu - 1) t}{n}$.Substituting the given values: $\mu = 1.6$,$t = 7 \times 10^{-6} \ m$,and $n = 7$.$\lambda = \frac{(1.6 - 1) \times 7 \times 10^{-6}}{7} = 0.6 \times 10^{-6} \ m = 6 \times 10^{-7} \ m$.

Class 12 Physics Wave Optics Thin film Interference, fresnel biprism, lloyd's mirror

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When a mica sheet of thickness $7 \ \mu m$ and refractive index $\mu = 1.6$ is placed in the path of one of the interfering beams in a biprism experiment,the central fringe shifts to the position of the $7^{th}$ bright fringe. Find the wavelength of the light used.

A
$6.5 \times 10^{-8} \ m$
B
$2 \times 10^{-5} \ m$
C
$4 \times 10^{-7} \ m$
D
$6 \times 10^{-7} \ m$

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Wave Optics

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Thin film Interference, fresnel biprism, lloyd's mirror

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When a mica sheet of thickness $7 \ \mu m$ and refractive index $\mu = 1.6$ is placed in the path of one of the interfering beams in a biprism experiment,the central fringe shifts to the position of the $7^{th}$ bright fringe. Find the wavelength of the light used.

Similar Questions

In a biprism experiment,if the $5^{\text{th}}$ bright band with wavelength $\lambda_1$ coincides with the $6^{\text{th}}$ dark band with wavelength $\lambda_2$,then the ratio $(\lambda_1 / \lambda_2)$ is:

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