In a double slit experiment, when one of the slits is covered by a transparent mica sheet of refractive index 1.56 , the central fringe shifts to the position of 7^ th bright fringe, obtained with both slits uncovered.

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Correct answer: 0

The shift in the central fringe due to the introduction of a transparent sheet of thickness

t

and refractive index

\mu

is given by

\Delta y = \dfrac{D}{d}(\mu - 1)t

.

The position of the

n^{th}

bright fringe is

y_n = n \dfrac{D\lambda}{d}

.

Equating the two, we get:

\dfrac{D}{d}(\mu - 1)t = n \dfrac{D\lambda}{d}

(\mu - 1)t = n\lambda

Substituting the given values:

(1.56 - 1)t = 7 \times 450 \times 10^{-9}

0.56 t = 3150 \times 10^{-9}

t = \dfrac{3150}{0.56} \times 10^{-9}

t = 5625 \times 10^{-9}

m

Comparing with

t = \alpha \times 10^{-9}

m, we get

\alpha = 5625

.

Answer:

5625

In a double slit experiment, when one of the slits is covered by a...

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