During a Young's double slit experiment with monochromatic light, introducing a mica sheet of refractive index 1.6 and thickness 1.964 micron ( 1 micron = 10^ -6 m ) into the path of one of the interfering waves causes

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Quantrex Academy · Accepted Answer

Correct answer: B. 589.2 nm

The fringe shift produced by introducing a mica sheet of thickness

t

and refractive index

\mu

in the path of one of the interfering waves is given by:

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

where

D

is the distance between the slits and the screen, and

d

is the distance between the slits.

When the mica sheet is removed and the distance between the slits and the screen is doubled to

2D

, the new fringe width (distance between successive maxima) becomes:

W' = \dfrac{\lambda(2D)}{d}

According to the given condition, the new fringe width equals the initial fringe shift:

W' = \Delta y

\dfrac{\lambda(2D)}{d} = \dfrac{D}{d}(\mu - 1)t

2\lambda = (\mu - 1)t

\lambda = \dfrac{(\mu - 1)t}{2}

Substituting the given values,

\mu = 1.6

and

t = 1.964 \times 10^{-6} \text{ m}

:

\lambda = \dfrac{(1.6 - 1) \times 1.964 \times 10^{-6}}{2}

\lambda = \dfrac{0.6 \times 1.964 \times 10^{-6}}{2}

\lambda = 0.3 \times 1.964 \times 10^{-6} \text{ m}

\lambda = 0.5892 \times 10^{-6} \text{ m} = 589.2 \text{ nm}

During a Young's double slit experiment with monochromatic light,...

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