In Young's double-slit experiment using monochromatic light the fringe pattern shifts by a certain distance on the screen when a mica sheet of refractive index 1.6 and thickness 1.964 microns is introduced in the path of one of the interfering waves. The mica sheet is then removed and the distance between the plane of the slits and the screen is doubled. It is found that the distance between the successive maximum now is the same as the observed fringe shift upon the introduction of mica sheet. Calculate the wavelength of the light.

Solution

Given, a mica sheet is introduced in between the path of the interfering waves.
Refractive index of the mica sheet,

μ

= 1.6
Thickness of the mica sheet, t = 1.964 microns = 1.964

\times 10^{- 6} m

Due to introduction of mica sheet, the shift on the screen is,

Y_{0} = \frac{D}{d} (μ - 1) t

Now, when the distance between the plane of slits and screen is changed from D to 2D, fringe width will become,

β = \frac{2 D}{d} (λ)

According to given problem,

β = \frac{λ (2 D)}{d} = 2 \frac{λ D}{d}

and,

(μ - 1) t . \frac{D}{d} = 2 λ \frac{D}{d}

\Rightarrow

λ = \frac{(μ - 1) t}{2}

= \frac{(1.6 - 1) \times 1.964 \times 10^{- 6}}{2} = 5892 Å

is the required wavelength of the light.

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