In a Young's experiment, the width of the fringes obtained with light of wavelength 6000A is 2.0 mm. What will be the fringe width, if the entire apparatus is immersed in a liquid of refractive index 1.33?

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Solution

Given,

Width of fringes, ω = 2 \times 10^{- 3} m Wavelength of light used, λ = 6000 Å = 6 \times 10^{- 7} m

The formula for fringe width is,

ω = \frac{Dλ}{d}

\Rightarrow

\frac{D}{d} = \frac{ω}{λ} = \frac{2 \times 10^{- 3}}{6 \times 10^{- 7}} = \frac{1}{3} \times 10^{4}

When the apparatus is immersed in liquid

Wavelength,

λ' = \frac{λ}{μ} = \frac{6 \times 10^{- 7}}{1.33} m

Fringe width,

ω' = \frac{D}{d} λ'

\Rightarrow

ω' = \frac{1}{3} \times 10^{4} \times \frac{6 \times 10^{- 7}}{1.33} m

= 1.5 \times 10^{- 3} m = 1.5 mm .

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