Wave Optics

JEE Main level, numerical answer, medium

32 of 191

Question 32 of 191

In a double slit experiment, when one of the slits is covered by a transparent mica sheet of refractive index 1.561.56, the central fringe shifts to the position of 7th7^{th} bright fringe, obtained with both slits uncovered. If the light source wavelength is 450450 nm, the thickness of mica sheet is α×10−9\alpha \times 10^{-9} m. The value of α\alpha is ______.

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Answer: 5625

The shift in the central fringe due to the introduction of a transparent sheet of thickness tt and refractive index μ\mu is given by Δy=Dd(μ−1)t\Delta y = \dfrac{D}{d}(\mu - 1)t.

The position of the nthn^{th} bright fringe is yn=nDλdy_n = n \dfrac{D\lambda}{d}.

Equating the two, we get:

Dd(μ−1)t=nDλd\dfrac{D}{d}(\mu - 1)t = n \dfrac{D\lambda}{d}

(μ−1)t=nλ(\mu - 1)t = n\lambda

Substituting the given values:

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

0.56t=3150×10−90.56 t = 3150 \times 10^{-9}

t=31500.56×10−9t = \dfrac{3150}{0.56} \times 10^{-9}

t=5625×10−9t = 5625 \times 10^{-9} m

Comparing with t=α×10−9t = \alpha \times 10^{-9} m, we get α=5625\alpha = 5625.

Answer: 56255625

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