In a Young's double slit experiment, the fringes are displaced by a distance $x$ when a glass plate of refractive index $1.5$ is introduced in the path of one of the beams. When this plate is replaced by another plate of same thickness, the shift of fringes is $(3 / 2) x$ . The refractive index of second plate is

1.75

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1.40

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1.25

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1.67

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Solution

The correct option is A $1.75$
Shift $= \frac{(μ - 1) t D}{d}$
$x = \frac{(1.5 - 1) t D}{d} . . . (1)$
and
$\frac{3}{2} x = \frac{(μ - 1) t D}{d} . . . (2)$
Dividing Eq. $(1)$ by Eq. $(2)$
$\frac{2}{3} = \frac{0.5}{μ - 1}$
$2 μ - 2 = 1.5$
$2 μ = 3.5$
$∴ μ = 1.75$

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The correct option is A 1.75Shift =(μ−1)tDd x=(1.5−1)tDd ...(1) and 32x=(μ−1)tDd ...(2) Dividing Eq. (1) by Eq. (2) 23=0.5μ−1 2μ−2=1.5 2μ=3.5 ∴ μ=1.75

The correct option is A $1.75$
Shift $= \frac{(μ - 1) t D}{d}$
$x = \frac{(1.5 - 1) t D}{d} . . . (1)$
and
$\frac{3}{2} x = \frac{(μ - 1) t D}{d} . . . (2)$
Dividing Eq. $(1)$ by Eq. $(2)$
$\frac{2}{3} = \frac{0.5}{μ - 1}$
$2 μ - 2 = 1.5$
$2 μ = 3.5$
$∴ μ = 1.75$