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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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 $(\frac{3}{2}) x$ . The refractive index of second plate is

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Light of wavelength 500 nm is used to form interference pattern in Young’s double slit experiment. A uniform glass plate of refractive index 1.5 and thickness 0.1 mm is introduced in the path of one of the interfering beams. The number of fringes which will shift the cross wire due to this is

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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. $(2)$
$\frac{2}{3} = \frac{0.5}{μ - 1}$
$2 μ - 2 = 1.5$
$2 μ = 3.5$
$∴ μ = 1.75$