In 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 os replaced by another plate of the same thickness, the shift of fringes is $(3 / 2) x$ . The refractive index of the second plate is

1.75

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1.50

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1.25

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1.00

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Solution

The correct option is A $1.75$
case I: $x = \frac{β}{λ} (1.5 - 1) t$
case II: $\frac{3 x}{2} = \frac{β}{λ} (μ - 1) t$
$\frac{(μ - 1)}{0.5} = \frac{3}{2} \Rightarrow μ = \frac{3}{4} + 1 = \frac{7}{4} = 1.75$

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Q. In Young's double slit experiment, the source is red light of wavelength

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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 1.75case I: x=βλ(1.5−1)tcase II: 3x2=βλ(μ−1)t(μ−1)0.5=32⇒μ=34+1=74=1.75

The correct option is A $1.75$
case I: $x = \frac{β}{λ} (1.5 - 1) t$
case II: $\frac{3 x}{2} = \frac{β}{λ} (μ - 1) t$
$\frac{(μ - 1)}{0.5} = \frac{3}{2} \Rightarrow μ = \frac{3}{4} + 1 = \frac{7}{4} = 1.75$