Wave Optics

JEE Main level, numerical answer, medium

37 of 191

Question 37 of 191

In a Young's double slit experiment set up, the two slits are kept 0.4 mm apart and screen is placed at 1 m from slits. If a thin transparent sheet of thickness 20μ m20 \mu \mathrm{~m} is introduced in front of one of the slits then center bright fringe shifts by 20 mm on the screen. The refractive index of transparent sheet is given by α10\frac{\alpha}{10}, where α\alpha is ____\_\_\_\_.

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Answer: 14

In Young's Double Slit Experiment, when a thin transparent sheet of thickness tt and refractive index μ\mu is introduced in front of one slit, the central bright fringe shifts by a distance Δy\Delta y.
The formula for the shift is given by Δy=Dd(μ−1)t\Delta y = \frac{D}{d}(\mu - 1)t.
Given values:
Distance between slits d=0.4 mm=0.4×10−3 md = 0.4\text{ mm} = 0.4 \times 10^{-3}\text{ m}
Distance to screen D=1 mD = 1\text{ m}
Thickness of sheet t=20μm=20×10−6 mt = 20\mu\text{m} = 20 \times 10^{-6}\text{ m}
Shift Δy=20 mm=20×10−3 m\Delta y = 20\text{ mm} = 20 \times 10^{-3}\text{ m}
Substituting the values into the formula:
20×10−3=10.4×10−3×(μ−1)×20×10−620 \times 10^{-3} = \frac{1}{0.4 \times 10^{-3}} \times (\mu - 1) \times 20 \times 10^{-6}
20×10−3=20×10−60.4×10−3×(μ−1)20 \times 10^{-3} = \frac{20 \times 10^{-6}}{0.4 \times 10^{-3}} \times (\mu - 1)
20×10−3=50×10−3×(μ−1)20 \times 10^{-3} = 50 \times 10^{-3} \times (\mu - 1)
μ−1=2050=0.4\mu - 1 = \frac{20}{50} = 0.4
μ=1.4\mu = 1.4
The refractive index is given as α10\frac{\alpha}{10}.
So, 1.4=α101.4 = \frac{\alpha}{10}
α=14\alpha = 14

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