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JEE Main 2023 (29 Jan, Shift 2) · Refraction, Glass Slab & Apparent Depth

Refraction, Glass Slab & Apparent Depth question from JEE Main 2023 (29 Jan, Shift 2)

Q32JEE Main 2023 · 29 JAN · SHIFT 2Easy

Refraction, Glass Slab & Apparent Depth

In an experiment of measuring the refractive index of a glass slab using travelling microscope in physics lab, a student measures real thickness of the glass slab as 5.25 mm and apparent thickness of the glass slab as 5.00 mm. Travelling microscope has 20 divisions in one cm on main scale and 50 divisions on vernier scale is equal to 49 divisions on main scale. The estimated uncertainty in the measurement of refractive index of the slab is x10×10−3\frac{x}{10} \times 10^{-3}, where xx is ___________

Numerical answer type. Work it out, then open the answer.

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Answer: 41

  1. 1 MSD =120 cm=0.5 mm= \frac{1}{20}\text{ cm} = 0.5\text{ mm}; LC =1 MSD−1 VSD=0.5(1−4950)=0.01 mm= 1\text{ MSD} - 1\text{ VSD} = 0.5\left(1 - \frac{49}{50}\right) = 0.01\text{ mm}
  2. μ=5.255.00=1.05\mu = \frac{5.25}{5.00} = 1.05
  3. Δμμ=0.015.25+0.015.00=0.001905+0.002=0.003905\frac{\Delta \mu}{\mu} = \frac{0.01}{5.25} + \frac{0.01}{5.00} = 0.001905 + 0.002 = 0.003905
  4. Δμ=1.05×0.003905≈4.1×10−3=4110×10−3\Delta \mu = 1.05 \times 0.003905 \approx 4.1 \times 10^{-3} = \frac{41}{10} \times 10^{-3}
  5. So x=41x = 41.
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