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JEE Main 20254 AprilEvening Shift
Exam
JEE Main
Year
2025
Date
4 April
Shift
Evening Shift
Subject
Physics
Chapter
Experimental Physics

JEE Main 2025 · 4 April · Evening Shift · Physics · Experimental Physics

For the determination of refractive index of glass slab, a...

For the determination of refractive index of glass slab, a travelling microscope is used whose main scale contains 300 equal divisions equals to 15 cm. The vernier scale attached to the microscope has 25 divisions equals to 24 divisions of main scale. The least count (LC) of the travelling microscope is (in cm) :

Options

  1. A0.0010.001
  2. B0.0020.002
  3. C0.00050.0005
  4. D0.00250.0025

Correct answer

B. 0.0020.002

Step-by-step solution

300msd=15 cm1msd=15300 cm=0.05 cm25vsd=24msd1vsd=2425msdLC=1msd−1vsd\begin{aligned} & 300 \mathrm{msd}=15 \mathrm{~cm} \\ & 1 \mathrm{msd}=\frac{15}{300} \mathrm{~cm}=0.05 \mathrm{~cm} \\ & 25 \mathrm{vsd}=24 \mathrm{msd} \\ & 1 \mathrm{vsd}=\frac{24}{25} \mathrm{msd} \\ & \mathrm{LC}=1 \mathrm{msd}-1 \mathrm{vsd}\end{aligned}
<br/>LC=1msd−2425msd=125msdLC=125×0.05=0.002 cm<br/>\begin{aligned}<br/>& \mathrm{LC}=1 \mathrm{msd}-\frac{24}{25} \mathrm{msd}=\frac{1}{25} \mathrm{msd} \\ & \mathrm{LC}=\frac{1}{25} \times 0.05=0.002 \mathrm{~cm}<br/>\end{aligned}
correct option (2)
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