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Physics Question 26 – JEE-MAIN 2025

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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) :

The least count of a vernier scale is the smallest measurement that can be made accurately with the instrument.

Step 1: Calculate the value of one Main Scale Division (MSD).✦ Active

The main scale has 300 divisions equal to 15 cm. Therefore, the value of one main scale division is:

1 MSD=15 cm300=0.05 cm
Step 2: Calculate the value of one Vernier Scale Division (VSD).○ Expand

The vernier scale has 25 divisions equal to 24 divisions of the main scale. So, one vernier scale division is:

1 VSD=2425 MSD=2425×0.05 cm=0.048 cm
Step 3: Calculate the Least Count (LC).○ Expand

The least count of the travelling microscope is the difference between one main scale division and one vernier scale division:

LC=1 MSD−1 VSD=0.05 cm−0.048 cm=0.002 cm
💡 Teacher's Secret Hint

Alternatively, for a vernier scale where N VSD = (N−1) MSD, the least count can be calculated as LC=1 MSDN. In this case, N=25, so LC=0.05 cm25=0.002 cm.

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