JEE MAIN - Physics (2025 - 4th April Evening Shift - No. 11)

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 ) :
0.002
0.0025
0.0005
0.001

Explanation

Determine one main scale division (msd):

$ 300 \, \text{msd} = 15 \, \text{cm} $

Therefore,

$ 1 \, \text{msd} = \frac{15}{300} \, \text{cm} = 0.05 \, \text{cm} $

Determine one vernier scale division (vsd):

$ 25 \, \text{vsd} = 24 \, \text{msd} $

Therefore,

$ 1 \, \text{vsd} = \frac{24}{25} \, \text{msd} $

Calculate the least count (LC):

The least count is given by the difference between one main scale division and one vernier scale division:

$ \text{LC} = 1 \, \text{msd} - 1 \, \text{vsd} $

Substitute the expression for 1 vsd:

$ \text{LC} = 1 \, \text{msd} - \frac{24}{25} \times 1 \, \text{msd} $

This simplifies to:

$ \text{LC} = \frac{1}{25} \times 1 \, \text{msd} $

Calculate the LC in cm:

Substituting the value of 1 msd:

$ \text{LC} = \frac{1}{25} \times 0.05 \, \text{cm} = 0.002 \, \text{cm} $

Hence, the least count of the traveling microscope is 0.002 cm.

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