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Question

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 correct answer is
0.002

Main Scale Division Calculation

The main scale has 300 divisions covering a total length of 15 cm.

The value of one main scale division (MSD) is calculated as:

\(\text{MSD} = \frac{\text{Total Length}}{\text{Number of Divisions}} = \frac{15 \text{ cm}}{300}\)

\(\text{MSD} = 0.05 \text{ cm}\)

Vernier Scale Least Count Calculation

The vernier scale has 25 divisions that are equal to 24 main scale divisions.

This means:

\(25 \text{ VSD} = 24 \text{ MSD}\)

\(1 \text{ VSD} = \frac{24}{25} \text{ MSD}\)

\(1 \text{ VSD} = \frac{24}{25} \times 0.05 \text{ cm} = 0.96 \times 0.05 \text{ cm} = 0.048 \text{ cm}\)

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

\(\text{LC} = \text{MSD} - \text{VSD}\)

\(\text{LC} = 0.05 \text{ cm} - 0.048 \text{ cm}\)

\(\text{LC} = 0.002 \text{ cm}\)

Alternative Least Count Calculation

Alternatively, the least count can be calculated directly using the formula:

\(\text{LC} = \frac{\text{Value of smallest main scale division}}{\text{Number of divisions on the vernier scale}}\)

\(\text{LC} = \frac{0.05 \text{ cm}}{25}\)

\(\text{LC} = 0.002 \text{ cm}\)

Therefore, the least count of the travelling microscope is 0.002 cm.

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Important Questions from Practical Physics

  1. In a metre bridge experiment, the positions of the cell, E, and galvanometer, G, are interchanged. We shall observe in the galvanometer :

  2. Savitha, a XI standard student, while conducting an experiment to determine the effective length of a simple pendulum L, notes down the data of time taken to complete 30 oscillations as $60 \text{ s}$ and hence calculates the length of the simple pendulum as :
    (Take $\pi^2 = 9.8$, and $g = 9.8 \text{ m/s}^2$)
  3. In a vernier callipers, 20 VSD coincide with 16 MSD (each division of length $1 \text{ mm}$). The least count of the vernier callipers is :
  4. One main scale division of a Vernier calliper is equal to $1\text{ mm}$ and the number of divisions on the Vernier scale is $10$. When both the jaws touch each other, the Vernier scale shifts to the left of zero of the main scale in such a way that $4^{\text{th}}$ Vernier division coincides with a division of the main scale. If this Vernier calliper measures the length of a wire to be $1\text{ cm}$, the actual length of the wire is :
  5. Which of the following measurements require "index correction" ?
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