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):

• A. 0.001
• B. 0.002
• C. 0.0005
• D. 0.0025

Hint:

The least count of a vernier scale is the difference between the value of one main scale division and the value of one vernier scale division.

Answer 1

The Correct Option is B

To determine the least count of the travelling microscope, we need to understand the concept of least count and how it applies in this context.

The least count (LC) of an instrument is the smallest measurement that can be taken accurately with the instrument. For a travelling microscope, the least count can be calculated using the formula:

\(LC = \frac{\text{Value of 1 main scale division}}{\text{Number of vernier scale divisions}}\)

Given Data:

• The main scale of the travelling microscope has 300 divisions, which equal 15 cm.
• The vernier scale has 25 divisions, which equal 24 divisions of the main scale.

Calculations:

1. First, calculate the value of 1 main scale division:

\(\text{Value of 1 main scale division} = \frac{15 \text{ cm}}{300} = 0.05 \text{ cm}\)

1. Now, calculate the least count using the formula:

The total length covered by 25 divisions on the vernier scale is equivalent to 24 divisions on the main scale. Therefore, one division on the vernier scale is:

\(1 \text{ vernier division} = \frac{24 \text{ main scale divisions}}{25}\\)

1. Substitute into the least count formula:

\(LC = \frac{0.05 \text{ cm}}{25} = 0.002 \text{ cm}\)

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

Conclusion:

The correct option is 0.002.

Answer 2

To determine the least count of the travelling microscope, we need to understand the relationship between the main scale and the vernier scale.

1. First, we find the length of one division on the main scale. The problem states that 300 divisions on the main scale are equal to 15 cm. Therefore, one division on the main scale is:

\[ \text{One division on main scale} = \frac{15 \, \text{cm}}{300} = 0.05 \, \text{cm} \]

1. The vernier scale has 25 divisions that equal 24 main scale divisions. This means the total length of 25 vernier scale divisions is:

\[ 24 \times 0.05 \, \text{cm} = 1.2 \, \text{cm} \]

1. Thus, the length of one division on the vernier scale is:

\[ \text{One division on vernier scale} = \frac{1.2 \, \text{cm}}{25} = 0.048 \, \text{cm} \]

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

\[ \text{LC} = \text{Main scale division} - \text{Vernier scale division} = 0.05 \, \text{cm} - 0.048 \, \text{cm} = 0.002 \, \text{cm} \]

Therefore, the least count of the travelling microscope is 0.002 cm, making the correct option:

0.002

This approach allows us to calculate the precision of measurements that can be made using the travelling microscope.