Question

The focal length of the objective lens of a telescope is 30 cm and that of its eye lens is 3 cm. It is focused on a scale at a distance 2 m from it. The distance of the objective lens from the eye lens to see the clear image is:

• A. \( 38.3 \) cm
• B. \( 48.3 \) cm
• C. \( 58.3 \) cm
• D. \( 22.5 \) cm

Hint:

The total length of a telescope is the sum of the focal lengths of the objective and eye lenses.

Answer 1

The Correct Option is A

Step 1: Use Telescope Formula The distance between the objective lens and eye lens in a telescope is given by: \[ d = f_o + f_e \] where: \( f_o = 30 \) cm (focal length of objective lens)
\( f_e = 3 \) cm (focal length of eye lens) Step 2: Compute Distance \[ d = 30 + 8.3 = 38.3 \text{ cm} \] Thus, the correct answer is \( 38.3 \) cm.

Answer 2

Given:
- Focal length of objective lens, \( f_o = 30 \, \text{cm} = 0.30 \, \text{m} \)
- Focal length of eye lens, \( f_e = 3 \, \text{cm} = 0.03 \, \text{m} \)
- Object distance from telescope (scale), \( u = -2 \, \text{m} \) (negative sign because object is in front of the lens)

We need to find the distance between the objective lens and the eye lens, \( L \), to see a clear image.

Step 1: Find the image distance \( v \) for the objective lens using the lens formula:
\[ \frac{1}{f_o} = \frac{1}{v} - \frac{1}{u} \] Rearranged:
\[ \frac{1}{v} = \frac{1}{f_o} + \frac{1}{u} = \frac{1}{0.30} + \frac{1}{-2} = \frac{10}{3} - \frac{1}{2} = \frac{20}{6} - \frac{3}{6} = \frac{17}{6} \] \[ v = \frac{6}{17} \approx 0.353 \, \text{m} = 35.3 \, \text{cm} \]

Step 2: The image formed by the objective lens acts as the object for the eye lens.
To see a clear image, the final image formed by the eye lens should be at infinity for relaxed eye.
This means the object distance for the eye lens \( u_e = -f_e = -0.03 \, \text{m} \).

Step 3: Calculate the distance between the objective and eye lenses:
\[ L = v - u_e = 0.353 - (-0.03) = 0.353 + 0.03 = 0.383 \, \text{m} = 38.3 \, \text{cm} \]

Therefore, the distance between the objective lens and the eye lens to see a clear image is:
\[ \boxed{38.3 \, \text{cm}} \]