Question

The magnifying power of a telescope with tube length 60 cm is 5. Then the focal length of its eye piece is:

• A. 20 cm
• B. 40 cm
• C. 30 cm
• D. 10 cm

Hint:

The magnifying power of a telescope is the ratio of the focal length of the objective to the focal length of the eye piece.

Answer 1

The Correct Option is D

The magnifying power \( M \) of a telescope is given by: \[ M = \frac{f}{D} \] where \( f \) is the focal length of the eye piece and \( D \) is the focal length of the objective lens (the tube length of the telescope). From the problem, we are given that: \[ M = 5, \quad D = 60 \, \text{cm} \] Substituting into the equation: \[ 5 = \frac{f}{60} \] Solving for \( f \): \[ f = 5 \times 60 = 300 \, \text{cm} \] Thus, the focal length of the eye piece is \( \boxed{10 \, \text{cm}} \).

Answer 2

Step 1: Understanding the problem
The magnifying power \( M \) of a telescope is given by:
\[ M = \frac{f_o}{f_e} \]
where \( f_o \) is the focal length of the objective lens and \( f_e \) is the focal length of the eye piece.

Step 2: Given data
- Tube length \( L = 60 \, \text{cm} \)
- Magnifying power \( M = 5 \)

Step 3: Relationship between tube length and focal lengths
For a telescope, tube length is approximately:
\[ L = f_o + f_e \]

Step 4: Express \( f_o \) in terms of \( f_e \)
From magnifying power:
\[ M = \frac{f_o}{f_e} \implies f_o = M f_e = 5 f_e \]

Step 5: Substitute into tube length equation
\[ L = f_o + f_e = 5 f_e + f_e = 6 f_e \]
Therefore,
\[ f_e = \frac{L}{6} = \frac{60}{6} = 10 \, \text{cm} \]

Step 6: Conclusion
The focal length of the eye piece is 10 cm.