Question

The magnifying power of a telescope is $9.$ When it is adjusted for parallel rays the distance between the objective and eyepiece is $20\, cm.$ The focal length of lenses are

• A. 10 cm, 10 cm
• B. 15 cm, 5 cm
• C. 18 cm, 2 cm
• D. 11 cm, 9 cm

Answer 1

The Correct Option is C

To find the focal lengths of the lenses in the telescope, we need to use the formula for the magnifying power of a telescope adjusted for parallel rays:

M = \frac{f_o}{f_e}

where f_o is the focal length of the objective lens, f_e is the focal length of the eyepiece, and M is the magnifying power.

According to the problem statement:

• The magnifying power M = 9
• The distance between the objective and eyepiece, d = 20 \, \text{cm}

The relationship between the focal lengths and the distance between lenses is given by:

d = f_o + f_e

Plug in the given values to form two equations:

1. M = \frac{f_o}{f_e} = 9
2. f_o + f_e = 20

From the first equation, we can express f_o in terms of f_e:

f_o = 9f_e

Substituting this expression into the second equation:

9f_e + f_e = 20

10f_e = 20

Solve for f_e:

f_e = \frac{20}{10} = 2 \, \text{cm}

Use this value to find f_o:

f_o = 9 \times 2 = 18 \, \text{cm}

Therefore, the focal lengths of the lenses are:

• Objective lens, f_o = 18 \, \text{cm}
• Eyepiece lens, f_e = 2 \, \text{cm}

The correct answer is 18 cm, 2 cm, which matches option 3 in the list of options provided.