MHT CET · Physics · Ray Optics

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

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

Verified Solution

Answer & Solution

Correct Answer

(B) 18 cm, 2 cm

Step-by-step Solution

Detailed explanation

For final image at infinity, magnifying power of a telescope is given by

m = \frac{f_{o}}{f e} = 9

where, m = magnification,

{a n d f}_{o}

= focal length of eyepiece

\Rightarrow f_{o} = 9 f_{e}

…. (i)
Also, distance between objective and eyepiece

= f_{o} + f_{e} = 20

(given)

\Rightarrow 9 f_{e} + f_{e} = 20 \Rightarrow f_{e} = 2 c m

f_{o} = 9 f_{e} = 18 c m

Apply the relevant identity from the chapter and substitute the values established in the previous step, keeping the original constraint in view.

Use the simplified expression to isolate the unknown, then plug the result back into the question setup to confirm the working is internally consistent.

Cross-check against a boundary or limiting case. Both the dimensional balance and the qualitative behaviour at the extreme should agree with the candidate answer.

Combine the steps above to identify the correct option. The remaining choices each fail at least one of the consistency, dimensional, or boundary checks.

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