The magnifying power of a telescope is
9. When it is adjusted for parallel rays, the distance between the objective and eye-piece is 20 cm. The focal length of objective and eye-piece are:
Show Hint
- 18 cm, 2 cm
- 15 cm, 5 cm
- 10 cm, 10 cm
- 11 cm, 9 cm
The Correct Option is A
Solution and Explanation
Magnifying power $m = f_{o}/f_{e}$ and length $L = f_{o} + f_{e}$ for normal adjustment.
Step 2: Analysis
$f_{o}/f_{e} = 9 \Rightarrow f_{o} = 9f_{e}$. $f_{o} + f_{e} = 20 \Rightarrow 9f_{e} + f_{e} = 20 \Rightarrow 10f_{e} = 20$.
Step 3: Conclusion
$f_{e} = 2$ cm and $f_{o} = 18$ cm.
Final Answer: (A)
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