Question

The magnifying power of a telescope is 9. When it is adjusted for parallel rays, the distance between the objective and the eye piece is found to be 20 cm. The focal length of lenses are:

• A. \(18\,\text{cm},\,2\,\text{cm}\)
• B. \(11\,\text{cm},\,9\,\text{cm}\)
• C. \(10\,\text{cm},\,10\,\text{cm}\)
• D. \(15\,\text{cm},\,5\,\text{cm}\)

Hint:

For parallel rays, separation of lenses equals sum of focal lengths.

Answer 1

The Correct Option is A

Step 1: For telescope in normal adjustment: \[ M=\frac{f_o}{f_e}=9 \]
Step 2: \[ f_o+f_e=20 \]
Step 3: Solving, \[ f_e=2\,\text{cm},\quad f_o=18\,\text{cm} \]