Question

An astronomical telescope has focal lengths 100 cm and 10 cm of objective and eyepiece lens respectively when final image is formed at least distance of distinct vision, magnification power of telescope will be :

• A. -10
• B. -11
• C. -14
• D. -15

Answer 1

Answer: (C)

-14

Answer 2

The magnification power (M) of an astronomical telescope when the final image is formed at the least distance of distinct vision (D) is given by the formula: $M = - \frac{f_o}{f_e} \left( 1 + \frac{f_e}{D} \right)$

Given: Focal length of objective lens, $f_o = 100$ cm Focal length of eyepiece lens, $f_e = 10$ cm Least distance of distinct vision, $D = 25$ cm

Substitute the given values into the formula: $M = - \frac{100 \text{ cm}}{10 \text{ cm}} \left( 1 + \frac{10 \text{ cm}}{25 \text{ cm}} \right)$ $M = - 10 \left( 1 + \frac{2}{5} \right)$ $M = - 10 \left( \frac{5+2}{5} \right)$ $M = - 10 \left( \frac{7}{5} \right)$ $M = - 2 \times 7$ $M = -14$