Q198 / 200 · MHT CET - Medical 2026 (25 April Shift 2) · Physics
The focal length of the objective of astronomical telescope is $2\ \text{m}$. Under normal adjustment, length of telescope is $2.10\ \text{m}$. The magnifying power of the telescope under normal adjustment is
- Correct answerA$20$
- B$2$
- C$200$
- D$15$
Solution
Given focal length of the objective, $f_o = 2\ \text{m}$.
Length of the telescope under normal adjustment, $L = 2.10\ \text{m}$.
For normal adjustment, the length of the telescope is given by $L = f_o + f_e$.
Substituting the given values:
$2.10 = 2 + f_e$
$f_e = 0.10\ \text{m}$
The magnifying power of the telescope under normal adjustment is given by $M = \dfrac{f_o}{f_e}$.
$M = \dfrac{2}{0.10} = 20$
Answer: $20$
Length of the telescope under normal adjustment, $L = 2.10\ \text{m}$.
For normal adjustment, the length of the telescope is given by $L = f_o + f_e$.
Substituting the given values:
$2.10 = 2 + f_e$
$f_e = 0.10\ \text{m}$
The magnifying power of the telescope under normal adjustment is given by $M = \dfrac{f_o}{f_e}$.
$M = \dfrac{2}{0.10} = 20$
Answer: $20$
Hence, the correct answer is option: 'A'.