BITSAT 2024
- Exam
- BITSAT
- Year
- 2024
- Subject
- Physics
- Chapter
- Ray Optics
BITSAT 2024 · Physics · Ray Optics
The magnifying power of a telescope is 9. When it is adjusted for...
The magnifying power of a telescope is 9. When it is adjusted for parallel rays, the distance between the objective and eyepiece is 20 cm. The ratio of focal length of objective lens to focal length of eyepiece is found to be , then the value of is
Options
- A8
- B7
- C9
- D12
Correct answer
C. 9
Step-by-step solution
Magnifying power of telescope,m = 9
Distance between objective and $\begin{aligned}& \therefore \text{ From Eq. (i), } f_0=9 \times2=18 \mathrm{ cm}\\ & \therefore \quad \frac{f_0}{f_e}=\frac{18}{2}=9\end{aligned}