Target Exam

CUET

Subject

Physics

Chapter

Ray Optics

Question:

The focal length of objective of an astronomical telescope is 1 m. If the magnifying power of telescope is 20, then what is length of telescope for relaxed eye?

Options:

85 cm

95 cm

105 cm

115 cm

Correct Answer:

105 cm

Explanation:

Magnification of astronomical telescope for relaxed eye is m = $\frac{f_0}{f_e}$

$f_0= 100cm$

m = 20 

$\Rightarrow f_e = 5 cm$

Length of the telescope L = $f_0+f_e$= 105cm