Practicing Success

Target Exam

CUET

Subject

Physics

Chapter

Ray Optics

Question:

Options:

a

b

c

d

Correct Answer:

d

Explanation:

$\text{Power of Objective lens of telescope = 0.5 D}$ 

$\Rightarrow \text{focal length of objective lens of telescope }f_0 = 200 cm$

$\text{Power of Eye lens of telescope = 20 D }$

$\Rightarrow \text{ focal length of eyepiece of telescope } f_e = 5 cm$

$\text{Magnification of the telescope M } = \frac{f_0}{f_e} = 40$