An astronomical telescope has a magnifying po | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) The magnifying power $m$ of an astronomical telescope in normal adjustment is given by the formula:$m = \frac{f_o}{f_e}$where $f_o$ is the focal l

Vedclass · Accepted Answer

$200$

Vedclass · Answer

(B) The magnifying power $m$ of an astronomical telescope in normal adjustment is given by the formula:$m = \frac{f_o}{f_e}$where $f_o$ is the focal length of the objective lens and $f_e$ is the focal length of the eyepiece.Given:Magnifying power $m = 10$Focal length of eyepiece $f_e = 20 \, cm$Substituting the values into the formula:$10 = \frac{f_o}{20 \, cm}$$f_o = 10 	imes 20 \, cm$$f_o = 200 \, cm$Therefore,the focal length of the objective lens is $200 \, cm$.

An astronomical telescope has a magnifying power $10$. The focal length of the eyepiece is $20 \, cm$. The focal length of the objective is.....$cm$.

Similar Questions

In a reflecting telescope,a secondary mirror is used to:

$A$ telescope has an objective lens of focal length $144\, cm$ and an eyepiece of focal length $6\, cm$. Find the magnifying power of the telescope for normal adjustment.

What is a telescope?

The magnifying power of a telescope is high if its objective and eyepiece have respectively

$A$ reflecting telescope utilizes

Vedclass Test Series

Exam Paper Generator

Online Exam Module