If the tube length of an astronomical telesco | vedclass

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

Question

(A) For an astronomical telescope in normal adjustment,the magnifying power is given by $m = \frac{f_o}{f_e} = 20$,which implies $f_o = 20f_e$.The len

Vedclass · Accepted Answer

$100$

Vedclass · Answer

(A) For an astronomical telescope in normal adjustment,the magnifying power is given by $m = \frac{f_o}{f_e} = 20$,which implies $f_o = 20f_e$.The length of the telescope tube is given by $L = f_o + f_e = 105 \, cm$.Substituting $f_o = 20f_e$ into the length equation: $20f_e + f_e = 105$.$21f_e = 105$,so $f_e = \frac{105}{21} = 5 \, cm$.Now,calculate the focal length of the objective: $f_o = 20 	imes 5 = 100 \, cm$.

If the tube length of an astronomical telescope is $105 \, cm$ and the magnifying power is $20$ for normal setting,calculate the focal length of the objective in $cm$.

Similar Questions

What is the primary purpose of the eyepiece in a telescope?

The angular magnification of an astronomical telescope is $10$. The final image is formed at infinity. The length of the telescope is $44 \, cm$. The focal length of the objective lens is .......... $cm$.

In a Galilean telescope,the final image formed is:

In an astronomical telescope in normal adjustment,a straight black line of length $L$ is drawn on the inside part of the objective lens. The eyepiece forms a real image of this line. The length of this image is $I$. The magnification of the telescope is:

When the tube length of an astronomical telescope is increased,its magnifying power will .......

Vedclass Test Series

Exam Paper Generator

Online Exam Module