In normal adjustment,for a refracting telesco | vedclass

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

Question

(A) In normal adjustment,the length of the telescope $L$ is given by $L = f_o + f_e$,where $f_o$ is the focal length of the objective and $f_e$ is the

Vedclass · Accepted Answer

$20$

Vedclass · Answer

(A) In normal adjustment,the length of the telescope $L$ is given by $L = f_o + f_e$,where $f_o$ is the focal length of the objective and $f_e$ is the focal length of the eyepiece.Given $L = 30\,cm$,so $f_o + f_e = 30$ (Equation $1$).The angular magnification $m$ for a telescope in normal adjustment is given by $m = \frac{f_o}{f_e}$.Given $m = 2$,so $\frac{f_o}{f_e} = 2$,which implies $f_o = 2f_e$ (Equation $2$).Substituting Equation $2$ into Equation $1$:$2f_e + f_e = 30$$3f_e = 30$$f_e = 10\,cm$.Now,substituting $f_e$ back into Equation $2$:$f_o = 2 	imes 10 = 20\,cm$.Thus,the focal length of the objective is $20\,cm$.

In normal adjustment,for a refracting telescope,the distance between the objective and the eyepiece is $30\,cm$. The focal length of the objective,when the angular magnification of the telescope is $2$,will be $.....\,cm$.

Similar Questions

An astronomical refracting telescope has an objective of focal length $20 \, m$ and an eyepiece of focal length $2 \, cm$. Then:

The large aperture of an astronomical telescope is used to:

What is the angular magnification of a telescope if the focal lengths of the objective and eye lenses are $10 \ cm$ and $10 \ mm$ respectively,and the tube length is $11 \ cm$?

An astronomical telescope has an objective of focal length $100 \, cm$ and an eyepiece of focal length $5 \, cm$. The final image of a star is seen $25 \, cm$ from the eyepiece. The magnifying power of the telescope is ...........

Vedclass Test Series

Exam Paper Generator

Online Exam Module