The magnifying power of a telescope is 9. Whe | vedclass

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

Question

(A) For a telescope adjusted for parallel rays (normal adjustment),the magnifying power is given by $m = \frac{f_o}{f_e} = 9$,which implies $f_o = 9f_

Vedclass · Accepted Answer

$18\, cm, 2\, cm$

Vedclass · Answer

(A) For a telescope adjusted for parallel rays (normal adjustment),the magnifying power is given by $m = \frac{f_o}{f_e} = 9$,which implies $f_o = 9f_e$ .........$(i)$The length of the telescope tube is the sum of the focal lengths of the objective and the eye-piece: $L = f_o + f_e = 20\, cm$ .............$(ii)$Substituting equation $(i)$ into equation $(ii)$:$9f_e + f_e = 20\, cm$$10f_e = 20\, cm$$f_e = 2\, cm$Now,substituting $f_e$ back into equation $(i)$:$f_o = 9 	imes 2\, cm = 18\, cm$Therefore,the focal lengths are $18\, cm$ and $2\, cm$.

The magnifying power of a telescope is $9$. When it is adjusted for parallel rays,the distance between the objective and the eye-piece is found to be $20\, cm$. The focal lengths of the two lenses are

Similar Questions

All wires have the same resistance $r$. The equivalent resistance between $A$ and $B$ is $R$ when the keys $K_1$ and $K_2$ are open. Now,if the keys are closed,what will be the new equivalent resistance?

The least value of the natural number $n$ satisfying $C(n, 5) + C(n, 6) > C(n+1, 5)$ is

$5.6 \ L$ of helium gas at $STP$ is adiabatically compressed to $0.7 \ L$. Taking the initial temperature to be $T_1$,the work done in the process is

For a situation shown in figure, find the refrective index of glass so that it will suffer total internal reflection at the vertical surface.

$A$ concave mirror gives an image three times as large as the object placed at a distance of $20 \ cm$ from it. For the image to be real,the focal length should be ...... $cm$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module