The focal lengths of the objective and eye le | 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 for the final image formed at the near point (least distance of distinct vision $D = 25 \, c

Vedclass · Accepted Answer

$-48$

Vedclass · Answer

(B) The magnifying power $m$ of an astronomical telescope for the final image formed at the near point (least distance of distinct vision $D = 25 \, cm$) is given by the formula:$m = - \frac{f_o}{f_e} \left( 1 + \frac{f_e}{D} ight)$Given:Focal length of objective lens $f_o = 200 \, cm$Focal length of eye lens $f_e = 5 \, cm$Least distance of distinct vision $D = 25 \, cm$Substituting the values:$m = - \frac{200}{5} \left( 1 + \frac{5}{25} ight)$$m = - 40 \left( 1 + 0.2 ight)$$m = - 40 	imes 1.2 = - 48$Therefore,the maximum magnifying power is $-48$.

The focal lengths of the objective and eye lenses of a telescope are respectively $200 \, cm$ and $5 \, cm$. The maximum magnifying power of the telescope will be:

Similar Questions

If the focal lengths of the objective and eyepiece of an astronomical telescope are $200 \, cm$ and $4 \, cm$ respectively,then the length of the telescope for normal adjustment is ......... $cm$.

To increase the magnifying power of a telescope ($f_o$ = focal length of the objective and $f_e$ = focal length of the eye lens):

If both the object and the image are at infinite distances from a refracting telescope,its magnifying power will be equal to:

An astronomical telescope has a large aperture to

$A$ reflecting telescope utilizes

Vedclass Test Series

Exam Paper Generator

Online Exam Module