A telescope has an objective lens of focal le | vedclass

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

Question

(A) The magnification $m$ produced by the objective lens is given by the ratio of the image height $I$ to the object height $O$,which is also equal to

Vedclass · Accepted Answer

$5$

Vedclass · Answer

(A) The magnification $m$ produced by the objective lens is given by the ratio of the image height $I$ to the object height $O$,which is also equal to the ratio of the focal length $f_0$ to the object distance $u_0$ for distant objects.$m = \frac{I}{O} = \frac{f_0}{u_0}$Given:Object height $O = 50 \ m = 5000 \ cm$Focal length $f_0 = 200 \ cm$Object distance $u_0 = 2 \ km = 2000 \ m = 200000 \ cm$Substituting the values:$\frac{I}{5000} = \frac{200}{200000}$$I = 5000 	imes \frac{200}{200000}$$I = 5000 	imes \frac{1}{1000} = 5 \ cm$Thus,the height of the image formed by the objective lens is $5 \ cm$.

$A$ telescope has an objective lens of focal length $200 \ cm$ and an eyepiece with focal length $2 \ cm$. If this telescope is used to see a $50 \ m$ tall building at a distance of $2 \ km$,what is the height of the image of the building formed by the objective lens in $cm$?

Similar Questions

The equation of the radical axis of the pair of circles $7x^2+7y^2-7x+14y+18=0$ and $4x^2+4y^2-7x+8y+20=0$ is

$\int \frac{dx}{(x+1) \sqrt{4x+3}}$ is equal to

Vitamin $K$ is required for:

The mass defect in a particular nuclear reaction is $0.3 ~g$. The amount of energy liberated in kilowatt hour is: (Velocity of light $= 3 \times 10^8 ~m/s$)

If $3 \sin x + 4 \cos x = 5$,then $6 \tan \frac{x}{2} - 9 \tan^2 \frac{x}{2}$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module