In an astronomical telescope in normal adjust | vedclass

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

Question

(A) In normal adjustment,the distance between the objective lens and the eyepiece is $d = f_0 + f_e$.The object (the line of length $L$) is placed on

Vedclass · Accepted Answer

$L/I$

Vedclass · Answer

(A) In normal adjustment,the distance between the objective lens and the eyepiece is $d = f_0 + f_e$.The object (the line of length $L$) is placed on the objective lens. Its distance from the eyepiece is $u = -(f_0 + f_e)$.The magnification of the eyepiece is given by $m_e = \frac{f_e}{f_e + u}$.Substituting $u = -(f_0 + f_e)$,we get $m_e = \frac{f_e}{f_e - (f_0 + f_e)} = \frac{f_e}{-f_0} = -\frac{f_e}{f_0}$.The magnitude of magnification is $|m_e| = \frac{I}{L} = \frac{f_e}{f_0}$.Since the magnification of the telescope is $M = \frac{f_0}{f_e}$,we have $M = \frac{L}{I}$.

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:

Similar Questions

$A$ small telescope has an objective lens of focal length $144\;cm$ and an eyepiece of focal length $6.0\;cm$. What is the magnifying power of the telescope? What is the separation between the objective and the eyepiece?

$A$ telescope has an objective lens of focal length $150\,cm$ and an eyepiece of focal length $5\,cm$. If a $50\,m$ tall tower at a distance of $1\,km$ is observed through this telescope in normal setting,the angle formed by the image of the tower is $\theta$,then $\theta$ is close to.....$^o$

An astronomical telescope has an eyepiece of focal length $5 \ cm$. The angular magnification in normal adjustment is $10$. When the final image is formed at the least distance of distinct vision $(25 \ cm)$ from the eyepiece,then the angular magnification will be:

The magnifying power of a telescope with tube length $60 \; cm$ is $5$. What is the focal length of its eyepiece in $cm$?

The focal length of the objective and eye lens of an astronomical telescope are $2 \, m$ and $5 \, cm$ respectively. The final image is formed at $(i)$ the least distance of distinct vision and $(ii)$ infinity. The magnifying power in both cases will be:

Vedclass Test Series

Exam Paper Generator

Online Exam Module