In an astronomical telescope in normal adjust | vedclass

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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

If the tube length of an astronomical telescope is $105 \, cm$ and the magnifying power is $20$ for normal setting,calculate the focal length of the objective in $cm$.

The astronomical telescope consists of an objective lens and an eye-piece. The focal length of the objective is:

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:

If the focal length of the objective lens is increased,then:

The magnifying power of an astronomical telescope for normal adjustment is given by:

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