(D) For an astronomical telescope with the final image at infinity,the angular magnification is given by $m = -\frac{f_o}{f_e}$.Given the magnitude of

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(D) For an astronomical telescope with the final image at infinity,the angular magnification is given by $m = -\frac{f_o}{f_e}$.Given the magnitude of magnification is $5$,we have $|m| = \frac{f_o}{f_e} = 5$,which implies $f_o = 5f_e$.The separation between the objective and the eyepiece (tube length) is $L = f_o + f_e = 36 \, cm$.Substituting $f_o = 5f_e$ into the equation: $5f_e + f_e = 36 \, cm$.$6f_e = 36 \, cm \Rightarrow f_e = 6 \, cm$.Now,calculating $f_o$: $f_o = 5 \times 6 \, cm = 30 \, cm$.Thus,the focal lengths are $f_o = 30 \, cm$ and $f_e = 6 \, cm$.

Class 12 Physics Ray Optics and Optical Instruments Telescope

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An astronomical telescope has an angular magnification of magnitude $5$ for distant objects. The separation between the objective and the eyepiece is $36 \, cm$. The final image is formed at infinity. The focal length $f_o$ of the objective and $f_e$ of the eyepiece are

A
$45 \, cm$ and $-9 \, cm$
B
$50 \, cm$ and $10 \, cm$
C
$7.2 \, cm$ and $5 \, cm$
D
$30 \, cm$ and $6 \, cm$

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Class 12

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Ray Optics and Optical Instruments

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Telescope

In a simple telescope,the focal length of the objective lens is $f_o = 60 \ cm$ and the focal length of the eyepiece is $f_e = 5 \ cm$. If the rays coming from the object make an angle of $2^\circ$ at the objective,then the angular magnification (angular size of the image) is ........... $^\circ$.

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$A$ telescope has an objective of focal length $50 \text{ cm}$ and an eyepiece of focal length $5 \text{ cm}$. The least distance of distinct vision is $25 \text{ cm}$. The telescope is focused for distinct vision on a scale $200 \text{ cm}$ away. The separation between the objective and the eyepiece is.......$\text{cm}$.

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The minimum magnifying power of a telescope is $M$. If the focal length of its eye lens is halved, the magnifying power will become:

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$A$ small telescope has an objective lens of focal length $140\, cm$ and an eyepiece of focal length $5.0\, cm$. What is the magnifying power of the telescope for viewing distant objects when the telescope is in normal adjustment?

Medium

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$A$ reflecting telescope has a large mirror as its objective with a radius of curvature equal to $80 \, cm$. What is the magnifying power of this telescope if the eyepiece used has a focal length of $1.6 \, cm$?

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An astronomical telescope has an angular magnification of magnitude $5$ for distant objects. The separation between the objective and the eyepiece is $36 \, cm$. The final image is formed at infinity. The focal length $f_o$ of the objective and $f_e$ of the eyepiece are

Similar Questions

In a simple telescope,the focal length of the objective lens is $f_o = 60 \ cm$ and the focal length of the eyepiece is $f_e = 5 \ cm$. If the rays coming from the object make an angle of $2^\circ$ at the objective,then the angular magnification (angular size of the image) is ........... $^\circ$.

The minimum magnifying power of a telescope is $M$. If the focal length of its eye lens is halved, the magnifying power will become:

$A$ small telescope has an objective lens of focal length $140\, cm$ and an eyepiece of focal length $5.0\, cm$. What is the magnifying power of the telescope for viewing distant objects when the telescope is in normal adjustment?

$A$ reflecting telescope has a large mirror as its objective with a radius of curvature equal to $80 \, cm$. What is the magnifying power of this telescope if the eyepiece used has a focal length of $1.6 \, cm$?

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