Write a note on Cassegrain telescope.
(N/A) Cassegrain telescope uses a concave primary mirror and a convex secondary mirror to focus incident light. The light reflects off the primary mirror,hits the convex secondary mirror,and is then directed through a central hole in the primary mirror to the eyepiece.
This design is named after its inventor,Laurent Cassegrain.
The primary advantage of this configuration is that it provides a large focal length within a compact,short telescope tube.
The largest telescope in India is located in Kavalur,Tamil Nadu. It is a $2.34 \ m$ diameter reflecting telescope.
It was ground,polished,set up,and is being used by the Indian Institute of Astrophysics,Bengaluru.
The largest reflecting telescopes in the world are the pair of Keck telescopes in Hawaii,$USA$,with reflectors of $10 \ m$ in diameter.
This design is named after its inventor,Laurent Cassegrain.
The primary advantage of this configuration is that it provides a large focal length within a compact,short telescope tube.
The largest telescope in India is located in Kavalur,Tamil Nadu. It is a $2.34 \ m$ diameter reflecting telescope.
It was ground,polished,set up,and is being used by the Indian Institute of Astrophysics,Bengaluru.
The largest reflecting telescopes in the world are the pair of Keck telescopes in Hawaii,$USA$,with reflectors of $10 \ m$ in diameter.
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Similar Questions
$(a)$ $A$ giant refracting telescope at an observatory has an objective lens of focal length $15 \; m$. If an eyepiece of focal length $1.0 \; cm$ is used,what is the angular magnification of the telescope?
$(b)$ If this telescope is used to view the moon,what is the diameter of the image of the moon formed by the objective lens? The diameter of the moon is $3.48 \times 10^{6} \; m$,and the radius of the lunar orbit is $3.8 \times 10^{8} \; m$.
$(b)$ If this telescope is used to view the moon,what is the diameter of the image of the moon formed by the objective lens? The diameter of the moon is $3.48 \times 10^{6} \; m$,and the radius of the lunar orbit is $3.8 \times 10^{8} \; m$.
Medium
View SolutionThe magnifying power of an astronomical telescope in the normal adjustment position is $100$. The distance between the objective and the eye piece is $101 \,cm$. Find the focal length of the objective lens in $cm$.
Medium
View SolutionIf the tube length of an astronomical telescope is $96 \ cm$ and the magnifying power is $15$ for normal setting,then the focal length of the objective is . . . . . . $cm$.
In normal adjustment,for a refracting telescope,the distance between the objective and the eyepiece is $30\,cm$. The focal length of the objective,when the angular magnification of the telescope is $2$,will be $.....\,cm$.
To increase the magnifying power of a telescope ($f_o$ = focal length of the objective and $f_e$ = focal length of the eye lens):
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