(a) With the help of a labelled ray diagram, explain the construction and working of a Cassegrain reflecting telescope.

(b) An amateur astronomer wishes to estimate roughly the size of the Sun using his crude telescope consisting of an objective lens of focal length 200 cm and an eyepiece of focal length 10 cm. By adjusting the distance of the eyepiece from the objective, he obtains an image of the Sun on a screen 40 cm behind the eyepiece. The diameter of the Sun’s image is measured to be 6·0 cm. Estimate the Sun’s size, given that the average Earth-Sun distance is 1.5\times 10^{11}m.

 

 

 
 
 
 
 

Answers (1)
S safeer

(a) Ray diagram;-

When the parallel rays from distance object are reflected by the large primary mirror. Then these rays fall on the secondary mirror which reflects these ray outside the hole, Thus, the final magnified image is formed.

(b) To calculate the sun's size,

We have

V_{e}=40\; cm and f_{e}=10\; cm

 for eyepiece ;

\frac{1}{u_e}=\frac{1}{40}-\frac{1}{10}

{u_e}=-\frac{40}{3}

Now, the magnification produced by eyepiece (m_{e}) is given as :

m_{e}=\frac{\nu _{0}}{\left | u_{e} \right |}=\frac{40}{\frac{40}{3}}=3

\therefore The diameter of the image formed by the objective is 

d=\frac{6}{3}

d=2\; cm

If D be the diameter of the sun then the angle subtended by it on the objective will be \alpha

\therefore \alpha =\frac{D}{1.5\times 10^{11}}rad _____(1)

Since Angle subtended by the image at the objective  = angle subtended by the sun.

Therefore, \alpha =\frac{size \; of\; image}{f_{0}}=\frac{2}{200}=\frac{1}{100}rad

Hence, on comparing with the equation. (1) we get,

D=1.5\times 10^{9}m 

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