Question

A Cassegrain telescope uses two mirrors as shown in figure. Such a telescope is built with the mirrors $20\text{mm}$ apart. If the radius of curvature of the large mirror is $220\text{mm}$ and the small mirror is $140\text{mm}$, where will the final image of an object at infinity be?

Answer 1

Step 1: Find focal length of objective mirror and secondary mirror.
Formula used: $f=\frac{R}{2}$
Given,
Radius of curvature of objective mirror, $R_1=220\text{mm}$
Radius of curvature of secondary mirror, $R_2=140\text{mm}$
Focal length of the objective mirror,
$f_1=\frac{R_1}{2}$
$f_1=\frac{220}{2}=110\text{mm}$
Focal length of the secondary mirror,
$f_2=\frac{R_2}{2}$
$f_2=\frac{140}{2}=70\text{mm}$

Step 2: Find the final image distance.
Formula used: $\frac{1}{v}+\frac{1}{u}=\frac{1}{f}$
When object is at infinity, parallel rays falling on objective mirror, on reflection, would collect at its focus at $f_1$.
Then, they fall on secondary mirror which is at distance of $20\text{mm}$ from objective mirror.
For secondary mirror,
Object distance, $u=f_1-d=110-20=90\text{mm}$

Now, using formula,
$\frac{1}{v}+\frac{1}{u}=\frac{1}{f_2}$
$\frac{1}{v}=\frac{1}{f_2}-\frac{1}{u}$
$\frac{1}{v}=\frac{1}{70}-\frac{1}{90}$
$\frac{1}{v}=\frac{(9-7)}{630}=\frac{2}{630}$
$v=\frac{630}{2}=315\text{mm}$
Final answer: Final image will be formed at $315\text{mm}$ from the smaller mirror.