Question

A Cassegrain telescope uses two mirrors of radii of curvature 25 cm and 16 cm separated by a distance of 2.5 cm. The position of the final image of an object at infinity is

• A. 40 cm from convex mirror
• B. 4.44 cm from concave mirror
• C. 4.44 cm from convex mirror
• D. 40 cm from concave mirror

Hint:

For optical systems with multiple components (mirrors or lenses), the image formed by the first component acts as the object for the second component. Be careful with sign conventions and the location of this intermediate object relative to the second component.

Answer 1

The Correct Option is A

In a Cassegrain telescope, the primary mirror is a concave mirror and the secondary mirror is a convex mirror.
Object is at infinity, so the primary (concave) mirror forms an image at its focus.
For the primary mirror: Radius of curvature $R_1 = -25$ cm (concave).
Focal length $f_1 = R_1/2 = -12.5$ cm.
The image $I_1$ is formed at the focus, 12.5 cm from the primary mirror.
This image $I_1$ acts as a virtual object for the secondary (convex) mirror.
The distance between the mirrors is $d=2.5$ cm.
The object distance for the secondary mirror ($u_2$) is the distance from the secondary mirror to $I_1$. The secondary mirror is between the primary mirror and its focus. $u_2 = f_1 - d = 12.5 - 2.5 = 10$ cm. (The object is virtual, so $u_2$ is positive).
For the secondary mirror: Radius of curvature $R_2 = +16$ cm (convex).
Focal length $f_2 = R_2/2 = +8$ cm.
Now, we use the mirror formula for the secondary mirror to find the final image position ($v_2$).
$\frac{1}{v_2} + \frac{1}{u_2} = \frac{1}{f_2}$.
$\frac{1}{v_2} + \frac{1}{10} = \frac{1}{8}$.
$\frac{1}{v_2} = \frac{1}{8} - \frac{1}{10} = \frac{5-4}{40} = \frac{1}{40}$.
$v_2 = 40$ cm.
Since $v_2$ is positive, the final image is formed 40 cm to the right of the secondary (convex) mirror.