Q.A Cassegrain telescope uses two mirrors as shown in Fig. 9.26.

The key idea is that the large concave mirror forms an intermediate image, which then acts as a virtual object for the small convex mirror. The final image location is found by applying the mirror formula twice.

Step 1: Image from the primary mirror.

For an object at infinity, the concave primary mirror (radius $R_1 = 220\ \text{mm}$, so focal length $f_1 = R_1/2 = 110\ \text{mm}$) forms an image at its focus.

Thus, the first image is at $110\ \text{mm}$ in front of the primary mirror.

Step 2: Object distance for the secondary mirror.

The secondary mirror is $20\ \text{mm}$ in front of the primary. The first image lies $110\ \text{mm}$ behind the primary, so relative to the secondary mirror, this image is at a distance

$u_2 = -(110 - 20) = -90\ \text{mm}$ (negative because it is a virtual object — rays are converging toward a point behind the secondary).

Step 3: Apply mirror formula to the secondary mirror.

The secondary mirror is convex with $R_2 = 140\ \text{mm}$, so $f_2 = -R_2/2 = -70\ \text{mm}$.

Using $\frac{1}{v_2} + \frac{1}{u_2} = \frac{1}{f_2}$:

$\frac{1}{v_2} + \frac{1}{-90} = \frac{1}{-70}$ …

Why This Approach Works

A Cassegrain telescope is essentially a clever combination of two mirrors: a large concave primary and a small convex secondary. The primary mirror collects parallel light from a distant object and would normally form an image at its focal point. But instead of letting that image form, we place a convex secondary mirror in the path to intercept the converging rays before they meet. The secondary then reflects these rays back through a hole in the primary, forming a final image behind the primary mirror — right where an eyepiece can view it.

The key insight: we treat this as two successive reflections, each governed by the mirror formula $\frac{1}{f} = \frac{1}{u} + \frac{1}{v}$, with careful attention to sign conventions. The image from the first mirror becomes the object for the second mirror. The trickiest part is getting the signs right for the secondary mirror — especially when the object is virtual (rays are converging toward a point behind the mirror).

Step-by-Step Solution

1. Find the focal length of the primary (large concave) mirror.

The radius of curvature $R_1 = 220\ \text{mm}$. For a spherical mirror, focal length $f = R/2$. Since it's concave, the focal length is negative by the Cartesian sign convention (light travels from left to right; the centre of curvature is to the left of the mirror).

2. Locate the image formed by the primary mirror alone.

For an object at infinity ($u_1 = \infty$), the mirror formula gives:

So $v_1 = -110\ \text{mm}$. The negative sign means the image is formed in front of the primary mirror (to its left), at a distance of 110 mm from it. This is the primary focus $F_1$.

3. Determine the position of this image relative to the secondary mirror.

The two mirrors are 20 mm apart. The primary is at the back, the secondary is 20 mm in front of it. So the primary's image at 110 mm in front of the primary is at a distance of $110 - 20 = 90\ \text{mm}$ in front of the secondary mirror.

Since the secondary mirror is convex and faces the primary, the incoming rays from the primary are converging toward a point 90 mm behind the secondary (i.e., to its right). This means the object for the secondary mirror is virtual — the rays would meet at that point if the secondary weren't there.

4. Find the focal length of the secondary (small convex) mirror. …

Method: Reflection at Spherical Mirrors (Mirror Formula)

This problem involves two successive reflections — first from the large concave mirror, then from the small convex mirror. We use the mirror formula:

where $f = \frac{R}{2}$ (focal length), $u$ = object distance, $v$ = image distance (all measured from the mirror pole, with sign convention).

Step 1: First reflection — Large mirror

• Radius of curvature, $R_1 = +220\ \text{mm}$ (concave → $R$ positive by convention)
• Focal length, $f_1 = \frac{R_1}{2} = +110\ \text{mm}$
• Object at infinity, so $u_1 = \infty$

Using mirror formula:

Image after first mirror forms at $110\ \text{mm}$ in front of the large mirror (on the same side as the incoming light).

Step 2: Locate object for the small mirror

• Mirrors are 20 mm apart (distance between their poles)
• The image from the large mirror is $110\ \text{mm}$ in front of it
• So, distance of this image from the small mirror = $110 - 20 = 90\ \text{mm}$

Since the small mirror is convex and faces the large mirror, this image lies behind the small mirror. By sign convention (real object in front → $u$ negative), here the object is virtual for the small mirror:

Step 3: Second reflection — Small mirror

• Radius of curvature, $R_2 = -140\ \text{mm}$ (convex → $R$ negative)
• Focal length, $f_2 = \frac{R_2}{2} = -70\ \text{mm}$ …

Common Mistakes & How to Avoid Them

Mistake 1: Confusing Sign Conventions for Mirrors

The Error: Students often apply the lens sign convention (Cartesian sign convention) to mirrors, or mix up the sign of $R$ for concave vs convex mirrors.

Why it happens: The problem involves two mirrors — the large mirror is concave (used as the primary), and the small mirror is convex (used as the secondary). Students forget that for mirrors:

• Concave mirror: $R$ is negative (by Cartesian convention, centre of curvature is on the same side as the incoming light)
• Convex mirror: $R$ is positive

How to avoid: Always draw a quick ray diagram first. Mark the direction of incident light (from left to right). Then:

• If the centre of curvature is in front of the mirror (towards the object), $R$ is negative (concave)
• If the centre of curvature is behind the mirror, $R$ is positive (convex)

For this problem:

• Large mirror: $R_1 = -220\ \text{mm}$ (concave)
• Small mirror: $R_2 = +140\ \text{mm}$ (convex)

Mistake 2: Forgetting the Object for the Secondary Mirror

The Error: Students directly use the object at infinity for the small mirror, instead of realising that the image from the first mirror becomes the object for the second mirror.

Why it happens: The problem says "object at infinity" — students assume this applies to both mirrors.

How to avoid: Always break the problem into two independent steps:

1. Step 1: Find where the large mirror forms the image of the object at infinity.

   • For an object at infinity, the image is at the focus of the concave mirror.
   • $f_1 = \frac{R_1}{2} = \frac{-220}{2} = -110\ \text{mm}$
   • So image $I_1$ is $110\ \text{mm}$ in front of the large mirror.

2. Step 2: This image $I_1$ becomes the object for the small mirror. Now find its position relative to the small mirror.

Mistake 3: Incorrect Object Distance for the Secondary Mirror

The Error: Students take the object distance for the small mirror as $110\ \text{mm}$ directly, without accounting for the separation between mirrors.

Why it happens: They forget that distances must be measured from the mirror being used.

How to avoid: Draw the setup with distances clearly labelled.

• Mirrors are $20\ \text{mm}$ apart
• $I_1$ is $110\ \text{mm}$ in front of the large mirror
• So $I_1$ is $110 - 20 = 90\ \text{mm}$ in front of the small mirror

For the small mirror (convex), the object is in front of it, so:

• $u_2 = -90\ \text{mm}$ (negative, as per sign convention — object is on the incident side)

Mistake 4: Using Wrong Mirror Formula Sign

The Error: Plugging values into $\frac{1}{f} = \frac{1}{v} + \frac{1}{u}$ without checking signs.

How to avoid: Always write the mirror formula as:

For the small mirror:

• $f_2 = \frac{R_2}{2} = \frac{+140}{2} = +70\ \text{mm}$ (convex mirror has positive focal length)
• $u_2 = -90\ \text{mm}$

Substitute: …