An object is placed at a distance of 12 cm from a convex lens. A convex mirror of focal length 15 cm

Q: Why do the rays have to strike the mirror normally?

For the final image to coincide with the object, the light rays must retrace their exact path backwards through the lens. By the laws of reflection, a ray retraces its path only if the angle of incidence is zero, which means it strikes the surface normally (perpendicularly).

Q: Why do normal rays point to the center of curvature?

In any spherical mirror, the normal to the surface at any point is the line joining that point to the center of curvature. Therefore, any ray directed towards the center of curvature will strike the mirror normally.

Q: What happens if the mirror is removed?

The convex mirror was intercepting the converging rays from the lens. If removed, these rays will continue to travel and actually converge at the point they were aiming for—the center of curvature of the mirror.

Q: Do we need to use the lens formula here?

Interestingly, no! The problem is cleverly designed so that understanding the geometry of autocollimation gives you the image position directly without needing to calculate the focal length of the lens.

Q: Is the final image real or virtual when the mirror is removed?

When the mirror is removed, the rays actually converge at the center of curvature. Since the light rays physically meet at this point, the image formed is real and inverted.

Question

An object is placed at a distance of $12	ext{ cm}$ from a convex lens. A convex mirror of focal length $15	ext{ cm}$ is placed on other side of lens at $8	ext{ cm}$ as shown in the figure. Image of object coincides with the object.
When the convex mirror is removed, a real and inverted image is formed at a position. The distance of the image from the object will be ........ cm.

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Let's visualize the optical arrangement presented in the problem. We have an object placed $12	ext{ cm}$ in front of a convex lens. On the other side of this lens, at a distance of $8	ext{ cm}$, a convex mirror is positioned. The problem states a very specific and crucial condition: the final image of the object coincides exactly with the object itself.

### The Principle of Autocollimation

The key to unlocking this problem lies entirely in the phrase: **"Image of object coincides with the object."**

In optics, this phenomenon is known as autocollimation. For the final image to form exactly where the object is, the light rays must retrace their entire path backward after reflecting from the mirror. According to the laws of reflection, a light ray will retrace its path only if it strikes the mirror's surface normally (perpendicularly), meaning the angle of incidence is zero.

For a spherical convex mirror, any ray that strikes the surface normally must be directed straight towards its **center of curvature ($C$)**. Therefore, the converging rays emerging from the convex lens must be aiming exactly at the center of curvature of the convex mirror.

### Finding the Center of Curvature

If we were to remove the convex mirror, these converging rays would no longer be intercepted. They would continue their journey and physically converge at the point they were aiming for—the center of curvature of the mirror.

We are given the focal length of the convex mirror, $f = 15	ext{ cm}$. The radius of curvature ($R$) is simply twice the focal length:

$$R = 2f = 2(15) = 30	ext{ cm}$$

This tells us that the center of curvature is located $30	ext{ cm}$ behind the pole of the convex mirror.

### Final Calculation

The question asks for the distance between the object and the image formed when the convex mirror is completely removed. As we established, without the mirror, the lens forms a real, inverted image exactly at the center of curvature of the mirror.

To find the total distance from the object to this new image position, we simply add up the individual segments along the principal axis:

1. Distance from the object to the lens: $12	ext{ cm}$
2. Distance from the lens to the mirror's position: $8	ext{ cm}$
3. Distance from the mirror's position to the center of curvature: $30	ext{ cm}$

Adding these together gives the total distance:

$$	ext{Total Distance} = 12 + 8 + 30 = 50	ext{ cm}$$

Thus, the real and inverted image is formed **$50	ext{ cm}$** away from the object.

Enter Numerical Value:

Visualized Solution

Visualizing the Setup

The Principle of Autocollimation

Locating the Virtual Image

Radius of Curvature of Mirror

Calculating $R$

Total Distance Setup

Adding the Segments

Final Answer

The Way Forward

The Sigma Insight: Lens

Analyzing the Setup

The Principle of Autocollimation

Finding the Center of Curvature

Final Calculation

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Enter Numerical Value:

Visualized Solution

Visualizing the Setup

The Principle of Autocollimation

Locating the Virtual Image

Radius of Curvature of Mirror

Calculating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } R

Total Distance Setup

Adding the Segments

Final Answer

The Way Forward

The Sigma Insight: Lens

Analyzing the Setup

The Principle of Autocollimation

Finding the Center of Curvature

Final Calculation

Similar Questions

A biconvex lens of focal length 15 cm is in front of a plane mirror. The distance between the lens and the mirror is 10 cm. A small object is kept at a distance of 30 cm from the lens. The final image is

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Calculating $R$

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