Q.Suppose that the lower half of the concave mirror's reflecting surface in Fig. 9.6 is covered with an opaque (non-reflective) material.

The key idea is that each point on the mirror contributes to the entire image — covering part of the mirror reduces the light intensity but does not cut off part of the image.

Reasoning:

1. Every point on the mirror's surface reflects light from every point on the object. The image is formed where all reflected rays (or their extensions) converge.
2. Covering the lower half removes some of these rays, but the remaining upper half still sends enough rays to the same image point.
3. The image location, size, orientation, and nature (real/virtual) depend only on the mirror's geometry and the object's position — not on how much of the mirror is used.

The key insight is that every point on a mirror obeys the law of reflection independently. When you cover the lower half, you are simply removing some of the rays that would have reached the image point. The remaining rays from the upper half still converge (or appear to diverge from) exactly the same image location.

Think of it this way: a concave mirror forms an image because rays from each object point, after reflection, meet at a corresponding image point. Each ray path is determined solely by the geometry of the mirror surface at the point where it hits. If you block half the mirror, you lose half the rays — but the rays that do reflect still follow the same paths and still meet at the same image point. The image is simply formed by fewer rays, so it is less bright.

A common misconception is that covering half the mirror will cut off half the image — like covering half a camera lens. But a mirror is not a lens. A lens focuses light by refraction through its entire aperture; covering half a lens can indeed distort or shift the image. A mirror, however, is a reflective surface where each point acts independently. The image location depends only on the mirror's curvature and the object position, not on which part of the mirror is used.

Let's work through the specific case shown in Figure 9.6(a) — a concave mirror with the object between the pole and the focus.

1. Identify the image formed by the full mirror.

   For an object $AB$ placed between $P$ and $F$ of a concave mirror, the ray diagram shows that reflected rays diverge. Their backward extensions meet behind the mirror, forming an erect, magnified, virtual image $A'B'$. This is the standard result.

2. Now cover the lower half.

   Consider a specific object point, say $A$ at the top of the arrow. In the full mirror, rays from $A$ strike both the upper and lower halves of the mirror. After covering the lower half, only rays hitting the upper half remain. But each of these remaining rays still obeys the law of reflection at its point of incidence. Their backward extensions still meet at the same point $A'$ as before — because the geometry of the mirror surface at those points hasn't changed.

3. Repeat for every object point.

   The same reasoning applies to point $B$ and every point in between. The image location $A'B'$ is unchanged. The only difference is that fewer rays contribute to forming each image point, so the image is dimmer.

4. What about the nature of the image?

   Since the mirror's curvature and the object's position are unchanged, the image remains virtual, erect, and magnified. The covering does not alter the focal length or the mirror equation.

Method: Ray Diagram Analysis for Partial Mirror Obstruction

Concept: When part of a mirror's reflecting surface is covered, the image does not disappear — it only becomes dimmer (less bright). The full image is still formed because each point on the object sends rays to all parts of the mirror.

Steps:

1. Draw the principal axis and the concave mirror

   Mark the pole (P), focus (F), and centre of curvature (C).

2. Place the object at a given position (say, beyond C for a real, inverted image).

3. Identify the obstructed region

   Here, the lower half of the mirror is covered. Only the upper half reflects light.

4. Trace at least two rays from the top of the object

   • Ray 1: Parallel to principal axis → reflects through F (using the upper half of mirror).
   • Ray 2: Through F → reflects parallel to principal axis (again using the upper half).
   • These rays meet at the same image point as before.

5. Repeat for the bottom of the object

   • Even though the lower half is covered, rays from the bottom can still strike the upper half and reflect to form the image.

6. Observe the result

   • Image location, size, and orientation remain unchanged.
   • Image brightness decreases because fewer rays reach the image point.

Key Result:

The image is still formed completely, but it is less bright.

No part of the image is missing — the mirror simply uses fewer reflecting rays.

Why this happens (the "why"):

Each point on the object sends rays in all directions toward the mirror. Even if half the mirror is blocked, the remaining half still receives rays from every point on the object. These rays converge to form the full image — just with less light energy.

Exam Tip:

In numerical or diagram-based questions, always state:

• No change in image position, size, or nature.
• Only brightness reduces — because the effective aperture (area collecting light) is halved.

Here are the most common mistakes students make with this classic partial mirror obstruction problem, and how to avoid each one.

Mistake 1: Thinking "Half the Image Disappears"

The Error: Students often assume that if the lower half of the mirror is covered, the image will lose its lower half (e.g., you’d only see the top half of the person’s face).

Why it’s wrong: Every point on the object sends rays to all parts of the mirror. Even if half the mirror is blocked, the remaining half still receives rays from the entire object. The image is formed by the intersection of reflected rays from the uncovered portion.

How to Avoid: Remember the ray diagram rule: each point on the object emits rays in all directions. The mirror only needs some rays from each point to converge. Covering half the mirror reduces the brightness (intensity) of the image, not its extent.

Mistake 2: Confusing "Image Size" with "Brightness"

The Error: Students say the image becomes smaller or is only half the size.

Why it’s wrong: The mirror’s focal length and radius of curvature remain unchanged. The image location and magnification depend only on the mirror’s geometry (focal length) and the object’s position — not on how much of the mirror is reflecting.

How to Avoid: Use the mirror formula:

and magnification:

Since $f$ and $u$ are unchanged, $v$ and $m$ are unchanged. The image is full size but dimmer.

Mistake 3: Assuming the Image Shifts Position

The Error: Students think covering part of the mirror shifts the image up/down or changes its orientation.

Why it’s wrong: The principal axis and focal point are fixed. The uncovered portion still reflects rays according to the same laws. The image forms at the same location as before.

How to Avoid: Draw a ray diagram using only the top half of the mirror. You’ll see the rays still converge at the same image point. The image is complete and at the same place.

Mistake 4: Forgetting to Mention "Reduced Brightness"

The Error: Students answer only “the image is unchanged” without noting the dimming effect.

Why it’s important: The question asks for the effect on the image. The primary effect is that less light reaches the image, making it fainter or less intense.

How to Avoid: Always include the brightness change in your answer. For example: “The image remains full-sized and at the same location, but its intensity is reduced because fewer rays contribute to its formation.”

Quick Summary Table

Final Exam Tip: If asked this question, write:

“The image is complete and at the same position, but its brightness decreases because only half the mirror’s surface reflects light.”