A point source of light, S is placed at a distance L in front of the centre of plane mirror of width

Q: Why do we measure the distance from the virtual image instead of the actual source?

Because the reflected rays that reach the man's eyes appear to diverge directly from the virtual image. Geometrically, it's much simpler to treat the virtual image as the origin of these rays to form similar triangles.

Q: Does the field of view depend on the distance $L$?

In this specific setup, the ratio of the distances ($3L / L$) causes $L$ to cancel out. So, the final width of the field of view ($3d$) is independent of $L$, as long as the path is at $2L$.

Q: What if the man walks closer to the mirror, say at a distance $L$?

If the man walks at distance $L$, the height of the large triangle becomes $L + L = 2L$. The ratio would be $2L/L = 2$, so his field of view would be $2d$.

Q: Why are the triangles similar?

The mirror and the man's path are parallel lines. The rays from the image $S'$ act as transversals intersecting these parallel lines, creating equal corresponding angles, which guarantees the triangles are similar.

Q: What if the mirror was tilted?

If the mirror was tilted, the image $S'$ would shift, and the reflected rays would no longer form a simple symmetric triangle with the man's path. You would need to use coordinate geometry to find the new intercept.

Question

A point source of light, $S$ is placed at a distance $L$ in front of the centre of plane mirror of width $d$ which is hanging vertically on a wall. A man walks in front of the mirror along a line parallel to the mirror, at a distance $2L$ as shown below.
The distance over which the man can see the image of the light source in the mirror is

Options:

(A) $\frac{d}{2}$

(B) $d$

(C) $3d$

(D) $2d$

Sigma Percentile · Accepted Answer

### The Magic of the Field of View

Imagine you are standing in a room with a small mirror on the wall. You can see the reflection of a lamp, but as you walk across the room, the reflection suddenly disappears. Why does this happen? This is the beautiful concept of the **field of view**.

In this problem, we are going to explore exactly how much of the room allows you to see that reflection. We have a point source of light, $S$, placed at a distance $L$ in front of a plane mirror of width $d$. A man is walking along a path parallel to the mirror, at a distance $2L$. Our goal is to find the exact distance over which he can see the image of the light source.

### The Virtual World Behind the Mirror

Before we worry about the man, we need to understand what the mirror is doing. A plane mirror creates a virtual world. For every object in front of it, it creates a perfect, virtual twin behind it.

The rule is simple: the image is formed exactly at the same distance behind the mirror as the object is in front. Since our source $S$ is at a distance $L$ in front of the mirror, its virtual image $S'$ will form at a distance $L$ behind the mirror.

This virtual image $S'$ is the key to everything. To the man walking in the room, the light doesn't seem to be bouncing off a mirror. Instead, it appears to be shining directly from $S'$, straight through the wall!

### Tracing the Extreme Rays

Now, how do we find the field of view? The mirror is like a window into that virtual world. The edges of the mirror act as the frame of this window.

To find the boundaries of what the man can see, we draw the **extreme rays**. These are the rays of light that leave the source $S$ and hit the very top and bottom edges of the mirror, which we can call points $C$ and $D$.

When these extreme rays reflect off the mirror, they travel outwards towards the man's path. If we extend these reflected rays backwards, they meet perfectly at the virtual image $S'$.

The region between these two reflected rays on the man's path is his field of view. Let's call the points where these rays intersect his path $A$ and $B$. As long as the man is walking between $A$ and $B$, he is bathed in the reflected light and can see the image. Our mission is to find the length of this segment $AB$.

### The Elegance of Similar Triangles

This is where physics hands the baton to geometry. We have a beautiful setup of triangles that will solve the problem for us.

Look at the small triangle formed by the virtual image $S'$ and the mirror edges $C$ and $D$. This is $	riangle S'CD$.

Now, look at the larger triangle formed by the same virtual image $S'$ and the intercept $AB$ on the man's path. This is $	riangle S'AB$.

Because the mirror is perfectly parallel to the man's path, these two triangles share the same angles. They are **similar triangles**. This is a powerful realization because it means their proportions are locked together.

The ratio of their bases must equal the ratio of their heights:

$$ \frac{AB}{CD} = \frac{	ext{Height of } 	riangle S'AB}{	ext{Height of } 	riangle S'CD} $$

### Calculating the Heights

Let's find these heights. The height of a triangle here is just the perpendicular distance from the tip ($S'$) to the base.

For the small triangle $	riangle S'CD$, the base is the mirror itself. The height is the distance from the image $S'$ to the mirror. We already know this is $L$.

$$ 	ext{Height of } 	riangle S'CD = L $$

For the large triangle $	riangle S'AB$, the base is the field of view $AB$. The height is the total distance from the image $S'$ to the man's path.

How far is that? Well, it's the distance from $S'$ to the mirror (which is $L$), plus the distance from the mirror to the man's path (which is $2L$).

$$ 	ext{Height of } 	riangle S'AB = L + 2L = 3L $$

### The Final Computation

We have all the pieces of the puzzle. Let's plug them into our similarity equation. The base of the small triangle $CD$ is just the width of the mirror, $d$.

$$ \frac{AB}{d} = \frac{3L}{L} $$

The $L$ beautifully cancels out, showing that the field of view doesn't actually depend on the absolute distance $L$, but only on the ratios!

$$ \frac{AB}{d} = 3 $$

Multiplying both sides by $d$, we arrive at our grand conclusion:

$$ AB = 3d $$

The man can see the image over a distance that is exactly three times the width of the mirror.

### Beyond the Parallel Path

This problem is a classic example of how ray optics and simple geometry intertwine. But what if we changed the rules?

What if the man's path was not parallel to the mirror? The triangles would no longer be perfectly similar in this simple way. The field of view would become skewed, and we would need to rely on coordinate geometry to find the exact intersection points.

Or what if the source $S$ was not placed symmetrically in front of the mirror's center? The field of view would shift along the path. These variations are fantastic thought experiments to test your true understanding of the concepts. Keep visualizing, and keep questioning!

Select Answer:

Visualized Solution

$Visualizing the Setup$

$Image Formation$

$Extreme Rays$

$Virtual Rays$

$Field of View$

$Similar Triangles Setup$

$Similar Triangles Setup$

$Similarity Condition$

$Height of Small Triangle$

$Height of Large Triangle$

$Substituting Values$

$Simplifying the Ratio$

$Final Answer$

$The Way Forward$

The Sigma Insight: Plane Mirror

The Magic of the Field of View

The Virtual World Behind the Mirror

Tracing the Extreme Rays

The Elegance of Similar Triangles

Calculating the Heights

The Final Computation

Beyond the Parallel Path

Similar Questions

Two plane mirrors $M_{1}$ and $M_{2}$ are at right angle to each other shown. A point source $P$ is placed at $a$ and $2 a$ meter away from $M_{1}$ and $M_{2}$ , respectively. The shortest distance between the images thus formed is (Take $5 = 2.3$ )

A ray of light travelling in the direction $\frac{1}{2} (\hat{i} + 3 \hat{j})$ is incident on a plane mirror. After reflection, it travels along the direction $\frac{1}{2} (\hat{i} - 3 \hat{j})$ . The angle of incidence is

The plane mirrors ( $M_{1}$ and $M_{2}$ ) are inclined to each other such that a ray of light incident on mirror $M_{1}$ and parallel to the mirror $M_{2}$ is reflected from mirror $M_{2}$ parallel to the mirror $M_{1}$ . The angle between the two mirror is

If two mirrors are kept at $6 0^{\circ}$ to each other, then the number of images formed by them is

To get three images of a single object, one should have two plane mirrors at an angle of

Select Answer:

Visualized Solution

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

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

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

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

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

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

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

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

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Height of Small Triangle

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Height of Large Triangle

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

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

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

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

The Sigma Insight: Plane Mirror

The Magic of the Field of View

The Virtual World Behind the Mirror

Tracing the Extreme Rays

The Elegance of Similar Triangles

Calculating the Heights

The Final Computation

Beyond the Parallel Path

Similar Questions

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If two mirrors are kept at 60∘ to each other, then the number of images formed by them is

To get three images of a single object, one should have two plane mirrors at an angle of

$Visualizing the Setup$

$Image Formation$

$Extreme Rays$

$Virtual Rays$

$Field of View$

$Similar Triangles Setup$

$Similar Triangles Setup$

$Similarity Condition$

$Height of Small Triangle$

$Height of Large Triangle$

$Substituting Values$

$Simplifying the Ratio$

$Final Answer$

$The Way Forward$

If two mirrors are kept at $6 0^{\circ}$ to each other, then the number of images formed by them is