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

Q: Why do we draw rays from the virtual image instead of the actual source?

Drawing rays from the virtual image simplifies the geometry. The reflected rays appear to diverge exactly from the virtual image, making it easier to use similar triangles to find the field of view.

Q: How do we know $CD = d/2$?

By looking at the similar triangles $\triangle S'OA$ and $\triangle S'SC$, the ratio of their bases is $1:2$. Since $OA = d/2$, $SC$ must be $d$. Subtracting $SD = d/2$ leaves $CD = d/2$.

Q: Does the field of view depend on the distance of the source from the mirror?

Yes, the field of view depends on both the distance of the source ($L$) and the distance of the observer ($2L$). If the source moves closer, the field of view widens.

Q: What if the mirror was not placed symmetrically?

If the mirror is asymmetric relative to the source, the field of view will also be asymmetric. You would calculate the upper and lower visible segments separately using the same similar triangles method.

Q: Why is the distance $HI$ equal to $d$?

The lines defining $H$ and $I$ are drawn horizontally from the top and bottom edges of the mirror. Since the mirror is vertical and has a width $d$, the vertical distance between these horizontal lines is also $d$.

Question

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

Options:

(A) $d/2$

(B) $d$

(C) $2d$

(D) $3d$

Sigma Percentile · Accepted Answer

### Setting the Stage

Imagine you are standing in front of a mirror, and a light source is placed between you and the mirror. The question asks us to find the greatest distance over which a man can see the image of the light source as he walks parallel to the mirror. This concept is known as the **field of view**.

To visualize this, we set up a coordinate system. Let the mirror $AB$ have a width $d$. The point source $S$ is placed at a distance $L$ in front of the mirror. The man walks along a line parallel to the mirror at a distance $2L$ from it.

### Tracing the Field of View

The first step in finding the field of view is to locate the virtual image of the source. For a plane mirror, the image is formed at the exact same distance behind the mirror as the object is in front of it. Therefore, the virtual image $S'$ is located at a distance $L$ behind the mirror.

Next, we trace the extreme rays from the virtual image $S'$ that pass through the edges of the mirror, $A$ and $B$. These rays, when extended forward, define the boundaries of the region where the reflected light can be seen. Let these rays intersect the man's path at points $G$ and $J$.

### The Geometry of Similar Triangles

To calculate the total visible distance $GJ$, we can use the properties of similar triangles. Let's draw horizontal lines from the mirror edges $A$ and $B$ to intersect the man's path at points $H$ and $I$. The vertical distance between these lines, $HI$, is simply the width of the mirror, which is $d$.

Now, let's analyze the upper section. We have a small triangle $	riangle S'AD$ and a larger similar triangle $	riangle S'AH$. The horizontal distance from the image $S'$ to the mirror is $L$, and the horizontal distance from the mirror to the man's path is $2L$. Therefore, the total horizontal distance $S'H$ is $3L$.

By the properties of similar triangles, the ratio of the vertical segments is equal to the ratio of the horizontal segments. Since the horizontal distance $AH$ is twice the distance $AD$, the vertical segment $GH$ must be twice the segment $CD$.

$$GH = 2CD = 2\left(\frac{d}{2}ight) = d$$

### The Final Calculation

By symmetry, the lower section behaves exactly the same way. The vertical segment $IJ$ is also equal to $d$.

Now, we simply add up all the segments to find the total distance $GJ$ over which the man can see the image:

$$GJ = GH + HI + IJ$$

$$GJ = d + d + d = 3d$$

Thus, the greatest distance over which the man can see the image of the light source is **$3d$**.

Select Answer:

Visualized Solution

Visual Anchor

Locating the Virtual Image

Tracing the Extreme Rays

Analyzing the Central Region

Using Similar Triangles (Part 1)

Using Similar Triangles (Part 2)

Final Calculation

The Way Forward

The Sigma Insight: Plane Mirror

Setting the Stage

Tracing the Field of View

The Geometry of Similar Triangles

The Final Calculation

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