Three plane mirrors form an equilateral triangle with each side of length L. There is a small hole a

Q: Why does the ray retrace its path in Option A?

The right mirror is tilted at $120^\circ$ to the horizontal. A ray fired at $30^\circ$ strikes this mirror exactly at a $90^\circ$ angle (normal incidence). By the laws of reflection, it bounces straight back!

Q: How do we know the path forms an inscribed equilateral triangle in Option B?

When you fire a ray from the midpoint of one side at $60^\circ$, it strikes the exact midpoint of the adjacent side. The angle of reflection ensures it continues to the midpoint of the third side, forming a smaller equilateral triangle inside the larger one.

Q: Is there a faster way to trace the rays without calculating coordinates?

Yes! You can use the 'method of images' by reflecting the triangle itself across its edges to form a grid of triangles. A straight line drawn through this grid represents the path of the ray, making it much easier to count reflections and find exit points.

Q: Why does Option C say it NEVER comes out?

Option C is a trap! It assumes that because the starting position $L/3$ lacks the perfect symmetry of the midpoint, the ray might get trapped in an infinite loop. However, tracing the path reveals it perfectly exits after 5 reflections.

Q: Does the ray always come out eventually for any angle?

Not necessarily! For some irrational angles or specific starting positions, the ray might bounce infinitely without ever hitting the exact point of the hole. But for these specific rational angles ($30^\circ$, $60^\circ$), the paths are closed loops or exit paths.

Question

Three plane mirrors form an equilateral triangle with each side of length L. There is a small hole at a distance $l > 0$ from one of the corners as shown in the figure. A ray of light is passed through the hole at an angle $	heta$ and can only come out through the same hole. The cross section of the mirror configuration and the ray of light lie on the same plane.

Options:

(A) The ray of light will come out for $	heta = 30^\circ$, for $0 < l < L$.

(B) There is an angle for $l = \frac{L}{2}$ at which the ray of light will come out after two reflections.

(C) The ray of light will NEVER come out for $	heta = 60^\circ$ and $l = \frac{L}{3}$.

(D) The ray of light will come out for $	heta = 60^\circ$, and $0 < l < \frac{L}{2}$ after six reflections.

Sigma Percentile · Accepted Answer

Imagine you are standing inside a giant equilateral triangle constructed entirely out of perfect plane mirrors. At the bottom edge, there is a small hole located at a distance $l$ from the right corner. We are going to shoot a laser beam through this hole at a specific angle $	heta$, and our mission is to determine under what conditions the laser beam will bounce around the mirrors and eventually escape back through the exact same hole. This is a classic problem of geometric optics that tests our understanding of the laws of reflection and spatial symmetry.

### Analyzing Option A: The Normal Incidence

Let's test the first option. We set our launch angle $	heta$ to $30^\circ$ and fire the ray towards the right mirror. To understand what happens, we need to think about the geometry of the setup. The right mirror itself is tilted at an angle of $120^\circ$ relative to the positive x-axis (the bottom mirror).

Notice how the ray hits the mirror! The ray is traveling at an angle of $30^\circ$. The normal (perpendicular line) to the right mirror points inwards at an angle of $120^\circ + 90^\circ = 210^\circ$. Because $210^\circ$ and $30^\circ$ are exactly opposite directions (they differ by $180^\circ$), the ray is traveling exactly along the normal line.

This means the angle of incidence is exactly $0^\circ$! It's a **normal incidence**. According to the laws of reflection, a ray striking a mirror normally will simply retrace its path. It bounces straight back and exits the hole after just **one reflection**. This holds true regardless of where the hole is located ($0 < l < L$). Therefore, Option A is absolutely correct.

### Analyzing Option B: The Inscribed Triangle

Now let's clear the board and test Option B. We move the hole exactly to the midpoint of the bottom mirror, so $l = \frac{L}{2}$. This time, we fire the laser at $	heta = 60^\circ$ towards the right.

Because of the perfect symmetry of firing from the midpoint at an angle equal to the triangle's internal angles, the ray strikes the right mirror exactly at its midpoint. Watch the path carefully. Upon striking the right mirror, it reflects horizontally to hit the midpoint of the left mirror. That's reflection number one and two.

From the left mirror, it bounces straight back to our starting hole at the bottom midpoint. The path of the ray forms a perfect **inscribed equilateral triangle**! It exits the system after exactly **two reflections**. So, Option B is also a correct statement.

### Analyzing Options C and D: Breaking the Symmetry

Let's move on to Options C and D. We keep the angle at $	heta = 60^\circ$, but move the hole to $l = \frac{L}{3}$. This breaks the perfect symmetry we had earlier. Let's trace the ray's journey now.

Count the bounces with me. The ray hits the right mirror (bounce 1), reflects horizontally to the left mirror (bounce 2), reflects down to the bottom mirror (bounce 3), reflects up to the right mirror (bounce 4), and reflects horizontally to the left mirror (bounce 5).

After the fifth reflection, geometry dictates that it lands exactly back at the hole! Since it successfully comes out after **five reflections**, Option C, which says it NEVER comes out, is fundamentally wrong. Furthermore, Option D claims it takes six reflections for $l < \frac{L}{2}$, but we just saw it takes exactly five. Thus, Option D is also incorrect.

In conclusion, the elegant symmetries of the equilateral triangle allow the ray to escape under the specific conditions outlined in Options A and B.

Select Answer:

Visualized Solution

\text{The Equilateral Mirror System}

\text{Testing Option A: } \theta = 30^\circ

\text{Normal Incidence}

\text{Testing Option B: } l = L/2, \theta = 60^\circ

\text{The Inscribed Triangle}

\text{Testing Options C \& D: } \theta = 60^\circ

\text{Counting the Reflections}

\text{Final Answer}

The Sigma Insight: Plane Mirror

Analyzing Option A

Analyzing Option B

Analyzing Options C and D

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