Reflection of Light Ray Inside Equilateral Triangular Mirror Box

To determine which statement(s) is(are) correct, we analyze the path of the light ray inside the equilateral triangular mirror configuration using coordinate geometry and the law of reflection.

Let the vertices of the equilateral triangle $ABC$ of side length $L$ be placed in a Cartesian coordinate system as:

• $B = (0, 0)$
• $C = (L, 0)$
• $A = \left(\frac{L}{2}, \frac{\sqrt{3}L}{2}\right)$

The equations of the three sides are:

1. Base $BC$: $y = 0$ for $x \in [0, L]$
2. Left side $AB$: $y = \sqrt{3}x$ for $x \in \left[0, \frac{L}{2}\right]$
3. Right side $AC$: $y = \sqrt{3}(L - x)$ for $x \in \left[\frac{L}{2}, L\right]$

The hole $H$ is located on the base mirror $BC$ at a distance $l$ from corner $C(L,0)$, so its position is: $H = (L - l, 0), \quad 0 < l < L$

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Analysis of Option A:

For $\theta = 30^\circ$, the ray enters through $H(L-l, 0)$ into the interior of the triangle at an angle of $30^\circ$ to the base $BC$ (directed towards side $AB$).

The angle of the ray with the positive x-axis is $150^\circ$, so its direction vector is: $\hat{u} = \left(-\cos 30^\circ, \sin 30^\circ\right) = \left(-\frac{\sqrt{3}}{2}, \frac{1}{2}\right)$

The line equation of the ray is: $y - 0 = \tan(150^\circ)(x - (L-l)) \implies x + \sqrt{3}y = L - l$

To find the intersection point $P_1$ with the left mirror $AB$ ($y = \sqrt{3}x$): $x + \sqrt{3}(\sqrt{3}x) = L - l \implies 4x = L - l \implies x = \frac{L-l}{4}$ $y = \frac{\sqrt{3}(L-l)}{4}$

Since $0 < l < L$, $P_1 = \left(\frac{L-l}{4}, \frac{\sqrt{3}(L-l)}{4}\right)$ lies strictly on the side $AB$.

Now, check the angle of incidence on side $AB$:

• Slope of mirror $AB$: $m_{AB} = \sqrt{3} = \tan(60^\circ)$
• Slope of incoming ray: $m_{\text{ray}} = -\frac{1}{\sqrt{3}} = \tan(150^\circ)$

The product of their slopes is: $m_{AB} \cdot m_{\text{ray}} = \sqrt{3} \cdot \left(-\frac{1}{\sqrt{3}}\right) = -1$

Since the product of the slopes is $-1$, the ray strikes mirror $AB$ at normal incidence ($90^\circ$ to the mirror surface).

Upon normal incidence, the ray retraces its path completely back along the same line: $x + \sqrt{3}y = L - l$ It returns directly to $(L-l, 0) = H$ and exits through the hole after $1$ reflection for any $0 < l < L$.

Thus, Option A is correct.

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Analysis of Option B:

For $l = \frac{L}{2}$, the hole is located at the midpoint of the base: $H = \left(\frac{L}{2}, 0\right)$

Let a ray enter through $H$ at an angle $\theta = 60^\circ$ to the base (directed towards side $AB$). Its direction vector inside the triangle is: $\hat{u}_1 = \left(-\cos 60^\circ, \sin 60^\circ\right) = \left(-\frac{1}{2}, \frac{\sqrt{3}}{2}\right)$

The line equation of the incoming ray is: $\sqrt{3}x + y = \frac{\sqrt{3}L}{2}$

1. First Reflection (on side $AB$): Intersection of ray with side $AB$ ($y = \sqrt{3}x$): $\sqrt{3}x + \sqrt{3}x = \frac{\sqrt{3}L}{2} \implies x_1 = \frac{L}{4}, \quad y_1 = \frac{\sqrt{3}L}{4}$ Point $P_1 = \left(\frac{L}{4}, \frac{\sqrt{3}L}{4}\right)$ is the midpoint of side $AB$. The reflected ray direction becomes horizontal to the right: $\hat{u}_2 = (1, 0)$

2. Second Reflection (on side $AC$): The horizontal ray $y = \frac{\sqrt{3}L}{4}$ intersects side $AC$ ($y = \sqrt{3}(L-x)$) at: $\frac{\sqrt{3}L}{4} = \sqrt{3}(L-x) \implies x_2 = \frac{3L}{4}, \quad y_2 = \frac{\sqrt{3}L}{4}$ Point $P_2 = \left(\frac{3L}{4}, \frac{\sqrt{3}L}{4}\right)$ is the midpoint of side $AC$. By the law of reflection, the reflected ray direction is directed downwards at $60^\circ$: $\hat{u}_3 = \left(-\frac{1}{2}, -\frac{\sqrt{3}}{2}\right)$

3. Return to the Hole: The ray travels along the line: $y - \frac{\sqrt{3}L}{4} = \sqrt{3}\left(x - \frac{3L}{4}\right)$ Setting $y = 0$ gives $x = \frac{L}{2}$, which is the exact location of the hole $H\left(\frac{L}{2}, 0\right)$.

Thus, the ray comes out after two reflections.

Therefore, Option B is correct.

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Analysis of Options C and D:

Consider a ray entering at $\theta = 60^\circ$ for any distance $0 < l < L$. The hole position is $H(L-l, 0)$.

Tracing the trajectory step-by-step:

1. Segment 1 ($H \to P_1$): Starts at $H(L-l,0)$, hits side $AB$ at $P_1 = \left(\frac{L-l}{2}, \frac{\sqrt{3}(L-l)}{2}\right)$ $\rightarrow$ 1st reflection.
2. Segment 2 ($P_1 \to P_2$): Reflects horizontally to side $AC$, hitting at $P_2 = \left(\frac{L+l}{2}, \frac{\sqrt{3}(L-l)}{2}\right)$ $\rightarrow$ 2nd reflection.
3. Segment 3 ($P_2 \to P_3$): Reflects downwards at $60^\circ$ to base $BC$, hitting at $P_3 = (l, 0)$. Since $l \neq L-l$ (for $l \neq L/2$), $P_3 \neq H$, so it reflects off the bottom mirror $\rightarrow$ 3rd reflection.
4. Segment 4 ($P_3 \to P_4$): Reflects upwards at $60^\circ$ to side $AC$, hitting at $P_4 = \left(\frac{L+l}{2}, \frac{\sqrt{3}(L-l)}{2}\right)$ $\rightarrow$ 4th reflection.
5. Segment 5 ($P_4 \to P_5$): Reflects horizontally to side $AB$, hitting at $P_5 = \left(\frac{L-l}{2}, \frac{\sqrt{3}(L-l)}{2}\right)$ $\rightarrow$ 5th reflection.
6. Segment 6 ($P_5 \to P_6$): Reflects downwards at $60^\circ$ to base $BC$, hitting at $P_6 = (L-l, 0) = H$.

Since $P_6$ is the hole $H$, the ray exits through the hole after traversing 6 path segments and undergoing 5 reflections.

• For $l = \frac{L}{3}$, the ray will exit through the hole after 5 reflections. Thus, statement (C) claiming it will "NEVER come out" is incorrect.
• Statement (D) claims the ray comes out "after six reflections", but it exits after five reflections. Thus, statement (D) is incorrect.

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Conclusion:

The correct statements are A and B.