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Animated Solution for Physics - Optics: Two plane mirrors $A$ and $B$ are aligned parallel to each other, as shown in the figure. A light ray is incident at an angle $3 0^{\circ}$ at a point just inside one end of $A$ . The plane of incidence coincides with the plane of the figure. The maximum number of times the ray undergoes reflections (including the first one) before it emerges out is

Q: Why do we use $\tan 30^\circ$ instead of $\sin$ or $\cos$?

We use $\tan$ because we know the adjacent side (the distance between the mirrors) and we want to find the opposite side (the horizontal distance covered). $\tan \theta = \frac{\text{Opposite}}{\text{Adjacent}}$ perfectly relates these two.

Q: What if the total length wasn't perfectly divisible by $x$?

If the division resulted in a decimal, say $30.5$, the ray would undergo 30 full reflections and then emerge halfway through the 31st segment. The number of reflections would still be 30.

Q: Would the answer change if the angle was given with respect to the mirror surface?

Yes! If the $30^\circ$ angle was with the mirror surface, the angle of incidence (with the normal) would be $60^\circ$. We would then use $\tan 60^\circ$, resulting in a different horizontal distance per reflection.

Select Answer:

Visualized Solution

\text{Visualizing the Setup}

\text{Two parallel mirrors } A \text{ and } B \text{ of length } l = 2\sqrt{3} \text{ m are separated by } h = 0.2 \text{ m.}
\text{A ray is incident at } 30^\circ \text{ to the normal.}

\text{Horizontal Distance per Reflection}

\text{Let } x \text{ be the horizontal distance covered by the ray between successive reflections.}
\text{From the geometry of the triangle:}
\tan \theta = \frac{x}{h} \implies x = h \tan \theta

\text{Calculating } x

\text{Substitute } h = 0.2 \text{ m and } \theta = 30^\circ:
x = 0.2 \tan 30^\circ
x = \frac{0.2}{\sqrt{3}} \text{ m}

\text{Total Number of Reflections}

\text{The maximum number of reflections } n \text{ is the total length } l \text{ divided by } x:
n = \frac{l}{x}
n = \frac{2\sqrt{3}}{0.2 / \sqrt{3}}

\text{Final Calculation}

n = \frac{2 \times 3}{0.2}
n = \frac{6}{0.2} = 30
\text{The ray undergoes exactly 30 reflections before emerging.}

\text{Edge Case Analysis}

\text{Since } n \text{ is an exact integer, the 30th reflection happens exactly at the edge of the mirror, after which it escapes.}

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The Sigma Insight: Plane Mirror

Solution Diagram

Visualizing the Setup Imagine two parallel mirrors, $A$ and $B$ , facing each other

A light ray sneaks in just inside the left edge of mirror

A

at an angle of

3 0^{\circ}

with the normal. Our goal is to find out how many times this ray bounces before it escapes from the other end.

Horizontal Distance per Reflection To solve this, we need to figure out how much horizontal distance the ray covers in a single bounce

Let's call this distance

x

. If we look at the right-angled triangle formed by the ray, the normal, and the mirror, we can use basic trigonometry.

Using the tangent function,

tan 3 0^{\circ}

equals the opposite side

x

divided by the adjacent side, which is the distance between the mirrors,

0.2

meters. So,

x

equals

0.2 \times tan 3 0^{\circ}

. This gives us

x = \frac{0.2}{3}

meters. This is the horizontal step the ray takes with every single reflection.

Total Number of Reflections To find the total number of segments or steps, we divide the total length of the mirrors, $23$ , by the distance of one step, $x$

Let's do the math.

Dividing

23

\frac{0.2}{3}

, the

3

multiply to give

3

. Two times three is six. Six divided by

0.2

gives exactly

30

. So, there are

30

segments.

Edge Case Analysis Here is the catch! The ray enters 'just inside' the edge

This means the 30th reflection happens just before the right edge. The next segment will carry the ray outside the mirrors. So, the maximum number of reflections is exactly 30. A beautiful play of geometry!

Common Traps & FAQs

Why do we use

tan 3 0^{\circ}

instead of

sin

cos

We use

tan

because we know the adjacent side (the distance between the mirrors) and we want to find the opposite side (the horizontal distance covered).

tan θ = \frac{Opposite}{Adjacent}

perfectly relates these two.

Does the first point of incidence count as a reflection?

Yes, the question explicitly states to include the first reflection. The ray enters and immediately reflects off mirror

A

, which counts as reflection number 1.

What if the total length wasn't perfectly divisible by

x

If the division resulted in a decimal, say

30.5

, the ray would undergo 30 full reflections and then emerge halfway through the 31st segment. The number of reflections would still be 30.

Why does the ray emerge after exactly 30 reflections?

Because the ray enters 'just inside' the edge, the 30th reflection occurs infinitesimally close to the far edge. The subsequent path takes it beyond the mirror's length, allowing it to escape.

Would the answer change if the angle was given with respect to the mirror surface?

Yes! If the

3 0^{\circ}

angle was with the mirror surface, the angle of incidence (with the normal) would be

6 0^{\circ}

. We would then use

tan 6 0^{\circ}

, resulting in a different horizontal distance per reflection.

Select Answer:

Visualized Solution

\text{Visualizing the Setup}

\text{Horizontal Distance per Reflection}

\text{Calculating } x

\text{Total Number of Reflections}

\text{Final Calculation}

\text{Edge Case Analysis}

The Sigma Insight: Plane Mirror

Visualizing the Setup Imagine two parallel mirrors, $A$ and $B$ , facing each other

Horizontal Distance per Reflection To solve this, we need to figure out how much horizontal distance the ray covers in a single bounce

Total Number of Reflections To find the total number of segments or steps, we divide the total length of the mirrors, $23$ , by the distance of one step, $x$

Edge Case Analysis Here is the catch! The ray enters 'just inside' the edge

Similar Questions

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

Select Answer:

Visualized Solution

\text{Visualizing the Setup}

\text{Horizontal Distance per Reflection}

\text{Calculating } x

\text{Total Number of Reflections}

\text{Final Calculation}

\text{Edge Case Analysis}

The Sigma Insight: Plane Mirror

Horizontal Distance per Reflection To solve this, we need to figure out how much horizontal distance the ray covers in a single bounce

Edge Case Analysis Here is the catch! The ray enters 'just inside' the edge

Similar Questions

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