A planar structure of length L and width W is made of two different optical media of refractive indi

Q: Why does the maximum time correspond to the critical angle?

As the angle of incidence at the boundary approaches the critical angle, the ray bounces more frequently, increasing the total distance it travels to cover the same horizontal length $L$. If the angle is any smaller, the ray refracts out and escapes!

Q: What happens if the incident angle $\theta$ is changed?

Changing $\theta$ changes the angle at which the ray hits the $n_1$-$n_2$ interface. If it hits at an angle greater than $\theta_C$, it still undergoes TIR but takes a shorter path, exiting faster. If it hits at an angle less than $\theta_C$, it escapes the structure entirely.

Q: How did we get the relation $D = L / \sin \theta_C$?

If you look at a single "bounce" of the ray, it forms a right triangle where the horizontal distance is $x$ and the slanted path is $d$. From trigonometry, $\sin \theta_C = x / d$. Since this applies to every bounce, the total path $D$ relates to the total length $L$ in the exact same way: $\sin \theta_C = L / D$.

Q: Why is the angle $\theta_C$ measured from the normal?

In optics, all angles of incidence, reflection, and refraction are conventionally measured from the normal (the perpendicular line) to the surface, not from the surface itself. This ensures Snell's law works consistently!

Question

A planar structure of length L and width W is made of two different optical media of refractive indices $n_1 = 1.5$ and $n_2 = 1.44$ as shown in figure. If $L \gg W$, a ray entering from end AB will emerge from end CD only if the total internal reflection condition is met inside the structure. For $L = 9.6 	ext{ m}$, if the incident angle $	heta$ is varied, the maximum time taken by a ray to exit the plane CD is $t 	imes 10^{-9} 	ext{ s}$, where t is _______.
[Speed of light $c = 3 	imes 10^8 	ext{ m/s}$]

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine a ray of light entering an optical structure, much like a fiber optic cable. Our goal is to find the **maximum time** this light ray can spend inside the core medium before it finally exits from the other end.

How do we maximize the time? By maximizing the distance it travels! The light travels the longest possible zig-zag path when it undergoes Total Internal Reflection (TIR) at the boundaries, hitting them at exactly the **critical angle**. If the angle were any smaller, the light would refract out and escape. If it were larger, the path would be more direct and shorter.

### The Master Equation: Critical Angle

Let's find this critical angle, $	heta_C$. For total internal reflection at the interface of the two media, the sine of the critical angle is simply the ratio of the refractive indices of the rarer medium to the denser medium.

$$ \sin 	heta_C = \frac{n_2}{n_1} $$

Substituting the given values of $n_2 = 1.44$ and $n_1 = 1.50$:

$$ \sin 	heta_C = \frac{1.44}{1.50} = 0.96 $$

### Unlocking the Path Length

Now, look closely at the geometry of one single bounce. If the ray covers a horizontal distance $x$, the actual slanted path length it travels is $d$. From the right-angled triangle formed by the ray's path, we can relate these distances using trigonometry:

$$ \sin 	heta_C = \frac{x}{d} \implies d = \frac{x}{\sin 	heta_C} $$

This relationship holds true for every single bounce. Therefore, the total path length $D$ for the entire structure of length $L$ is simply:

$$ D = \frac{L}{\sin 	heta_C} $$

### Speed of Light in the Medium

Before we calculate the time, we need the speed of light inside this specific medium. Light slows down when it enters a medium with a refractive index greater than 1. The speed $v$ is the speed of light in vacuum, $c$, divided by the refractive index $n_1$.

$$ v = \frac{c}{n_1} $$

Plugging in the values:

$$ v = \frac{3 	imes 10^8}{1.5} = 2 	imes 10^8 	ext{ m/s} $$

### Final Calculation

We have everything we need! The maximum time $t$ is the total distance $D$ divided by the speed $v$.

$$ t = \frac{D}{v} = \frac{L}{v \sin 	heta_C} $$

Let's carefully substitute $L = 9.6 	ext{ m}$, $v = 2 	imes 10^8 	ext{ m/s}$, and $\sin 	heta_C = 0.96$:

$$ t = \frac{9.6}{2 	imes 10^8 	imes 0.96} $$

The denominator becomes $1.92 	imes 10^8$. Dividing $9.6$ by this gives us:

$$ t = 5 	imes 10^{-8} 	ext{ s} $$

To match the format requested in the question ($t 	imes 10^{-9} 	ext{ s}$), we convert this to nanoseconds:

$$ t = 50 	imes 10^{-9} 	ext{ s} $$

So, our final answer is **50**!

Enter Numerical Value:

Visualized Solution

$t_{max}$

$sin θ_{C}$

$D = \frac{L}{s i n θ _{C}}$

$v = \frac{c}{n _{1}}$

$t = \frac{D}{v}$

$t = 50 \times 1 0^{- 9} s$

The Sigma Insight: Refraction and Total Internal Reflection

Analyzing the Setup

The Master Equation

Unlocking the Path Length

Speed of Light in the Medium

Final Calculation

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