In figure, the optical fibre is l = 2 m long and has a diameter of d = 20 um. If a ray of light is i

Q: Why do we use Snell's Law at the entry point?

Because the light is moving from air (a rarer medium) into the glass fibre (a denser medium), it bends towards the normal. We need this exact angle to trace its path inside.

Q: Why is the angle of incidence at the top surface $90^\circ - r$?

The normal to the entry face is horizontal, while the normal to the top surface is vertical. Since the ray travels in a straight line between them, the angles it makes with the horizontal and vertical normals are complementary.

Q: What if the ray doesn't hit the top surface first?

It doesn't matter! Whether it hits the top or bottom first, the geometry of the triangles formed by the bounces remains identical due to symmetry.

Q: Why do we divide the total length by $x$?

Think of $x$ as the 'stride length' of the ray. If you know how far you go in one step, and you know the total distance, dividing the total distance by your stride length gives you the number of steps!

Q: Is it exactly 57735 reflections?

Mathematically, yes, based on the approximations given ($\sin 40^\circ = 0.64$). In reality, the ray might emerge slightly before completing the very last fractional bounce, but for such a large number, the integer approximation is perfectly accurate for the options provided.

Question

In figure, the optical fibre is $l =2$ m long and has a diameter of $d = 20\mu$m. If a ray of light is incident on one end of the fibre at angle $	heta_1 = 40^\circ$, the number of reflections it makes before emerging from the other end is close to (refractive index of fibre is 1.31 and $\sin 40^\circ = 0.64$)

Options:

(A) 55000

(B) 66000

(C) 45000

(D) 57000

Sigma Percentile · Accepted Answer

### The Setup: Entering the Fibre

Imagine a ray of light embarking on a journey through an optical fibre. As it strikes the flat entry face at an angle of $40^\circ$, it doesn't just go straight; it bends. This bending is governed by **Snell's Law**, which relates the angles and refractive indices of the two mediums.

Our first mission is to find out exactly how much it bends. We set up our equation:
$$n_1 \sin i = n_2 \sin r$$

### The First Bend: Calculating Refraction

We know the ray is coming from air, so $n_1 = 1$. The fibre's refractive index is given as $n_2 = 1.31$, and our angle of incidence $i$ is $40^\circ$. Plugging these in, we get:
$$1 \cdot \sin 40^\circ = 1.31 \cdot \sin r$$

The problem kindly provides $\sin 40^\circ = 0.64$. Substituting this gives us:
$$\sin r = \frac{0.64}{1.31} \approx 0.49$$

Notice how $0.49$ is incredibly close to $0.5$. In the world of physics problems, this is a massive hint! Since $\sin 30^\circ = 0.5$, we can confidently say that the angle of refraction $r$ is **$30^\circ$**.

### The Internal Geometry: A Single Bounce

Now that the ray is inside, it travels in a straight line until it hits the top boundary of the fibre. Because the angle is right, it undergoes **Total Internal Reflection**. But how far does it travel horizontally before it bounces?

Let's look at the geometry. The ray forms a right-angled triangle, let's call it $\Delta OAB$, with the walls of the fibre. The angle it makes with the vertical normal at the top surface is $	heta$. Since the normal to the entry face and the normal to the top surface are perpendicular, we can easily find $	heta$:
$$	heta = 90^\circ - r = 90^\circ - 30^\circ = 60^\circ$$

### The Stride Length: Horizontal Distance

In our triangle $\Delta OAB$, the vertical side is simply the diameter of the fibre, $d = 20 \mu	ext{m}$. The horizontal side is the distance $x$ we want to find. Using basic trigonometry:
$$	an 	heta = \frac{	ext{Opposite}}{	ext{Adjacent}} = \frac{x}{d}$$

Rearranging for $x$, we get:
$$x = d 	an 60^\circ$$

Substituting our known values:
$$x = 20\mu	ext{m} \cdot \sqrt{3} = 20\sqrt{3} \mu	ext{m}$$

This $x$ is the "stride length" of our ray—the horizontal distance it covers with every single bounce!

### The Grand Total: Counting the Reflections

We know the ray takes strides of $20\sqrt{3} \mu	ext{m}$, and it has to travel down a fibre that is $l = 2 	ext{ m}$ long. To find the total number of reflections $n$, we just divide the total length by the length of one stride:
$$n = \frac{l}{x}$$

Let's plug in the numbers, remembering to convert micrometers to meters:
$$n = \frac{2}{20\sqrt{3} 	imes 10^{-6}}$$

Calculating this gives:
$$n \approx 57735$$

Looking at our options, this is closest to **$57000$**. And just like that, we've traced the path of a single ray of light through tens of thousands of reflections!

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Visualized Solution

$O pt i c a l F ib r e S e t u p$

$S n e l l^{'} s L a w a t E n t r y$

$S u b s t i t u t in g V a l u es$

$C a l c u l a t in g R e f r a c t i o n A n g l e$

$G eo m e t r y o f a S in g l e B o u n ce$

$H or i z o n t a l D i s t an ce p er B o u n ce$

$C a l c u l a t in g x$

$T o t a l N u mb er o f R e f l ec t i o n s$

$F ina l C a l c u l a t i o n$

$F ina l A n s w er$

The Sigma Insight: Refraction and Total Internal Reflection

The Setup

The First Bend

The Internal Geometry

The Stride Length

The Grand Total

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Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Optical Fibre Setup

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Snell′s Law at Entry

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Substituting Values

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Calculating Refraction Angle

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Geometry of a Single Bounce

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Horizontal Distance per Bounce

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Calculating x

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Total Number of Reflections

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Final Calculation

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Final Answer