A ray of laser of a wavelength 630 nm is incident at an angle of 30 degrees at the diamond-air inter

Q: Why did we get $\sin r > 1$?

We got $\sin r > 1$ because the angle of incidence ($30^\circ$) is greater than the critical angle for the diamond-air interface. When this happens, Snell's Law yields an impossible result, indicating that refraction cannot occur.

Q: What is the critical angle for diamond?

The critical angle $\theta_c$ is found using $\sin \theta_c = \frac{\mu_r}{\mu_d}$. For diamond and air, $\sin \theta_c = \frac{1}{2.42} \approx 0.413$. This gives a critical angle of about $24.4^\circ$. Since our incident angle is $30^\circ$, it exceeds the critical angle.

Q: Does the wavelength (630 nm) matter in this calculation?

The wavelength is given as extra information. While refractive index does vary slightly with wavelength (dispersion), the value $\mu = 2.42$ is provided for this specific laser, so we use it directly. The 630 nm value itself isn't needed for the calculation.

Q: What happens to the light if refraction is not possible?

When refraction is impossible, the light undergoes Total Internal Reflection (TIR). It reflects back into the denser medium (diamond) just as if the boundary were a perfect mirror, with the angle of reflection equaling the angle of incidence ($30^\circ$).

Q: Can Total Internal Reflection happen when light goes from air to diamond?

No, TIR can only occur when light travels from an optically denser medium (higher refractive index) to an optically rarer medium (lower refractive index). When going from air to diamond, the light will always bend towards the normal and refract.

Question

A ray of laser of a wavelength $630	ext{ nm}$ is incident at an angle of $30^\circ$ at the diamond-air interface. It is going from diamond to air. The refractive index of diamond is $2.42$ and that of air is $1$. Choose the correct option.

Options:

(A) Angle of refraction is $24.41^\circ$

(B) Angle of refraction is $30^\circ$

(C) Refraction is not possible

(D) Angle of refraction is $53.4^\circ$

Sigma Percentile · Accepted Answer

### The Setup: Light at the Boundary

Imagine you are a photon of light, part of a brilliant laser beam, traveling through the dense, crystalline lattice of a diamond. You approach the boundary where the diamond ends and the open air begins. You strike this interface at an angle of $30^\circ$ with respect to the normal.

The question is: do you break free into the air, or are you trapped inside? To answer this, we must look at the properties of the two media. The refractive index of diamond is a whopping $\mu_d = 2.42$, making it highly optically dense. Air, on the other hand, is optically rare with a refractive index of $\mu_a = 1$.

### The Master Equation: Snell's Law

To predict the path of our laser beam, we call upon the fundamental law of refraction: **Snell's Law**. This elegant equation balances the refractive indices and the angles of the light ray on both sides of the boundary.

$$ \mu_d \sin i = \mu_a \sin r $$

Here, $i$ is our angle of incidence ($30^\circ$), and $r$ is the elusive angle of refraction we wish to find.

### The Mathematical Impossibility

Let's substitute our known values into Snell's Law and see what the math reveals.

$$ 2.42 	imes \sin 30^\circ = 1 	imes \sin r $$

We know that the sine of $30^\circ$ is exactly $0.5$. Substituting this in, we get:

$$ 2.42 	imes 0.5 = \sin r $$

$$ \sin r = 1.21 $$

Stop right there! Look closely at that result. We have arrived at $\sin r = 1.21$. But from basic trigonometry, we know that the sine function is strictly bounded between $-1$ and $1$. It is mathematically impossible for the sine of any real angle to be greater than $1$.

### The Physical Reality: Total Internal Reflection

When the math breaks down like this, it is nature's way of telling us that the physical phenomenon we assumed—refraction—is not happening.

Because the calculated $\sin r$ exceeds $1$, it means our angle of incidence ($30^\circ$) is actually greater than the **critical angle** for the diamond-air interface. The light ray cannot escape into the air. Instead, it is perfectly reflected back into the diamond, obeying the law of reflection. This phenomenon is known as **Total Internal Reflection (TIR)**.

This is no accident; it is the very reason diamonds are cut the way they are. Jewelers design the facets so that light entering the diamond hits the internal surfaces at angles greater than the critical angle, causing the light to bounce around inside before finally exiting at the top. This is what gives diamonds their signature, mesmerizing sparkle!

Animated Solution for Physics - Optics: A ray of laser of a wavelength $630 nm$ is incident at an angle of $3 0^{\circ}$ at the diamond-air interface. It is going from diamond to air. The refractive index of diamond is $2.42$ and that of air is $1$ . Choose the correct option.

Select Answer:

Visualized Solution

$Visualizing the Interface$

$Snell’s Law$

$Substituting Values$

$Calculating sin r$

$The Mathematical Impossibility$

$Conclusion$

The Sigma Insight: Refraction and Total Internal Reflection

The Setup

The Master Equation

The Mathematical Impossibility

The Physical Reality

Similar Questions

A ray $O P$ of monochromatic light is incident on the face $A B$ of prism $A B C D$ near vertex $B$ at an incident angle of $6 0^{\circ}$ (see figure). If the refractive index of the material of the prism is $3$ , which of the following is (are) correct?

A ray of light entering from air into a denser medium of refractive index $\frac{4}{3}$ , as shown in figure. The light ray suffers total internal reflection at the adjacent surface as shown. The maximum value of angle $θ$ should be equal to

A ray of light $A O$ in vacuum is incident on a glass slab at angle $6 0^{\circ}$ and refracted at angle $3 0^{\circ}$ along $O B$ as shown in the figure. The optical path length of light ray from $A$ to $B$ is

A transparent solid cylinder rod has a refractive index of $\frac{2}{3}$ . It is surrounded by air. A light ray is incident at the mid-point of one end of the rod as shown in the figure. The incident angle $θ$ for which the light ray grazes along the wall of the rod is

A light ray is incident perpendicular to one face of a $9 0^{\circ}$ prism and is totally internally reflected at the glass-air interface. If the angle of reflection is $4 5^{\circ}$ , we conclude that for the refractive index $n$ as

AB and CD are two slabs. The medium between the slabs has refractive index 2. Find the minimum angle of incidence at Q, so that the ray is totally reflected by both the slabs.