A light ray is incident on the surface of a sphere of refractive index n at an angle of incidence th

Q: Why is the deviation at the reflection $180^\circ - 2\phi_0$?

When a ray reflects, it changes direction. If it were to go straight, the deviation would be zero. Since it reflects at an angle $\phi_0$ with the normal, the angle between the incident and reflected ray is $2\phi_0$. The deviation from its original straight path is $180^\circ - 2\phi_0$.

Q: Why is the angle of incidence at the second surface also $\phi_0$?

This is due to the geometry of the sphere. The triangle formed by the center of the sphere and the two points of intersection (refraction and reflection) is isosceles because two of its sides are radii of the sphere. Thus, the base angles are equal.

Q: Can $\theta_0$ and $\phi_0$ really be zero?

Yes! If $\theta_0 = 0^\circ$, the ray strikes the sphere normally (perpendicular to the surface). It will pass straight through the center, reflect straight back, and emerge without any deviation. In this case, the total deviation is exactly $180^\circ$.

Q: Why do we add the deviations? Can they cancel out?

We add them because at each step (refraction, reflection, refraction), the ray bends in the same rotational direction (clockwise in our diagram) relative to its immediate previous path. Since all bends are in the same sense, their magnitudes add up.

Q: What happens if the refractive index is very large?

If $n$ is very large, the ray bends strongly towards the normal, making $\phi_0$ very small. The total deviation would then be dominated by the reflection and the initial $\theta_0$, approaching $180^\circ + 2\theta_0$.

Question

A light ray is incident on the surface of a sphere of refractive index $n$ at an angle of incidence $	heta_0$. The ray partially refracts into the sphere with angle of refraction $\phi_0$ and then partly reflects from the back surface. The reflected ray then emerges out of the sphere after a partial refraction. The total angle of deviation of the emergent ray with respect to the incident ray is $\alpha$. Match the quantities mentioned in List-I with their values in List-II and choose the correct option.

List-I:

P. If $n = 2$ and $\alpha = 180^\circ$, then all the possible values of $	heta_0$ will be

Q. If $n = \sqrt{3}$ and $\alpha = 180^\circ$, then all the possible values of $	heta_0$ will be

R. If $n = \sqrt{3}$ and $\alpha = 180^\circ$, then all the possible values of $\phi_0$ will be

S. If $n = \sqrt{2}$ and $	heta_0 = 45^\circ$, then all the possible values of $\alpha$ will be

List-II:

1. $30^\circ$ and $0^\circ$

2. $60^\circ$ and $0^\circ$

3. $45^\circ$ and $0^\circ$

4. $150^\circ$

5. $0^\circ$

Sigma Percentile · Accepted Answer

### The Journey of a Light Ray: Refraction, Reflection, and Deviation

Have you ever wondered how a simple drop of water can paint a magnificent rainbow across the sky? The secret lies in a beautiful sequence of optical events: refraction, internal reflection, and a second refraction. This problem takes us on that exact journey through a glass sphere, challenging us to track the bending of light at every step.

### The Geometry of the Path

Let's trace the journey of this light ray. It enters the sphere at an angle of incidence $	heta_0$ and refracts at an angle $\phi_0$. At this first surface, Snell's law gives us our foundational equation:

$$\sin	heta_0 = n \sin\phi_0$$

Because the sphere is perfectly symmetric, the geometry inside is incredibly elegant. The triangle formed by the center of the sphere and the two points where the ray hits the boundary is an isosceles triangle (since two sides are just the radius of the sphere). This guarantees that the angle of incidence at the back surface is exactly $\phi_0$. The ray reflects, maintaining this angle, and travels to the third point. By the same geometric logic, it strikes the final boundary at an angle $\phi_0$ and emerges at an angle $	heta_0$.

### Calculating the Total Deviation

Now, how much does the ray bend in total? We need to sum up the deviations at each of the three interactions.

First, as it enters the sphere, it bends towards the normal by an angle:
$$\delta_1 = 	heta_0 - \phi_0$$

Next, it hits the back wall and reflects. If it had gone straight, it would have zero deviation. Instead, it bounces back, creating a deviation of:
$$\delta_2 = 180^\circ - 2\phi_0$$

Finally, as it exits the sphere, it bends away from the normal by:
$$\delta_3 = 	heta_0 - \phi_0$$

Since all these bends happen in the same rotational sense (clockwise in our setup), we simply add them up to find the total deviation $\alpha$:
$$\alpha = (	heta_0 - \phi_0) + (180^\circ - 2\phi_0) + (	heta_0 - \phi_0) = 180^\circ + 2	heta_0 - 4\phi_0$$

### Solving the Matrix Match

With our master equations ready, let's conquer the four cases.

**Case (P):** We are given $n = 2$ and $\alpha = 180^\circ$. 
Plugging $\alpha$ into our deviation equation:
$$180^\circ = 180^\circ + 2	heta_0 - 4\phi_0 \implies 	heta_0 = 2\phi_0$$

Substitute this into Snell's law:
$$\sin(2\phi_0) = 2\sin\phi_0$$

Expanding the double angle:
$$2\sin\phi_0\cos\phi_0 = 2\sin\phi_0$$

This tells us that $\cos\phi_0 = 1$, which means $\phi_0 = 0^\circ$. Consequently, $	heta_0 = 0^\circ$. This maps (P) to (5).

**Case (Q):** Here, $n = \sqrt{3}$, but $\alpha$ is still $180^\circ$. 
So, $	heta_0 = 2\phi_0$ remains true. Our Snell's law equation becomes:
$$\sin(2\phi_0) = \sqrt{3}\sin\phi_0$$
$$2\sin\phi_0\cos\phi_0 = \sqrt{3}\sin\phi_0$$

This gives us two possibilities. Either $\sin\phi_0 = 0$ (meaning $	heta_0 = 0^\circ$), or $\cos\phi_0 = \frac{\sqrt{3}}{2}$. The latter means $\phi_0 = 30^\circ$, making $	heta_0 = 60^\circ$. So, the possible values for $	heta_0$ are $0^\circ$ and $60^\circ$. This maps (Q) to (2).

**Case (R):** The conditions are exactly the same as Case (Q), but we need the values of $\phi_0$ instead. We just calculated these! The possible values for the angle of refraction, $\phi_0$, are $0^\circ$ and $30^\circ$. This maps (R) to (1).

**Case (S):** Finally, let's look at the last case. Here, $n = \sqrt{2}$ and $	heta_0 = 45^\circ$. 
Plugging this into Snell's law:
$$\sin 45^\circ = \sqrt{2}\sin\phi_0$$
$$\frac{1}{\sqrt{2}} = \sqrt{2}\sin\phi_0 \implies \sin\phi_0 = \frac{1}{2}$$

This means $\phi_0 = 30^\circ$. Now, substitute both angles into our total deviation formula:
$$\alpha = 180^\circ + 2(45^\circ) - 4(30^\circ) = 180^\circ + 90^\circ - 120^\circ = 150^\circ$$

This maps (S) to (4). And just like that, we've decoded the entire matrix!

List-I

List-II

Select Matching Pairs:

Visualized Solution

$Snell’s Law at First Interface$

$Total Angle of Deviation$

$Evaluating Case (P)$

$Evaluating Case (Q)$

$Evaluating Case (R)$

$Evaluating Case (S)$

The Sigma Insight: Refraction at Spherical Surface

The Journey of a Light Ray

The Geometry of the Path

Calculating the Total Deviation

Solving the Matrix Match

Similar Questions

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List-I

List-II

Select Matching Pairs:

Visualized Solution

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

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

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Evaluating Case (P)

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Evaluating Case (Q)

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Evaluating Case (R)

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Evaluating Case (S)