A parallel beam of light travelling in water (refractive index = 4/3) is refracted by a spherical ai

Q: Why does the convex surface of the air bubble diverge the light rays?

The diverging or converging nature of a lens depends on the relative refractive indices. Since the air bubble ($\mu=1$) is optically rarer than the surrounding water ($\mu=4/3$), a convex surface bends the light rays away from the normal, causing them to diverge.

Q: How do we determine the sign of the radius of curvature $R$?

According to the Cartesian sign convention, the direction of incident light is taken as positive. For the first surface, the center of curvature lies in the direction of light, so $R$ is positive. For the second surface, the center lies against the direction of light, so $R$ is negative.

Q: Why is the object distance for the second surface $-10$ mm and not $-6$ mm?

The first image is formed $6$ mm to the left of the first surface. The second surface is $4$ mm (the diameter of the bubble) to the right of the first surface. Therefore, the total distance from the second surface to the first image is $6 + 4 = 10$ mm. Since it's to the left, it's $-10$ mm.

Q: What does a negative image distance ($v$) signify?

A negative image distance means the image is formed on the same side as the incident light (to the left of the refracting surface). This indicates that the image is virtual and cannot be projected onto a screen.

Q: Can we use the thin lens formula for this problem?

No, the thin lens formula assumes the thickness of the lens is negligible. Here, the bubble has a diameter of $4$ mm, which is significant compared to the image distances. We must treat it as two separate thick spherical refracting surfaces.

Question

A parallel beam of light travelling in water (refractive index = 4/3) is refracted by a spherical air bubble of radius 2 mm situated in water. Assuming the light rays to be paraxial.
(a) Find the position of the image due to refraction at the first surface and the position of the final image.
(b) Draw a ray diagram showing the positions of both the images.

Sigma Percentile · Accepted Answer

Imagine a tiny spherical air bubble trapped underwater. A parallel beam of light travels through the water and strikes this bubble. We need to track the journey of these rays as they refract twice—first when entering the bubble, and then when exiting it.

### The Master Equation for Refraction

To find where the images are formed, we will use the master equation for refraction at a curved surface. It connects the object distance $u$, image distance $v$, and the radius of curvature $R$, along with the refractive indices of the two media:

$$\frac{\mu_2}{v} - \frac{\mu_1}{u} = \frac{\mu_2 - \mu_1}{R}$$

### First Refraction at Surface P

Let's focus on the first surface, point $P$. The light is coming from infinity, so $u$ is $-\infty$. It travels from water, with a refractive index of $\frac{4}{3}$, into air, which has an index of $1$. The center of curvature is to the right, so $R$ is $+2 	ext{ mm}$.

Plugging these values into our formula, the term with infinity becomes zero. We are left with:

$$\frac{1}{v_1} - \frac{4/3}{-\infty} = \frac{1 - 4/3}{2}$$

$$\frac{1}{v_1} = \frac{-1/3}{2} = -\frac{1}{6}$$

So, $v_1 = -6 	ext{ mm}$. This means the first surface forms a virtual image, $I_1$, $6 	ext{ mm}$ to the left of $P$.

### Second Refraction at Surface Q

Now, the light travels through the air bubble and hits the second surface at $Q$. The virtual image $I_1$ acts as the object for this surface. Since the bubble's diameter is $4 	ext{ mm}$, the distance of $I_1$ from $Q$ is $6 + 4$, which is $10 	ext{ mm}$ to the left. So, $u_2$ is $-10 	ext{ mm}$. The light is now going from air back into water, and the center of curvature is to the left, making $R$ $-2 	ext{ mm}$.

Let's substitute these new values into our refraction formula:

$$\frac{4/3}{v_2} - \frac{1}{-10} = \frac{4/3 - 1}{-2}$$

$$\frac{4}{3v_2} + \frac{1}{10} = \frac{1/3}{-2} = -\frac{1}{6}$$

Moving $\frac{1}{10}$ to the other side and simplifying, we find that:

$$\frac{4}{3v_2} = -\frac{1}{6} - \frac{1}{10} = -\frac{4}{15}$$

Solving this gives us $v_2 = -5 	ext{ mm}$.

### Conclusion

And there we have it! The final image is formed $5 	ext{ mm}$ to the left of the second surface. Notice how the parallel rays diverge after passing through the bubble, making it act like a diverging lens in water. **The first image is at $-6 	ext{ mm}$ from the first surface, and the final image is at $-5 	ext{ mm}$ from the second surface.**

Visualized Solution

\text{Understanding the Setup}

\text{Refraction at a Spherical Surface}

\text{First Refraction at Surface } P

\text{Calculating } v_1

\text{Second Refraction at Surface } Q

\text{Calculating } v_2

\text{Final Image Position}

The Sigma Insight: Refraction at Spherical Surface

The Master Equation for Refraction

First Refraction at Surface P

Second Refraction at Surface Q

Conclusion

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