A point object is placed at the centre of a glass sphere of radius 6 cm and refractive index 1.5. Th

Q: Why is the angle of incidence zero?

The angle of incidence is measured between the incident ray and the normal to the surface. For a sphere, any line drawn from the center to the surface is the normal. Since the rays originate from the center, they travel along the normal, making the angle of incidence zero.

Q: Doesn't the refractive index of 1.5 affect the image position?

In this specific case, no. Because the rays strike the surface normally, they do not bend, regardless of the refractive index. The refractive index would only matter if the rays struck the surface at an angle.

Q: Is the image real or virtual?

The image is virtual. The rays diverge from the center and continue to diverge after leaving the sphere. They do not actually meet anywhere outside; they only appear to meet at the center when extended backwards.

Q: What if the object was slightly off-center?

If the object was not at the center, the rays would strike the surface at various angles of incidence. You would then need to use the spherical refraction formula $\frac{\mu_2}{v} - \frac{\mu_1}{u} = \frac{\mu_2 - \mu_1}{R}$ to find the image position.

Q: Why is the answer 6 cm and not 0 cm?

The question asks for the distance of the virtual image *from the surface* of the sphere, not from the center. Since the image is at the center, its distance from the surface is exactly the radius of the sphere, which is 6 cm.

Question

A point object is placed at the centre of a glass sphere of radius $6	ext{ cm}$ and refractive index $1.5$. The distance of the virtual image from the surface of the sphere is

Options:

(A) $2	ext{ cm}$

(B) $4	ext{ cm}$

(C) $6	ext{ cm}$

(D) $12	ext{ cm}$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are holding a solid glass sphere. Right at the very center of this sphere, there is a tiny point object. The sphere has a radius of $6	ext{ cm}$ and is made of glass with a refractive index of $1.5$. Our goal is to find out where the image of this object is formed when viewed from the outside, and specifically, how far this image is from the surface of the sphere.

At first glance, you might think we need to dive straight into the complex spherical refraction formula, $\frac{\mu_2}{v} - \frac{\mu_1}{u} = \frac{\mu_2 - \mu_1}{R}$. But let's take a breath and look at the geometry of the situation. Sometimes, physics is beautifully simple if we just visualize the rays.

### The Magic of Normal Incidence

Let's trace the path of the light rays originating from the point object at the center. These rays travel radially outward towards the surface of the sphere. Now, what is the angle at which they strike the surface?

In geometry, any line drawn from the center of a circle or sphere to its boundary is a radius, and the radius is always perpendicular (or normal) to the surface at that point. This means that every single ray originating from the center strikes the surface at an angle of incidence of exactly $0^\circ$.

According to **Snell's Law**, $\mu_1 \sin i = \mu_2 \sin r$. If the angle of incidence $i = 0^\circ$, then $\sin i = 0$. This forces $\sin r$ to also be zero, which means the angle of refraction $r = 0^\circ$. The light rays do not bend at all! They emerge from the glass sphere completely undeviated, continuing straight along their radial paths.

### Final Calculation

Because the emergent rays travel in straight lines directly away from the center, an observer looking at the sphere from the outside will trace these diverging rays backwards. Where do they appear to intersect? Exactly where they started: at the center of the sphere!

Therefore, the virtual image is formed at the exact same location as the object—at the center.

The question asks for the distance of this virtual image from the **surface** of the sphere. Since the image is at the center, its distance to the surface is simply the radius of the sphere.

**Distance from surface** $= R = 6	ext{ cm}$.

It's a wonderful conceptual trap! The refractive index of $1.5$ was just extra information designed to tempt you into a long calculation. By understanding the physical reality of normal incidence, we arrived at the answer instantly.

Animated Solution for Physics - Optics: A point object is placed at the centre of a glass sphere of radius $6 cm$ and refractive index $1.5$ . The distance of the virtual image from the surface of the sphere is

Select Answer:

Visualized Solution

\text{Visual Anchor}

\text{Rays from Center}

\text{Normal Incidence}

\text{Image Formation}

\text{Final Answer}

\text{The Way Forward}

The Sigma Insight: Refraction at Spherical Surface

Analyzing the Setup

The Magic of Normal Incidence

Final Calculation

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