A clear transparent glass sphere `(mu=1.5)` of radius R is immersed in a liquid of refractive index 1.25. A parallel beam of light incident on it will converge to a point. The distance of this point from the center will be

Options

A. `-3R`
B. `+3R`
C. `-R`
D. `+R`

Explanation

To solve the problem, we will use the lens maker's formula for a spherical interface between two media. The formula is given by:

\[ \frac{\mu_1}{v} - \frac{\mu_2}{u} = \frac{\mu_2 - \mu_1}{R} \]

Where:
- \(\mu_1\) is the refractive index of the medium from which light is coming (in this case, the liquid, \(\mu_1 = 1.25\))
- \(\mu_2\) is the refractive index of the glass sphere (\(\mu_2 = 1.5\))
- \(R\) is the radius of curvature of the spherical surface (for a sphere, \(R = R\))
- \(u\) is the object distance (for parallel rays, \(u = -\infty\))
- \(v\) is the image distance from the center of the sphere

### Step 1: Substitute the known values into the lens maker's formula

Given that the rays are parallel, we can set \(u = -\infty\). Therefore, the lens maker's formula simplifies to:

\[ \frac{1.25}{v} - \frac{1.5}{-\infty} = \frac{1.5 - 1.25}{R} \]

Since \(\frac{1.5}{-\infty} = 0\), the equation reduces to:

\[ \frac{1.25}{v} = \frac{0.25}{R} \]

### Step 2: Solve for \(v\)

Rearranging the equation gives:

\[ v = \frac{1.25 R}{0.25} = 5R \]

### Step 3: Consider the distance from the center of the sphere

Since \(v = 5R\) is the distance from the center of the sphere to the image point, we need to find the distance from the center of the sphere to the point where the light converges.

### Step 4: Calculate the total distance

The total distance from the center of the sphere to the point where the light converges is \(5R\).

### Final Answer

Thus, the distance of the point from the center of the sphere is:

\[ \text{Distance from center} = 5R \]

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