A hemispherical portion of the surface of a solid glass sphere (mu = 1.5) of radius r is silvered to make the inner side reflecting. An object is placed on the axis of the hemisphere at a distance 3r from the centre of the sphere. The light from the object is refracted at the unsilvered part, then reflected from the silvered part and again refracted at the unsilvered part. Locate the final image formed.

Explanation

To solve the problem step by step, we will analyze the refraction and reflection processes occurring in the given optical system.

### Step 1: Understand the Setup
We have a hemispherical glass surface with a radius \( r \) that is silvered on the inner side. An object is placed on the axis of the hemisphere at a distance \( 3r \) from the center of the sphere.

### Step 2: Refraction at the Unsilvered Part
The first step is to find the image formed by the refraction at the unsilvered part of the hemisphere. We will use the lens formula for refraction:

\[ \frac{\mu_2}{V} - \frac{\mu_1}{U} = \frac{\mu_2 - \mu_1}{R} \]

Where:
- \( \mu_2 = 1.5 \) (refractive index of glass)
- \( \mu_1 = 1 \) (refractive index of air)
- \( U = -3r \) (object distance, negative as per sign convention)
- \( R = 2r \) (radius of curvature, positive for the lens)

Substituting the values into the formula gives:

\[ \frac{1.5}{V} - \frac{1}{-3r} = \frac{1.5 - 1}{2r} \]

This simplifies to:

\[ \frac{1.5}{V} + \frac{1}{3r} = \frac{0.5}{2r} \]

Rearranging gives:

\[ \frac{1.5}{V} = \frac{0.5}{2r} - \frac{1}{3r} \]

Finding a common denominator (which is \( 6r \)):

\[ \frac{1.5}{V} = \frac{1.5 - 2}{6r} = \frac{-0.5}{6r} \]

Thus:

\[ \frac{1.5}{V} = -\frac{0.5}{6r} \]

Cross-multiplying gives:

\[ 1.5 \cdot 6r = -0.5V \implies V = -\frac{9r}{1} = -9r \]

### Step 3: Reflection from the Silvered Part
The image formed at \( V = -9r \) acts as the object for the silvered part of the hemisphere. The distance of this image from the center of the sphere is \( 9r \) (considering the sign convention).

Using the mirror formula:

\[ \frac{1}{V} + \frac{1}{U} = \frac{1}{F} \]

Where:
- \( U = -9r \) (the object distance for the mirror, negative)
- \( F = -\frac{r}{2} \) (focal length of the mirror, negative)

Substituting the values gives:

\[ \frac{1}{V} - \frac{1}{9r} = -\frac{2}{r} \]

Rearranging gives:

\[ \frac{1}{V} = -\frac{2}{r} + \frac{1}{9r} \]

Finding a common denominator (which is \( 9r \)):

\[ \frac{1}{V} = -\frac{18}{9r} + \frac{1}{9r} = -\frac{17}{9r} \]

Thus:

\[ V = -\frac{9r}{17} \]

### Step 4: Refraction at the Unsilvered Part Again
Now, this image acts as the object for the second refraction at the unsilvered part of the hemisphere. The object distance for this refraction is:

\[ U = -\left( r + \frac{9r}{17} \right) = -\left( \frac{17r + 9r}{17} \right) = -\frac{26r}{17} \]

Using the lens formula again:

\[ \frac{\mu_2}{V} - \frac{\mu_1}{U} = \frac{\mu_2 - \mu_1}{R} \]

Substituting the values gives:

\[ \frac{1}{V} - \frac{1.5}{-\frac{26r}{17}} = \frac{1 - 1.5}{2r} \]

This simplifies to:

\[ \frac{1}{V} + \frac{1.5 \cdot 17}{26r} = -\frac{0.5}{2r} \]

Finding a common denominator (which is \( 52r \)) gives:

\[ \frac{1}{V} + \frac{25.5}{52r} = -\frac{13}{52r} \]

Thus:

\[ \frac{1}{V} = -\frac{13 + 25.5}{52r} = -\frac{38.5}{52r} \]

Finally, we find:

\[ V = -\frac{52r}{38.5} \approx -\frac{104r}{77} \]

### Final Result
The final image is formed at approximately \( -2r \) from the center of the sphere.