An object is at a distance of `d=2.5 cm` from the surface of a glass sphere with a radius `R=10 cm.` Find the position of the final image produced by the sphere. The refractive index of glass is `mu_ = 1.5.`

Explanation

To find the position of the final image produced by the glass sphere, we will use the formula for refraction at a spherical surface. The formula is given as:

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

Where:
- \( \mu_1 \) is the refractive index of the medium from which light is coming (air, in this case, \( \mu_1 = 1 \)).
- \( \mu_2 \) is the refractive index of the glass (\( \mu_2 = 1.5 \)).
- \( U \) is the object distance (given as \( d = 2.5 \, \text{cm} \)).
- \( V \) is the image distance we want to find.
- \( R \) is the radius of curvature of the glass sphere (given as \( R = 10 \, \text{cm} \)).

### Step 1: Determine the object distance \( U \)
Since the object is located in front of the spherical surface, we take the object distance as negative:
\[ U = -2.5 \, \text{cm} \]

### Step 2: Apply the refraction formula
Using the formula, we can express the relationship as:
\[ \frac{1.5}{V} = \frac{1}{-2.5} = \frac{1.5 - 1}{10} \]

### Step 3: Solve for \( V \)
First, we calculate the right-hand side:
\[ \frac{1.5 - 1}{10} = \frac{0.5}{10} = 0.05 \]
Now, we can set up the equation:
\[ \frac{1.5}{V} = 0.05 \]
Cross-multiplying gives:
\[ 1.5 = 0.05V \]
Now, solving for \( V \):
\[ V = \frac{1.5}{0.05} = 30 \, \text{cm} \]

### Step 4: Determine the position of the image
The image formed by the first surface is at \( 30 \, \text{cm} \) to the right of the surface.

### Step 5: Consider the second surface of the sphere
The distance from the first image to the second surface of the sphere is:
\[ \text{Distance to second surface} = R = 10 \, \text{cm} \]
Thus, the total distance from the object to the second surface is:
\[ \text{Total distance} = 30 \, \text{cm} + 10 \, \text{cm} = 40 \, \text{cm} \]

### Step 6: Apply the refraction formula again for the second surface
Now we need to find the image formed by the second surface. We will again use the refraction formula:
\[ \frac{1}{V_2} = \frac{1.5}{40} = \frac{1.5 - 1}{-10} \]
Calculating the right-hand side:
\[ \frac{0.5}{-10} = -0.05 \]
Thus, we have:
\[ \frac{1.5}{V_2} = -0.05 \]
Cross-multiplying gives:
\[ 1.5 = -0.05V_2 \]
Solving for \( V_2 \):
\[ V_2 = \frac{1.5}{-0.05} = -30 \, \text{cm} \]

### Final Result
The final image is located at \( 30 \, \text{cm} \) to the left of the second surface of the sphere.

### Summary
The position of the final image produced by the sphere is:
\[ \text{Final Image Position} = -30 \, \text{cm} \text{ (to the left of the second surface)} \]