One end of a cylindrical glass rod (mu = 1.5) of radius 1.0 cm is rounded in the shape of a hemisphere. The rod is immersed in water (mu= 4/3) and an object is placed in the water along the axis of the rod at a distance of 8.0 cm from the rounded edge. Locate the image of the object.

Explanation

To solve the problem, we will use the lens maker's formula for refraction at a spherical surface. The formula we will use is:

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

Where:
- \( \mu_1 \) = refractive index of the medium where the object is located (water)
- \( \mu_2 \) = refractive index of the medium where the image is formed (glass)
- \( u \) = object distance (negative in this case since it is measured against the direction of incident light)
- \( v \) = image distance (to be determined)
- \( R \) = radius of curvature of the spherical surface (positive since it is convex)

### Step 1: Identify the given values
- \( \mu_1 = \frac{4}{3} \) (refractive index of water)
- \( \mu_2 = 1.5 \) (refractive index of glass)
- \( R = 1.0 \, \text{cm} \) (radius of the hemisphere)
- \( u = -8.0 \, \text{cm} \) (object distance)

### Step 2: Substitute the values into the formula
Using the formula, we substitute the known values:

\[ \frac{1.5}{v} - \frac{4/3}{-8} = \frac{1.5 - \frac{4}{3}}{1} \]

### Step 3: Simplify the right-hand side
First, calculate \( 1.5 - \frac{4}{3} \):

\[ 1.5 = \frac{3}{2} = \frac{9}{6}, \quad \text{and} \quad \frac{4}{3} = \frac{8}{6} \]
\[ 1.5 - \frac{4}{3} = \frac{9}{6} - \frac{8}{6} = \frac{1}{6} \]

Thus, the right-hand side becomes:

\[ \frac{1}{6} \]

### Step 4: Rewrite the equation
Now we rewrite the equation:

\[ \frac{1.5}{v} + \frac{4/3}{8} = \frac{1}{6} \]

Calculating \( \frac{4/3}{8} \):

\[ \frac{4/3}{8} = \frac{4}{24} = \frac{1}{6} \]

### Step 5: Substitute back into the equation
Now the equation becomes:

\[ \frac{1.5}{v} + \frac{1}{6} = \frac{1}{6} \]

### Step 6: Isolate \( \frac{1.5}{v} \)
Subtract \( \frac{1}{6} \) from both sides:

\[ \frac{1.5}{v} = 0 \]

### Step 7: Solve for \( v \)
This implies:

\[ v \rightarrow \infty \]

### Conclusion
The image of the object will be formed at infinity.

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