A glass hemisphere of radius 0.04 m and refractive index of the material 1.6 is placed centrally over cross mark on a paper (i) with the flat face, (ii) with the curved face in contact with the paper. In each case, the cross mark is viewed directly from above. The position of the images will be

Options

A. 0.04 m from the flat face, (ii) 0.025 m from the flat face
B. (i) At the same position of the cross mar, (ii)0.025 m below the flat face
C. (i) 0.025 m from the flat face, (ii) 0.04 m from the flat face
D. For both (i) and (ii) 0.025 m from the highest point of the hemisphere

Answer

To solve the question step by step, we will analyze both cases of the glass hemisphere:

### Case (i): Flat face in contact with the paper

1. **Identify the parameters**:
- Radius (R) = 0.04 m
- Refractive index of glass (μ1) = 1.6
- Refractive index of air (μ2) = 1 (since we are viewing from air)

2. **Use the lens maker's formula**:
The formula we will use is:
\[ \frac{\mu_2}{V} - \frac{\mu_1}{U} = \frac{\mu_2 - \mu_1}{R} \]
Here, U is the object distance (from the curved surface to the object), which is equal to -0.04 m (negative because it is measured in the opposite direction).

3. **Substituting values**:
\[ \frac{1}{V} - \frac{1.6}{-0.04} = \frac{1 - 1.6}{0.04} \]
This simplifies to:
\[ \frac{1}{V} + 40 = \frac{-0.6}{0.04} \]
\[ \frac{1}{V} + 40 = -15 \]
\[ \frac{1}{V} = -15 - 40 = -55 \]
\[ V = -\frac{1}{55} \approx -0.01818 \text{ m} \]

4. **Interpret the result**:
The negative sign indicates that the image is formed on the same side as the object, which means the image is at the position of the cross mark (0.04 m from the flat surface).

### Case (ii): Curved face in contact with the paper

1. **Identify the parameters**:
- Radius (R) = 0.04 m
- Refractive index of glass (μ1) = 1.6
- Refractive index of air (μ2) = 1

2. **Use the formula for apparent depth**:
The relationship between real depth (h) and apparent depth (h') is given by:
\[ \mu = \frac{h}{h'} \]
Rearranging gives:
\[ h' = \frac{h}{\mu} \]

3. **Substituting values**:
\[ h' = \frac{0.04}{1.6} = 0.025 \text{ m} \]

4. **Interpret the result**:
The apparent depth is 0.025 m, meaning the image of the cross mark will appear to be at a depth of 0.025 m from the curved surface.

### Summary of Results:
- For the flat face in contact with the paper, the image position is at 0.04 m.
- For the curved face in contact with the paper, the image position is at 0.025 m.