When a picture drawn on paper is seen through a slab of a transparent material of thickness 5 cm, it appears to be raised by 1.5 cm. The critical angle at the boundary of this transparent material and air is -

Options

A. `sin^(-1) (2/3)`
B. `sin^(-1) (5/7)`
C. `sin^(-1) (6/11)`
D. `sin^(-1) (7/10)`

Answer

To solve the problem, we need to find the critical angle at the boundary of a transparent material and air, given that a picture appears raised by a certain amount when viewed through a slab of this material.

### Step-by-Step Solution:

1. **Identify Given Values**:
- Thickness of the slab (T) = 5 cm
- Apparent shift (D) = 1.5 cm

2. **Use the Formula for Apparent Shift**:
The formula for the apparent shift when viewing through a slab is given by:
\[ D = T \left(1 - \frac{1}{\mu}\right) \]
where \(\mu\) is the refractive index of the material.

3. **Substituting the Known Values**:
Substitute the values of D and T into the formula:
\[ 1.5 = 5 \left(1 - \frac{1}{\mu}\right) \]

4. **Rearranging the Equation**:
Divide both sides by 5:
\[ \frac{1.5}{5} = 1 - \frac{1}{\mu} \]
Simplifying the left side:
\[ 0.3 = 1 - \frac{1}{\mu} \]

5. **Solving for \(\frac{1}{\mu}\)**:
Rearranging gives:
\[ \frac{1}{\mu} = 1 - 0.3 = 0.7 \]

6. **Finding \(\mu\)**:
Taking the reciprocal:
\[ \mu = \frac{1}{0.7} = \frac{10}{7} \]

7. **Using the Critical Angle Formula**:
The critical angle (\(I_c\)) can be found using Snell's law:
\[ \sin I_c = \frac{n_2}{n_1} \]
where \(n_1\) is the refractive index of the denser medium (the transparent material) and \(n_2\) is the refractive index of air (which is 1). Thus:
\[ \sin I_c = \frac{1}{\mu} \]

8. **Substituting \(\mu\)**:
Substitute \(\mu = \frac{10}{7}\):
\[ \sin I_c = \frac{1}{\frac{10}{7}} = \frac{7}{10} \]

9. **Finding the Critical Angle**:
Finally, calculate the critical angle:
\[ I_c = \sin^{-1}\left(\frac{7}{10}\right) \]

### Final Answer:
The critical angle at the boundary of this transparent material and air is \(I_c = \sin^{-1}\left(\frac{7}{10}\right)\).