A ray falls on a prism `ABC(AB=BC)` and travels as shown in adjoining figure.The minimum refraction index of the prism material should be -

Options

A. 43558
B. `sqrt2`
C. 1.5
D. `sqrt3`

Explanation

To find the minimum refractive index of the prism material given that the ray falls on the prism ABC (where AB = BC), we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Geometry of the Prism**:
- The prism ABC has equal sides AB and BC, which implies that it is an isosceles prism.
- The angles at A and C are equal, and we denote the angle at B as \( \angle ABC = \theta \).

2. **Identifying the Angles**:
- Since AB = BC, the angles at A and C are equal, and we can denote them as \( \angle A = \angle C = \frac{180^\circ - \theta}{2} \).
- For the minimum refractive index condition, we consider the critical angle for total internal reflection at point D.

3. **Applying Snell's Law**:
- At point D, where the ray undergoes total internal reflection, we can use Snell's law:
\[ n_1 \sin(\theta_i) = n_2 \sin(\theta_r) \]
- Here, \( n_1 \) is the refractive index of the prism, \( n_2 \) is the refractive index of air (approximately 1), \( \theta_i \) is the angle of incidence, and \( \theta_r \) is the angle of refraction.

4. **Critical Angle**:
- The critical angle \( \theta_c \) for total internal reflection is given by:
\[ \theta_c = \sin^{-1}\left(\frac{n_2}{n_1}\right) \]
- For total internal reflection to occur, the angle of incidence must be greater than the critical angle.

5. **Finding the Minimum Refractive Index**:
- For the minimum refractive index, we set the critical angle \( \theta_c = 45^\circ \):
\[ \sin(45^\circ) = \frac{1}{\sqrt{2}} \]
- Substituting into the critical angle equation:
\[ \frac{1}{\sqrt{2}} = \frac{1}{n_1} \]
- Rearranging gives us:
\[ n_1 = \sqrt{2} \]

6. **Conclusion**:
- Therefore, the minimum refractive index of the prism material should be \( \sqrt{2} \).

### Final Answer:
The minimum refractive index of the prism material should be \( \sqrt{2} \).