A ray of light is incident normally on one face of a right-angled isosceles prism and then it grazes the hypotenuse. The refractive index of the material of the prism is

Options

A. 1.33
B. 1.414
C. 1.5
D. 1.732

Answer

To solve the problem of finding the refractive index of a right-angled isosceles prism when a ray of light is incident normally on one face and grazes the hypotenuse, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Geometry of the Prism**:
- The prism is a right-angled isosceles prism, which means it has one right angle (90 degrees) and the other two angles are equal (45 degrees each).
- Let’s denote the angles: angle A = 90°, angle B = 45°, angle C = 45°.

2. **Incident Ray**:
- A ray of light is incident normally on one of the faces of the prism. Since it is incident normally, the angle of incidence (i) at this face is 0°.
- Therefore, the ray passes straight through without bending.

3. **Refraction at the Hypotenuse**:
- The ray then reaches the hypotenuse of the prism. At this point, it will refract according to Snell's Law.
- Since the ray grazes the hypotenuse, it means that the angle of incidence at the hypotenuse is equal to 90° (the ray is at the critical angle).

4. **Applying Snell's Law**:
- Snell's Law states:
\[ n_1 \sin(i) = n_2 \sin(r) \]
- Here, \( n_1 \) is the refractive index of the prism (let's denote it as \( \mu \)), \( n_2 \) is the refractive index of air (which is approximately 1), \( i \) is the angle of incidence (which we found to be 45°), and \( r \) is the angle of refraction (which is 90° since the ray grazes the hypotenuse).

5. **Substituting Values**:
- From Snell's Law, we have:
\[ \mu \sin(45°) = 1 \sin(90°) \]
- We know that \( \sin(90°) = 1 \) and \( \sin(45°) = \frac{1}{\sqrt{2}} \).

6. **Solving for the Refractive Index**:
- Substituting these values into the equation gives:
\[ \mu \cdot \frac{1}{\sqrt{2}} = 1 \cdot 1 \]
- Rearranging this gives:
\[ \mu = \sqrt{2} \]

7. **Final Answer**:
- The refractive index of the material of the prism is \( \sqrt{2} \) or approximately 1.414.