A ray of light is incident on the hypotenuse of a right-angled prism after travelling parallel to the base inside the prism. If `mu` is the refractive index of the material of the prism, the maximum value of the base angle for which light is totally reflected from the hypotenuse is

Options

A. `sin^(-1) ((I)/(mu))`
B. `tan^(-1) ((1)/(mu))`
C. `sin^(-1) ((mu-1)/(mu))`
D. `cos^(-1)((1)/(mu))`

Answer

To solve the problem, we need to determine the maximum base angle (α) of a right-angled prism for which light is totally reflected from the hypotenuse when the light ray travels parallel to the base of the prism.

### Step-by-Step Solution:

1. **Understanding the Geometry of the Prism:**
- We have a right-angled prism where one angle is 90 degrees (the right angle).
- Let the base angle be α. Therefore, the other angle (the angle at the hypotenuse) will be (90 - α).

2. **Ray of Light Incident on the Hypotenuse:**
- A ray of light travels parallel to the base of the prism and strikes the hypotenuse.
- When the light ray strikes the hypotenuse, we need to analyze the angles involved.

3. **Using Snell's Law:**
- According to Snell's law, at the boundary between two media, we have:
\[ \mu_1 \sin(\theta_1) = \mu_2 \sin(\theta_2) \]
- Here, µ is the refractive index of the prism, and we can assume the refractive index of air (µ1) is 1.
- The angle of incidence (θ1) at the hypotenuse will be (90 - α) because the light is traveling parallel to the base.

4. **Setting Up the Equation:**
- At the hypotenuse, we can write:
\[ 1 \cdot \sin(90^\circ) = \mu \cdot \sin(90 - \alpha) \]
- Since sin(90°) = 1, we have:
\[ 1 = \mu \cdot \cos(\alpha) \]

5. **Solving for α:**
- Rearranging the equation gives:
\[ \cos(\alpha) = \frac{1}{\mu} \]
- To find α, we take the inverse cosine:
\[ \alpha = \cos^{-1}\left(\frac{1}{\mu}\right) \]

6. **Conclusion:**
- The maximum value of the base angle α for which light is totally reflected from the hypotenuse is:
\[ \alpha = \cos^{-1}\left(\frac{1}{\mu}\right) \]

### Final Answer:
The maximum value of the base angle for which light is totally reflected from the hypotenuse is:
\[ \alpha = \cos^{-1}\left(\frac{1}{\mu}\right) \]