A ray of light is incident on a glass sphere of refractive index 3/2. What should be the angle of incidence so that the ray which enters the sphere does not come out of the sphere ?

Options

A. `tan^(-1)(2//3)`
B. `sin^(-1)(2//3)`
C. `90^(@)`
D. `cos^(-1) (1//3)`

Answer

To solve the problem, we need to determine the angle of incidence (I) such that a ray of light entering a glass sphere with a refractive index (n) of 3/2 does not emerge from the sphere. This involves understanding the concept of total internal reflection (TIR).

### Step-by-Step Solution:

1. **Understanding the Critical Angle**:
The critical angle (θ_c) is the angle of incidence above which total internal reflection occurs. It can be calculated using the formula:
\[ \sin(\theta_c) = \frac{1}{n} \]
where \( n \) is the refractive index of the material.

2. **Calculate the Critical Angle**:
Given that the refractive index \( n = \frac{3}{2} \):
\[ \sin(\theta_c) = \frac{1}{\frac{3}{2}} = \frac{2}{3} \]
Now, we can find \( \theta_c \) using the inverse sine function:
\[ \theta_c = \sin^{-1}\left(\frac{2}{3}\right) \]

3. **Condition for Total Internal Reflection**:
For the ray to not emerge from the sphere, the angle of incidence (I) at the point of incidence on the sphere must be greater than the critical angle:
\[ I > \theta_c \]

4. **Finding the Angle of Incidence**:
Since the maximum angle of incidence that can be used to ensure TIR is when \( I = 90^\circ \), we can conclude that if the angle of incidence is \( 90^\circ \), the ray will not exit the sphere.

5. **Conclusion**:
Therefore, the angle of incidence (I) should be \( 90^\circ \) to ensure that the ray does not come out of the sphere.

### Final Answer:
The angle of incidence should be \( 90^\circ \).

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