Light enters at an angle of incidence in a transparent rod of refractive index n. For what value of the refractive index of the material of the rod the light once entered into it will not leave it through its lateral face whatsoever be the value of angle of incidence.

Options

A. `n gt sqrt(2)`
B. n=1
C. n=1.1
D. n=1.3

Answer

To solve the problem of determining the refractive index \( n \) of a transparent rod such that light entering it will not exit through its lateral face, we will follow these steps:

### Step 1: Understanding Total Internal Reflection
When light travels from a medium with a higher refractive index to a lower refractive index, it can undergo total internal reflection if the angle of incidence exceeds a certain critical angle. For light to not exit through the lateral face of the rod, we need to ensure that total internal reflection occurs at that interface.

### Step 2: Define the Critical Angle
The critical angle \( \theta_C \) can be defined using Snell's Law:
\[ n_1 \sin \theta_C = n_2 \sin 90^\circ \]
Where:
- \( n_1 \) is the refractive index of the rod (which we denote as \( n \)),
- \( n_2 \) is the refractive index of air (which is approximately 1),
- \( \theta_C \) is the critical angle.

Since \( \sin 90^\circ = 1 \), we can simplify this to:
\[ n \sin \theta_C = 1 \]

### Step 3: Rearranging for Critical Angle
From the above equation, we can express the critical angle as:
\[ \sin \theta_C = \frac{1}{n} \]

### Step 4: Condition for Total Internal Reflection
For total internal reflection to occur at the lateral face (let's denote this face as BC), the angle of incidence must be greater than the critical angle:
\[ \theta > \theta_C \]

### Step 5: Finding the Minimum Refractive Index
To ensure that light does not exit through the lateral face regardless of the angle of incidence, we can assume the maximum angle of incidence is \( 90^\circ \). In this case, the critical angle must be \( 90^\circ \) or less. Therefore, we set:
\[ \theta_C = 90^\circ \]

This leads us to the condition:
\[ \sin \theta_C = 1 \implies \frac{1}{n} = 1 \implies n \geq \sqrt{2} \]

### Step 6: Conclusion
Thus, for the light to not exit through the lateral face of the rod, the refractive index \( n \) of the material of the rod must be greater than \( \sqrt{2} \).

### Final Answer
The value of the refractive index \( n \) must be greater than \( \sqrt{2} \).

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