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. `ngtsqrt2`
B. `n=1`
C. `n=1.1`
D. `n=1.3`

Explanation

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

### Step 1: Understand Total Internal Reflection
When light travels from a medium of higher refractive index to a medium of lower refractive index, it can undergo total internal reflection if the angle of incidence exceeds a certain critical angle. For the light to remain inside the rod, we need to ensure that total internal reflection occurs at the lateral surface of the rod.

### Step 2: Define the Refractive Indices
Let:
- The refractive index of air (outside the rod) be \( n_1 = 1 \).
- The refractive index of the rod be \( n \).

### Step 3: Apply Snell's Law
At the interface where light enters the rod, Snell's law states:
\[ n_1 \sin(\theta_1) = n \sin(\theta_2) \]
Where:
- \( \theta_1 \) is the angle of incidence in air.
- \( \theta_2 \) is the angle of refraction in the rod.

### Step 4: Determine Critical Angle for Total Internal Reflection
For total internal reflection to occur at the lateral surface of the rod, the angle of incidence \( \theta_2 \) must be greater than the critical angle \( \theta_c \). The critical angle can be given by:
\[ \sin(\theta_c) = \frac{n_1}{n} \]
Thus, the critical angle is:
\[ \theta_c = \arcsin\left(\frac{n_1}{n}\right) = \arcsin\left(\frac{1}{n}\right) \]

### Step 5: Condition for Total Internal Reflection
For total internal reflection to occur, the angle of incidence \( \theta_2 \) must be greater than \( \theta_c \):
\[ \theta_2 > \theta_c \]
This means that the angle \( \theta_2 \) must be such that:
\[ \sin(\theta_2) > \frac{1}{n} \]

### Step 6: Maximum Angle of Incidence
The maximum angle of incidence \( \theta_1 \) at which light can enter the rod is \( 90^\circ \). Therefore, at this angle:
\[ \sin(\theta_1) = 1 \]
Substituting into Snell's law gives:
\[ 1 \cdot 1 = n \sin(\theta_2) \implies \sin(\theta_2) = \frac{1}{n} \]

### Step 7: Ensure Total Internal Reflection
To ensure that light does not exit the rod, we need:
\[ \sin(\theta_2) \geq \sin(\theta_c) \implies \frac{1}{n} \geq \frac{1}{n} \implies n \geq \sqrt{2} \]

### Conclusion
Thus, for the light to remain inside the rod without exiting through the lateral face, the refractive index \( n \) of the rod must be greater than \( \sqrt{2} \).

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

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