A ray of light travelling in a transparent medium f refractive index `mu`, falls on a surface separating the medium from air at an angle of incidence of `45^(@)`. For which of the following value of `mu` the ray can undergo total internal reflection ?

Options

A. `mu = 1.33`
B. `mu = 1.40`
C. `mu = 1.50`
D. `mu = 1.25`

Explanation

To determine the value of the refractive index \( \mu \) for which a ray of light can undergo total internal reflection when it travels from a medium with refractive index \( \mu \) to air, we can follow these steps:

### Step 1: Understand the conditions for total internal reflection
Total internal reflection occurs when light travels from a medium with a higher refractive index to a medium with a lower refractive index. In this case, the light is moving from the medium with refractive index \( \mu \) to air, which has a refractive index of approximately 1.

### Step 2: Apply Snell's Law
According to Snell's Law, we have:
\[ \mu \sin(\theta_i) = n \sin(\theta_r) \]
where:
- \( \theta_i \) is the angle of incidence (which is given as \( 45^\circ \)),
- \( n \) is the refractive index of the second medium (air, which is 1),
- \( \theta_r \) is the angle of refraction.

### Step 3: Set up the equation for total internal reflection
For total internal reflection to occur, the angle of refraction \( \theta_r \) must be \( 90^\circ \). At this point, the sine of \( 90^\circ \) is 1. Therefore, we can rewrite Snell's Law as:
\[ \mu \sin(45^\circ) = 1 \cdot \sin(90^\circ) \]
This simplifies to:
\[ \mu \sin(45^\circ) = 1 \]

### Step 4: Substitute the value of \( \sin(45^\circ) \)
We know that:
\[ \sin(45^\circ) = \frac{\sqrt{2}}{2} \]
Substituting this value into the equation gives:
\[ \mu \cdot \frac{\sqrt{2}}{2} = 1 \]

### Step 5: Solve for \( \mu \)
To find \( \mu \), we rearrange the equation:
\[ \mu = \frac{1}{\frac{\sqrt{2}}{2}} = \frac{2}{\sqrt{2}} = \sqrt{2} \]

### Step 6: Determine the condition for total internal reflection
For total internal reflection to occur, the refractive index \( \mu \) must be greater than \( \sqrt{2} \). Thus, we conclude:
\[ \mu > \sqrt{2} \]

### Final Answer
The value of \( \mu \) must be greater than \( \sqrt{2} \) for the ray to undergo total internal reflection.

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