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 strikes the surface separating a transparent medium from air at an angle of incidence of \( 45^\circ \), we can follow these steps:

### Step 1: Understand Total Internal Reflection
Total internal reflection occurs when a light ray travels from a medium with a higher refractive index to a medium with a lower refractive index (in this case, from the medium with refractive index \( \mu \) to air, which has a refractive index of 1).

### Step 2: Define the Critical Angle
The critical angle \( c \) is defined as the angle of incidence above which total internal reflection occurs. The relationship between the refractive indices and the critical angle is given by:
\[ \sin c = \frac{n_2}{n_1} \]
where \( n_1 \) is the refractive index of the first medium (in this case, \( \mu \)) and \( n_2 \) is the refractive index of the second medium (in this case, air, which is 1).

### Step 3: Set Up the Inequality for Total Internal Reflection
For total internal reflection to occur, the angle of incidence \( i \) must be greater than the critical angle \( c \):
\[ i > c \]
Given that \( i = 45^\circ \), we can write:
\[ 45^\circ > c \]

### Step 4: Relate the Critical Angle to the Refractive Index
From the equation for the critical angle, we can express it in terms of \( \mu \):
\[ \sin c = \frac{1}{\mu} \]
Thus, we have:
\[ c = \arcsin\left(\frac{1}{\mu}\right) \]

### Step 5: Substitute and Solve for \( \mu \)
Since we need \( 45^\circ > c \), we can substitute:
\[ 45^\circ > \arcsin\left(\frac{1}{\mu}\right) \]
Taking the sine of both sides, we get:
\[ \sin(45^\circ) > \frac{1}{\mu} \]
Since \( \sin(45^\circ) = \frac{1}{\sqrt{2}} \), we can write:
\[ \frac{1}{\sqrt{2}} > \frac{1}{\mu} \]
Rearranging gives:
\[ \mu > \sqrt{2} \]

### Step 6: Calculate the Numerical Value
Calculating \( \sqrt{2} \):
\[ \sqrt{2} \approx 1.414 \]
Thus, for total internal reflection to occur, the refractive index \( \mu \) must be greater than approximately \( 1.414 \).

### Final Answer
The value of \( \mu \) must be greater than \( 1.414 \) for the ray to undergo total internal reflection.

---