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`

Answer

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 ray of light travels from a medium with a higher refractive index to a medium with a lower refractive index. The condition for total internal reflection is that the angle of incidence \( i \) must be greater than the critical angle \( c \).

### Step 2: Define the Critical Angle
The critical angle \( c \) can be defined using Snell's law:
\[ \sin c = \frac{n_2}{n_1} \]
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 (air, which is approximately 1).

### Step 3: Apply the Condition for Total Internal Reflection
For total internal reflection to occur, we need:
\[ i > c \]
Given that \( i = 45^\circ \), we can write:
\[ 45^\circ > c \]

### Step 4: Calculate the Critical Angle
Using the definition of the critical angle:
\[ \sin c = \frac{1}{\mu} \]
Thus, we can express the critical angle as:
\[ c = \arcsin\left(\frac{1}{\mu}\right) \]

### Step 5: Set Up the Inequality
From the condition \( 45^\circ > c \), we can substitute for \( c \):
\[ 45^\circ > \arcsin\left(\frac{1}{\mu}\right) \]

### Step 6: Solve for \( \mu \)
Taking the sine of both sides (and noting that sine is an increasing function in this range):
\[ \sin(45^\circ) > \frac{1}{\mu} \]
Since \( \sin(45^\circ) = \frac{1}{\sqrt{2}} \), we can write:
\[ \frac{1}{\sqrt{2}} > \frac{1}{\mu} \]
Rearranging gives:
\[ \mu > \sqrt{2} \]

### Step 7: Numerical Value
Calculating \( \sqrt{2} \) gives approximately:
\[ \mu > 1.41 \]

### Conclusion
For total internal reflection to occur at an angle of incidence of \( 45^\circ \), the refractive index \( \mu \) of the medium must be greater than \( \sqrt{2} \) or approximately \( 1.41 \).

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