A ray of light, travelling in a medium of refractive index 'mu_, is incident at an angle i on a composite transparent plate consisting of three plates of refractive indices `mu_1,mu_2 and mu_3.` The ray emerges from the composite plate into a medium of refractive index `mu_4,` at angle x. Then,

Options

A. `sin x=sin i`
B. `sin x=mu/mu_4 sin i`
C. `sin x=mu_4/mu sini`
D. `sin x=(mu_1mu_3mu)/(mu_2mu_2mu_4) sin i`

Explanation

To solve the problem of finding the relationship between the angles of incidence and emergence of a ray of light passing through a composite transparent plate, we can use Snell's law at each interface. Here’s a step-by-step breakdown of the solution:

### Step 1: Understand the Setup
We have a ray of light traveling from a medium with refractive index \( \mu \) and incident at an angle \( i \) on a composite plate made up of three different materials with refractive indices \( \mu_1, \mu_2, \) and \( \mu_3 \). The ray then emerges into a medium with refractive index \( \mu_4 \) at an angle \( x \).

### Step 2: Apply Snell's Law at Each Interface
Snell's law states that:
\[ \mu_1 \sin(\theta_1) = \mu_2 \sin(\theta_2) \]
\[ \mu_2 \sin(\theta_2) = \mu_3 \sin(\theta_3) \]
\[ \mu_3 \sin(\theta_3) = \mu_4 \sin(x) \]

### Step 3: Relate the Angles
From the first interface (from medium \( \mu \) to \( \mu_1 \)):
\[ \mu \sin(i) = \mu_1 \sin(\theta_1) \quad \text{(1)} \]

From the second interface (from \( \mu_1 \) to \( \mu_2 \)):
\[ \mu_1 \sin(\theta_1) = \mu_2 \sin(\theta_2) \quad \text{(2)} \]

From the third interface (from \( \mu_2 \) to \( \mu_3 \)):
\[ \mu_2 \sin(\theta_2) = \mu_3 \sin(\theta_3) \quad \text{(3)} \]

From the last interface (from \( \mu_3 \) to \( \mu_4 \)):
\[ \mu_3 \sin(\theta_3) = \mu_4 \sin(x) \quad \text{(4)} \]

### Step 4: Combine the Equations
We can express \( \sin(\theta_1) \), \( \sin(\theta_2) \), and \( \sin(\theta_3) \) in terms of \( \sin(i) \) and \( \sin(x) \).

From equation (1):
\[ \sin(\theta_1) = \frac{\mu}{\mu_1} \sin(i) \]

Substituting this into equation (2):
\[ \mu_1 \left(\frac{\mu}{\mu_1} \sin(i)\right) = \mu_2 \sin(\theta_2) \]
\[ \sin(\theta_2) = \frac{\mu}{\mu_2} \sin(i) \]

Substituting this into equation (3):
\[ \mu_2 \left(\frac{\mu}{\mu_2} \sin(i)\right) = \mu_3 \sin(\theta_3) \]
\[ \sin(\theta_3) = \frac{\mu}{\mu_3} \sin(i) \]

Substituting this into equation (4):
\[ \mu_3 \left(\frac{\mu}{\mu_3} \sin(i)\right) = \mu_4 \sin(x) \]
\[ \sin(x) = \frac{\mu}{\mu_4} \sin(i) \]

### Step 5: Final Relationship
Thus, we arrive at the final relationship:
\[ \sin(x) = \frac{\mu}{\mu_4} \sin(i) \]

### Conclusion
The relationship between the angle of emergence \( x \) and the angle of incidence \( i \) is given by:
\[ \sin(x) = \frac{\mu}{\mu_4} \sin(i) \]