A ray of sunlight enters a spherical water droplet`(n=4//3)`at an angle of incidence `53^(@)`measured with respect to the normal to the surface.It is reflected from the back surface of the droplet and re-enters into air.The angle between the incoming and outgoing ray is [Take `sin53^(@)=0.8`]

Options

A. `15^(@)`
B. `34^(@)`
C. `138^(@)`
D. `30^(@)`

Answer

To solve the problem of finding the angle between the incoming and outgoing ray of sunlight as it passes through a spherical water droplet, we can follow these steps:

### Step 1: Understand the Setup
We have a spherical water droplet with a refractive index \( n = \frac{4}{3} \). A ray of sunlight enters the droplet at an angle of incidence \( i = 53^\circ \) with respect to the normal.

### Step 2: Apply Snell's Law at the First Surface
Using Snell's Law, we can relate the angle of incidence and the angle of refraction:
\[ n_1 \sin(i) = n_2 \sin(r) \]
Here, \( n_1 = 1 \) (air), \( n_2 = \frac{4}{3} \), \( i = 53^\circ \), and \( r \) is the angle of refraction.

Substituting the values:
\[ 1 \cdot \sin(53^\circ) = \frac{4}{3} \sin(r) \]
Given \( \sin(53^\circ) = 0.8 \):
\[ 0.8 = \frac{4}{3} \sin(r) \]
Solving for \( \sin(r) \):
\[ \sin(r) = \frac{0.8 \cdot 3}{4} = 0.6 \]
Thus, \( r = \sin^{-1}(0.6) \approx 37^\circ \).

### Step 3: Calculate the First Deviation
The first deviation \( \delta_1 \) is given by:
\[ \delta_1 = i - r = 53^\circ - 37^\circ = 16^\circ \]

### Step 4: Reflection at the Back Surface
At the back surface, the ray undergoes total internal reflection. The angle of incidence at this surface is equal to the angle of refraction from the previous step, which is \( r = 37^\circ \). The angle of reflection \( r' \) will also be \( 37^\circ \).

### Step 5: Calculate the Second Deviation
The second deviation \( \delta_2 \) at the back surface is:
\[ \delta_2 = 2r = 2 \cdot 37^\circ = 74^\circ \]

### Step 6: Apply Snell's Law at the Exit Surface
When the ray exits the droplet back into the air, we again apply Snell's Law:
\[ \frac{4}{3} \sin(37^\circ) = 1 \cdot \sin(e) \]
Where \( e \) is the angle of emergence.

Calculating \( \sin(37^\circ) = 0.6 \):
\[ \frac{4}{3} \cdot 0.6 = \sin(e) \]
\[ \sin(e) = \frac{4 \cdot 0.6}{3} = 0.8 \]
Thus, \( e = \sin^{-1}(0.8) = 53^\circ \).

### Step 7: Calculate the Third Deviation
The third deviation \( \delta_3 \) is:
\[ \delta_3 = e - r = 53^\circ - 37^\circ = 16^\circ \]

### Step 8: Total Deviation
The total deviation \( \delta \) is the sum of all deviations:
\[ \delta = \delta_1 + \delta_2 + \delta_3 = 16^\circ + 74^\circ + 16^\circ = 106^\circ \]

### Step 9: Final Angle Between Incoming and Outgoing Ray
The angle between the incoming and outgoing ray is:
\[ \text{Angle} = 180^\circ - \delta = 180^\circ - 106^\circ = 74^\circ \]

### Conclusion
The angle between the incoming and outgoing ray is \( 74^\circ \).

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