In a swimming pool, a person is viewing outside objects by keeping an eye at a depth h inside water. If the critical angle for water is ‘ `theta_(c)`’, then the value of the diameter of the circle of view for outside objects will be

Options

A. `2hsintheta_(c)`
B. `2hcostheta_(c)`
C. `2htantheta_(c)`
D. `2hcottheta_(c)`

Explanation

To find the diameter of the circle of view for a person looking out of a swimming pool from a depth \( h \) in water, we can follow these steps:

### Step 1: Understand the Concept of Critical Angle
The critical angle \( \theta_c \) is the angle of incidence above which total internal reflection occurs when light travels from a denser medium (water) to a rarer medium (air). This means that light cannot escape the water if the angle of incidence exceeds \( \theta_c \).

### Step 2: Set Up the Geometry
When a person looks out of the water, they can see objects only within a certain angle defined by the critical angle. We can visualize this situation as a right triangle, where:
- The vertical side represents the depth \( h \) of the person in the water.
- The horizontal side represents the radius \( r \) of the circle of view at the water's surface.

### Step 3: Apply Trigonometry
Using the definition of tangent in the right triangle formed:
\[ \tan(\theta_c) = \frac{r}{h} \]
From this, we can express the radius \( r \):
\[ r = h \tan(\theta_c) \]

### Step 4: Calculate the Diameter
The diameter \( d \) of the circle of view is twice the radius:
\[ d = 2r = 2(h \tan(\theta_c)) \]
Thus, we can write:
\[ d = 2h \tan(\theta_c) \]

### Final Result
The diameter of the circle of view for outside objects is:
\[ d = 2h \tan(\theta_c) \]