Light of intensity I falls along the axis on a perfectly reflecting right circular cone having semi-vertical angle `theta` and base radius R. If E is the energy of one photon and c is the speed oh light, then find
the number of photons hitting the cone per second

Options

A. `(piR^2I)/(2E)`
B. `(2piR^2I)/(E)`
C. `(piR^2I)/(4E)`
D. `(piR^2I)/(E)`

Explanation

To find the number of photons hitting a perfectly reflecting right circular cone per second, we can follow these steps:

### Step 1: Understand the Concept of Intensity
Intensity (I) is defined as the power (P) per unit area (A) that is incident on a surface. Mathematically, this can be expressed as:
\[ I = \frac{P}{A} \]
Where:
- \( I \) = Intensity of the light
- \( P \) = Power incident on the surface
- \( A \) = Area of the surface

### Step 2: Determine the Projected Area of the Cone
The area of the base of the cone (which is the area that the light is incident upon) can be calculated using the formula for the area of a circle:
\[ A = \pi R^2 \]
Where:
- \( R \) = Radius of the base of the cone

### Step 3: Relate Power to Intensity and Area
Using the relationship from Step 1, we can express the power incident on the cone as:
\[ P = I \cdot A \]
Substituting the area from Step 2, we get:
\[ P = I \cdot (\pi R^2) \]

### Step 4: Relate Power to the Number of Photons
The power can also be expressed in terms of the number of photons (N) hitting the cone per second and the energy of each photon (E):
\[ P = N \cdot E \]
Where:
- \( N \) = Number of photons hitting the cone per second
- \( E \) = Energy of one photon

### Step 5: Set the Two Expressions for Power Equal
From Steps 3 and 4, we can set the two expressions for power equal to each other:
\[ I \cdot (\pi R^2) = N \cdot E \]

### Step 6: Solve for the Number of Photons
Rearranging the equation to solve for \( N \):
\[ N = \frac{I \cdot \pi R^2}{E} \]

### Final Answer
The number of photons hitting the cone per second is given by:
\[ N = \frac{I \cdot \pi R^2}{E} \]

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