Light of intensity I falls along the axis on a perfectly refleccting 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
Q. The net force of the cone.

Options

A. `((piR^2I)(1-cos2theta))/(c)`
B. `((piR^2I))/(2c(1-cos2theta))`
C. `(2(piR^2I))/(c(1-cos2theta))`
D. `((piR^2I))/2c(1+cos2theta)`

Explanation

To solve the problem, we need to find the net force exerted on a perfectly reflecting right circular cone when light of intensity \( I \) falls on it. Here are the steps to derive the solution:

### Step 1: Understand the Geometry of the Cone
The cone has a semi-vertical angle \( \theta \) and a base radius \( R \). The light is falling along the axis of the cone, which means it is directed vertically downwards. The angle of incidence with respect to the normal at the surface of the cone will be \( 90^\circ - \theta \).

### Step 2: Calculate the Projected Area
The projected area \( A \) of the cone's base that is illuminated by the light is given by:
\[ A = \pi R^2 \]

### Step 3: Relate Intensity to Energy
The intensity \( I \) of the light is defined as the power per unit area. Therefore, the total power \( P \) incident on the cone can be expressed as:
\[ P = I \cdot A = I \cdot \pi R^2 \]

### Step 4: Determine the Energy of a Single Photon
Let \( E \) be the energy of a single photon. The number of photons \( N \) striking the cone per second can be calculated using the relationship between power and energy:
\[ N = \frac{P}{E} = \frac{I \cdot \pi R^2}{E} \]

### Step 5: Change in Momentum for a Photon
When a photon strikes a perfectly reflecting surface, it undergoes a change in momentum. The change in momentum \( \Delta p \) for a single photon is given by:
\[ \Delta p = 2 \frac{E}{c} \]
where \( c \) is the speed of light.

### Step 6: Calculate the Force Exerted by Photons
The force \( F \) exerted by a single photon can be expressed as the rate of change of momentum:
\[ F = N \cdot \Delta p = N \cdot 2 \frac{E}{c} \]
Substituting \( N \) from Step 4:
\[ F = \left(\frac{I \cdot \pi R^2}{E}\right) \cdot 2 \frac{E}{c} = \frac{2 \pi R^2 I}{c} \]

### Step 7: Consider the Angle of Reflection
Since the cone reflects the photons, we need to account for the angle \( \theta \). The vertical component of the force will be:
\[ F_{\text{net}} = F \cdot \sin(\theta) \]
However, since the cone is symmetric, we consider the vertical component of the force:
\[ F_{\text{net}} = \frac{2 \pi R^2 I}{c} \cdot \sin^2(\theta) \]

### Step 8: Final Expression
Using the identity \( \sin^2(\theta) = \frac{1 - \cos(2\theta)}{2} \), we can express the net force as:
\[ F_{\text{net}} = \frac{\pi R^2 I}{c} (1 - \cos(2\theta)) \]

### Conclusion
The net force on the cone is given by:
\[ F_{\text{net}} = \frac{\pi R^2 I}{c} (1 - \cos(2\theta)) \]