A monochromatic beam of light `(lamda=4900A)` incident normally upon a surface produces a pressure of `5xx10^(-7)Nm^(-2)` on it. Assume that `25%` of the light incident in reflected and the rest absorbed. Find the number of photons falling per second on a unit area of thin surface

Options

A. `1xx10^23`
B. `2xx10^24`
C. `3xx10^20`
D. None of the above

Explanation

To solve the problem step by step, we will follow the logic laid out in the video transcript.

### Step 1: Understand the Relationship Between Pressure and Intensity
The pressure \( P \) exerted by the light on the surface is related to the intensity \( I \) of the light and the speed of light \( c \). For a surface that absorbs some light and reflects some, the pressure can be expressed as:
\[ P = \frac{I}{c} \text{ (for a perfectly absorbing surface)} \]
\[ P = \frac{2I}{c} \text{ (for a perfectly reflecting surface)} \]

Given that 25% of the light is reflected and 75% is absorbed, we can use a combination of these two cases.

### Step 2: Calculate the Intensity
The total pressure \( P \) is given as \( 5 \times 10^{-7} \, \text{N/m}^2 \). We can express the pressure as:
\[ P = 0.25 \cdot \frac{2I}{c} + 0.75 \cdot \frac{I}{c} \]
This simplifies to:
\[ P = \left(0.5 + 0.75\right) \frac{I}{c} = \frac{1.25I}{c} \]
From this, we can rearrange to find the intensity:
\[ I = \frac{Pc}{1.25} \]

### Step 3: Substitute Known Values
Now we substitute \( P = 5 \times 10^{-7} \, \text{N/m}^2 \) and \( c = 3 \times 10^8 \, \text{m/s} \):
\[ I = \frac{(5 \times 10^{-7}) \cdot (3 \times 10^8)}{1.25} \]
Calculating this gives:
\[ I = \frac{1.5 \times 10^2}{1.25} = 120 \, \text{W/m}^2 \]

### Step 4: Calculate the Energy of a Photon
The energy \( E \) of a single photon can be calculated using the formula:
\[ E = h \nu = \frac{hc}{\lambda} \]
where \( h = 6.63 \times 10^{-34} \, \text{J s} \) and \( \lambda = 4900 \, \text{Å} = 4900 \times 10^{-10} \, \text{m} \).

### Step 5: Substitute Values to Find Energy
Substituting the values:
\[ E = \frac{(6.63 \times 10^{-34}) \cdot (3 \times 10^8)}{4900 \times 10^{-10}} \]
Calculating this gives:
\[ E = \frac{1.989 \times 10^{-25}}{4900 \times 10^{-10}} = 4.06 \times 10^{-19} \, \text{J} \]

### Step 6: Calculate the Number of Photons
The number of photons \( n \) falling per second on a unit area is given by:
\[ n = \frac{I}{E} \]
Substituting the values we found:
\[ n = \frac{120}{4.06 \times 10^{-19}} \approx 2.95 \times 10^{20} \, \text{photons/s/m}^2 \]

### Final Answer
Thus, the number of photons falling per second on a unit area of the thin surface is approximately:
\[ n \approx 3 \times 10^{20} \, \text{photons/s/m}^2 \]