The minimum intensity of light to be detected by human eye is `10^(-10) W//m^(2)`. The number of photons of wavelength `5.6 xx 10^(-7) m` entering the eye , with pupil area `10^(-6) m^(2)` , per second for vision will be nearly

Options

A. 100
B. 200
C. 300
D. 400

Answer

To solve the problem, we need to calculate the number of photons entering the human eye per second based on the given intensity of light, the wavelength of the light, and the area of the pupil.

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Minimum intensity of light (I) = \(10^{-10} \, \text{W/m}^2\)
- Wavelength of light (\(\lambda\)) = \(5.6 \times 10^{-7} \, \text{m}\)
- Area of the pupil (A) = \(10^{-6} \, \text{m}^2\)

2. **Use the Formula for Intensity:**
The intensity (I) of light can be expressed as:
\[ I = \frac{P}{A} \]
where \(P\) is the power. Rearranging gives:
\[ P = I \cdot A \]

3. **Calculate the Power (P):**
Substitute the values into the equation:
\[ P = 10^{-10} \, \text{W/m}^2 \times 10^{-6} \, \text{m}^2 = 10^{-16} \, \text{W} \]

4. **Calculate the Energy of a Single Photon:**
The energy (E) of a single photon can be calculated using the formula:
\[ E = \frac{hc}{\lambda} \]
where \(h\) (Planck's constant) = \(6.63 \times 10^{-34} \, \text{J s}\) and \(c\) (speed of light) = \(3 \times 10^8 \, \text{m/s}\).

Substitute the values:
\[ E = \frac{(6.63 \times 10^{-34} \, \text{J s}) \cdot (3 \times 10^8 \, \text{m/s})}{5.6 \times 10^{-7} \, \text{m}} \]

5. **Calculate E:**
\[ E = \frac{1.989 \times 10^{-25} \, \text{J m}}{5.6 \times 10^{-7} \, \text{m}} \approx 3.55 \times 10^{-19} \, \text{J} \]

6. **Calculate the Number of Photons (N):**
The number of photons per second (N) can be calculated using:
\[ N = \frac{P}{E} \]
Substitute the values:
\[ N = \frac{10^{-16} \, \text{W}}{3.55 \times 10^{-19} \, \text{J}} \approx 282.5 \approx 300 \, \text{photons} \]

### Final Answer:
The number of photons entering the eye per second for vision will be nearly **300**.