Question

A digital camera basically has an array of tiny light detectors (2000×1500 = 3 MegaPixels = 3 million very tiny detectors, covering a cm2. In each of these detectors, photons that hit the detector excite electrons and these excited electrons are counted. In a typical picture, the detector array in the camera is exposed to about 4.5×10-6 watts of light for 10 ms. If you take 535 nm as a typical wavelength for the light, what is the average number of photons that hit each pixel in a typical picture (don't use scientific notation, or Canvas might get confused).

2. If you have very low intensity green light (4×10-11watts at 570 nm) evenly illuminating the entire array of detectors, what will the camera's detectors see during the exposure time of 10ms?
A. Random pixels will have several excited electrons, others will have only one excited electron, and still others will have no excited electrons.
B. All pixels in the array count about the same number of excited electrons.
C. The pixels in the centre of the array will count the largest number of excited electrons and this will drop off towards the edges.
D. Random pixels will have exactly one excited electron, while others will have no excited electrons.

Answer 1

The average number of photons hitting each pixel in a typical picture is around 4.04 million. Low intensity green light would likely cause random pixels to have either one excited electron or none, resulting in an uneven distribution of excited electrons across the array.

1. Average Number of Photons Hitting Each Pixel:

Given:

- Total number of pixels =
`\( 2000 * 1500 = 3 \)` million pixels

- Energy of light hitting the detector array =
`\( 4.5 * 10^(-6) \)` watts

- Wavelength of light
`(\( \lambda \))` = 535 nm

The energy of a single photon can be calculated using the formula:

`\[ \text{Energy of a photon} = (hc)/(\lambda) \]`

Where:

- h = Planck's constant
`(\( 6.626 * 10^(-34) \) J s)`

- c = speed of light
`(\( 3 * 10^8 \) m/s)`

-
`\( \lambda \)` = wavelength of light in meters

First, let's convert the wavelength from nanometers to meters:

`\[ \lambda = 535 \, \text{nm} = 535 * 10^(-9) \, \text{m} \]`

Now, calculate the energy of a single photon:

`\[ \text{Energy of a photon} = \frac{(6.626 * 10^(-34) \, \text{J s}) * (3 * 10^8 \, \text{m/s})}{535 * 10^(-9) \, \text{m}} \]`

`\[ \text{Energy of a photon} \approx 3.73 * 10^(-19) \, \text{J} \]`

The number of photons hitting the detector array can be calculated using the energy of light and the energy of a single photon:

`\[ \text{Number of photons} = \frac{\text{Energy of light}}{\text{Energy of a photon}} * \text{Exposure time} \]`

Given that the exposure time is
`\(10 \, \text{ms} = 10 * 10^(-3) \, \text{s}\)`:

`\[ \text{Number of photons} = \frac{4.5 * 10^(-6) \, \text{W} * 10 * 10^(-3) \, \text{s}}{3.73 * 10^(-19) \, \text{J}} \]`

`\[ \text{Number of photons} \approx 1.21 * 10^(13) \]`

The average number of photons hitting each pixel can be found by dividing the total number of photons by the total number of pixels:

`\[ \text{Average number of photons per pixel} = (1.21 * 10^(13))/(3 * 10^6) \approx 4.04 * 10^6 \]`

2. Effect of Very Low Intensity Green Light:

Given:

- Intensity of green light
`= \( 4 * 10^(-11) \)` watts at 570 nm

- Exposure time = 10 ms

The intensity of the light hitting the array is very low. In this scenario, individual pixels might experience a very low number of photons, close to zero or only one photon, resulting in some pixels having no excited electrons while others may have one or a few excited electrons. Thus, the correct option is:

D. Random pixels will have exactly one excited electron, while others will have no excited electrons.

Answer 2

In a typical picture, the average number of photons that hit each pixel is approximately 4.

The camera's detectors will see (D.) random pixels will have exactly one excited electron, while others will have no excited electrons.

How to find photons?

Part 1: Average Number of Photons per Pixel

Convert watts to joules:

Energy = Power × Time

Energy = 4.5 × 10⁻⁶ watts × 10 × 10⁻³ seconds

Energy = 4.5 × 10⁻⁸ joules

Calculate the energy of one photon:

Energy per photon = (hc) / λ

where:

h = Planck's constant = 6.626 × 10⁻³⁴ J s

c = speed of light = 3 × 10⁸ m/s

λ = wavelength of light = 535 × 10⁻⁹9 m

Energy per photon = (6.626 × 10⁻³⁴ J s) × (3 × 10⁸ m/s) / (535 × 10⁻⁹ m)

Energy per photon = 3.704 × 10⁻¹⁹ J

Calculate the number of photons:

Number of photons = Total energy / Energy per photon

Number of photons = 4.5 × 10⁻⁸ joules / 3.704 × 10⁻¹⁹ J/photon

Number of photons ≈ 1215

Calculate the average number of photons per pixel:

Total pixels = 2000 × 1500

= 3 × 10⁶

Average photons per pixel = 1215 photons / 3 × 10⁶ pixels

Average photons per pixel ≈ 4 photons

Therefore, in a typical picture, the average number of photons that hit each pixel is approximately 4.

Part 2: Detectors with Low Intensity Light

With very low intensity green light (4 × 10⁻¹¹ watts) illuminating the detector array, the number of photons hitting each pixel will be significantly lower than in the previous scenario.

Random pixels with one excited electron: This is the most likely scenario. With very low light intensity, the probability of a single photon hitting each pixel during the exposure time is much higher than multiple photons hitting the same pixel. Therefore, it's expected to see a random pattern of pixels with either one excited electron or none.

Therefore, the most likely observation for the camera's detectors under low intensity light is D - Random pixels will have exactly one excited electron, while others will have no excited electrons.