9.4.2 Rayleigh Criterion Calculations

Rayleigh Criterion Calculations

The Rayleigh Criterion can be mathematically described by considering angular separation and single-slit diffraction
Angular separation can be calculated using the equation:

$θ = \frac{s}{d}$

Where:
- θ = angular separation (rad)
- s = distance between the two sources (m)
- d = distance between the sources and the observer (m)

9-4-2-angular-separation-ib-hl

Angular separation, θ, is equal to the separation, s, of two sources divided by the distance, d, between the sources and the observer

In single slit diffraction, the first minimum occurs when the angle of diffraction is:

θ = \frac{λ}{b}

Where:
- θ = the angle of diffraction (radians)
- λ = the wavelength of the light (m)
- b = the slit width (m)

According to the Rayleigh criterion, the two sources through a single slit would be just resolvable when the angle is equal to that of the first diffraction minimum or larger
- With the circular aperture, the value is multiplied by a factor of 1.22

For a circular aperture, the Rayleigh criterion is:

θ = 1.22 \frac{λ}{b}

Where
- θ = the angle of diffraction (radians)
- λ = the wavelength of the light (m)
- b = the diameter of the circular aperture (m)
When the angular separation is larger, or equal to, the Rayleigh criterion, then the two sources can be resolved
Therefore, for a circular slit, resolution occurs when:
- The angular separation ≥ The angle of diffraction
Mathematically, the condition for the resolution of two sources can be written as:

\frac{s}{d} \geq 1.22 \frac{λ}{b}

Worked Example

A student looks at a helicopter in the night sky with one eye closed and can just resolve two lights as individual sources. The wavelength of both sources is 530 nm. The approximate diameter of the student’s pupil is 6.0 mm. The distance from the student to the helicopter is 6.0 km.

Determine the minimum distance between the lights.

Step 1: List the known quantities

- Wavelength of light sources, λ = 530 nm = 530 × 10⁻⁹ m
- Student pupil diameter, b = 6.0 mm = 6.0 × 10⁻³ m
- Distance from the helicopter light sources to the student's eye, d = 6.0 km = 6 × 10³ m

Step 2: Select the relevant equation

- Since the lights can just be resolved, this is Rayleigh's criterion
- The equation needed is:

\frac{s}{d} \geq 1.22 \times \frac{λ}{b}

- As the situation is when the lights can just be resolved, this can be written as:

\frac{s}{d} = 1.22 \times \frac{λ}{b}

Step 3: Rearrange equation and input values

- Rearrange for the distance between the sources, s:

s = 1.22 \times \frac{λ}{b} \times d

s = 1.22 \times \frac{530 \times 10^{- 9}}{6 \times 10^{- 3}} \times (6 \times 10^{3}) \approx 0.65 m

Step 4: State the final answer

- The approximate distance between the light sources on the helicopter, s = 65 cm

Worked Example

A student views a car in the distance. It has headlights which are 1.5 m apart. The wavelength of light from the car headlights is 500 nm and the pupil diameter of the student is 4.0 mm.

Estimate the maximum distance at which the two headlights could be resolved by the student.

Step 1: List the known values

- Wavelength of light sources, λ = 500 nm = 500 × 10^-9 m
- Student pupil diameter, b = 4.0 mm = 4.0 × 10^-3 m
- Distance of separation between headlights, s = 1.5 m

Step 2: Select the relevant equation

- Since the answer will occur when Rayleigh’s criterion is met, the equation needed is:

\frac{s}{d} = 1.22 \times \frac{λ}{b}

Step 3: Rearrange equation and input values

- Rearrange for the distance between the sources and the observer, d:

s = 1.22 \times \frac{λ}{b} \times d

d \times \frac{λ}{b} = \frac{s}{1.22}

d = \frac{s \times b}{1.22 \times λ}

d = \frac{1.5 \times (4.0 \times 10^{- 3})}{1.22 \times (500 \times 10^{- 9})} = 9836 m

Step 4: State the final answer

- The maximum distance where the student could resolve the headlights, d = 9800 m (2 s.f.)

Exam Tip

You might be curious where the factor of 1.22 comes from, however, the derivation of this is beyond the scope of the IB DP Physics course so just make sure you know how to use it in your calculations

DP IB Physics: HL

Revision Notes

9.4.2 Rayleigh Criterion Calculations

Rayleigh Criterion Calculations

Worked Example

Worked Example

Exam Tip

Author: Katie

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