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Chapter 4: Problem 3

Assume that the pupil of the human eye has a diameter of \(4.0 \mathrm{mm}\) and receives light of wavelength \(5.0 \times 10^{-7} \mathrm{m}\). (a) Calculate the smallest angular separation that can be resolved by the eye at this wavelength. (b) What is the least distance between features on the moon (a distance of \(3.8 \times 10^{11} \mathrm{m}\) away that can be resolved?

Short Answer

Expert verified
The smallest resolvable angular separation is \(1.525 \times 10^{-4} \mathrm{radians} \). The least distance between features on the moon that can be resolved is \(57.95 \mathrm{km} \).

Step by step solution

01

Understand the Rayleigh Criterion

The Rayleigh criterion formula for the angular resolution \( \theta \) of an optical system is given by: \[ \theta = 1.22 \frac{ \lambda } { D } \] where \( \lambda \) is the wavelength of light and \( D \) is the diameter of the aperture.
02

Apply the Rayleigh Criterion

Plug the values into the Rayleigh criterion formula: \[ \theta = 1.22 \frac{5.0 \times 10^{-7} \mathrm{m}}{4.0 \times 10^{-3} \mathrm{m}} = 1.22 \times 1.25 \times 10^{-4} \mathrm{radians} = 1.525 \times 10^{-4} \mathrm{radians} \] This is the smallest angular separation that can be resolved by the human eye.
03

Calculate Least Distance Between Features on the Moon

Use the formula for the least distance \( d \) that can be resolved, which is given by: \[ d = \theta \times R \] Here, \( \theta = 1.525 \times 10^{-4} \mathrm{radians} \) (from Step 2), and the distance to the moon \( R = 3.8 \times 10^{11} \mathrm{m} \). Thus, \[ d = 1.525 \times 10^{-4} \mathrm{radians} \times 3.8 \times 10^{11} \mathrm{m} = 5.795 \times 10^{7} \mathrm{m} = 57.95 \times 10^{6} \mathrm{m} = 57.95 \mathrm{km} \] This is the least distance between features on the moon that can be resolved.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Angular Resolution
Angular resolution is a measure of the ability of an optical instrument, like the human eye or a telescope, to distinguish two closely spaced objects as separate. It鈥檚 akin to looking at two distant stars and being able to tell them apart. This ability is critical for clarity and detail in observations.

The Rayleigh Criterion is often used to define the limit of angular resolution. It states that the smallest angular separation \( \theta \) that can be resolved is: \[ \theta = 1.22 \frac{\lambda}{D } \]

Where:
  • \( \lambda \) is the wavelength of light.
  • \( D \) is the diameter of the aperture (e.g., the pupil of the eye).
In simpler terms, smaller values of \( \theta \) mean better resolution, allowing you to see finer details.
Wavelength of Light
The wavelength of light is the distance between successive peaks of a wave. It is crucial because different wavelengths correspond to different colors of light. For instance, violet light has a shorter wavelength than red light.

When considering the wavelength in relation to angular resolution, it is vital to recognize that shorter wavelengths provide finer resolution. This means that blue or violet light (shorter wavelength) can resolve finer details compared to red light (longer wavelength).

For this particular problem, the wavelength of light used is \( 5.0 \times 10^{-7} \text{ meters} \). This corresponds to green light, commonly used for calculations because it represents a mid-range wavelength that the human eye is most sensitive to.
Diameter of Aperture
The diameter of the aperture is another critical factor that influences angular resolution. The aperture is the opening through which light enters an optical system, such as the pupil in the human eye or the lens in a telescope. Think of it as the 'window' that lets light in.

A larger aperture diameter allows more light to enter and generally improves resolution. This is why professional telescopes or camera lenses have larger diameters than their amateur counterparts.

In the given exercise, the diameter of the human eye's pupil is considered to be 4.0 mm, which is fairly typical for an average human pupil under normal lighting conditions. The larger this diameter, the smaller the value of \( \theta \), which means better angular resolution.

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