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Chapter 25: Problem 38

The pupil of a cat's eye narrows to a vertical slit of width \(0.500 \mathrm{~mm}\) in daylight. What is the angular resolution for a pair of horizontally separated mice? (Use \(500-\mathrm{nm}\) light in your calculation.)

Short Answer

Expert verified
The angular resolution for a pair of horizontally separated mice as seen by a cat in daylight is \(1.22 \times 10^{-3}\) radians.

Step by step solution

01

Convert the pupil width to meters

In order to perform calculations, all units should be in meters. The given width of the pupil is 0.500 mm, which converts to \(0.500 \times 10^{-3}\) m or \(0.0005\) m.
02

Convert the wavelength of light to meters

Similarly, the specified wavelength of light used for the calculation is \(500 nm\), which is equivalent to \(500 \times 10^{-9}\) m or \(5 \times 10^{-7}\) m.
03

Apply the formula for angular resolution

Now, apply the formula for angular resolution in radians. The formula is: \(\Delta\theta = \frac{1.22 \times \lambda}{D}\), where \(\lambda\) is the wavelength of light and \(D\) is the diameter (width in this case) of the aperture (pupil). Substituting the provided values into the equation, we find that \(\Delta\theta = \frac{1.22 \times 5 \times 10^{-7} m}{0.0005 m}\).
04

Solve for the angular resolution

Now, calculate the angular resolution by dividing \(1.22 \times 5 \times 10^{-7}\) by \(0.0005\). The result is \(1.22 \times 10^{-3}\) radians.

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Key Concepts

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

Pupil Width Conversion
The pupil width conversion is essential for solving problems involving angular resolution. The first step is to ensure all units are compatible. In our exercise, the pupil width is given as 0.500 mm. To convert this to meters, which is necessary for our calculations, you simply multiply:
  • The conversion factor from millimeters to meters is \(1 ext{ mm} = 1 imes 10^{-3} ext{ m}\).
  • Thus, \(0.500 ext{ mm} = 0.500 imes 10^{-3} ext{ m} = 0.0005 ext{ m}\).
Converting to meters allows us to correctly apply the relevant formulas and maintain unit consistency across calculations.
Wavelength Conversion
Wavelength conversion is another critical step to ensure accurate results. Initially, the wavelength of the light is provided as 500 nm (nanometers). We need this value in meters to proceed with our calculations:
  • The conversion factor from nanometers to meters is \(1 ext{ nm} = 1 imes 10^{-9} ext{ m}\).
  • Thus, \(500 ext{ nm} = 500 imes 10^{-9} ext{ m} = 5 imes 10^{-7} ext{ m}\).
This conversion is necessary because the angular resolution formula requires all inputs to be in meters. Doing so ensures that we arrive at an accurate and reliable solution.
Resolution Formula
Understanding the resolution formula is key to calculating the angular resolution. The formula is given by:\[\Delta\theta = \frac{1.22 \times \lambda}{D}\]where:
  • \(\Delta\theta\) represents the angular resolution in radians.
  • \(\lambda\) is the wavelength of light in meters.
  • \(D\) is the diameter or width of the aperture (pupil, in this case) in meters.
To find the angular resolution of the cat's eye viewing horizontally separated objects, substitute the values:
  • \(\lambda = 5 \times 10^{-7} \, \text{m}\)
  • \(D = 0.0005 \, \text{m}\)
By plugging these into the formula, we calculate:\[\Delta\theta = \frac{1.22 \times 5 \times 10^{-7}}{0.0005}\]This results in an angular resolution of \(1.22 \times 10^{-3}\) radians, showing us how fine the detail is that the cat can perceive. Understanding this process equips us to solve similar problems efficiently.

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Most popular questions from this chapter

A lens has a focal length of \(28 \mathrm{~cm}\) and a diameter of \(4.0 \mathrm{~cm}\). What is the \(f\) -number of the lens?

A vehicle with headlights separated by \(2.00 \mathrm{~m}\) approaches an observer holding an infrared detector sensitive to radiation of wavelength \(885 \mathrm{~nm}\). What aperture diameter is required in the detector if the two headlights are to be resolved at a distance of \(10.0 \mathrm{~km}\) ?

The near point of an eye is \(75.0 \mathrm{~cm} .\) (a) What should be the power of a corrective lens prescribed to enable the eye to see an object clearly at \(25.0 \mathrm{~cm} ?\) (b) If, using the corrective lens, the person can see an object clearly at \(26.0 \mathrm{~cm}\) but not at \(25.0 \mathrm{~cm}\), by how many diopters did the lens grinder miss the prescription?

A patient has a near point of \(45.0 \mathrm{~cm}\) and far point of \(85.0 \mathrm{~cm}\). (a) Can a single lens correct the patient's vision? Explain the patient's options. (b) Calculate the power lens needed to correct the near point so that the patient can see objects \(25.0 \mathrm{~cm}\) away. Neglect the eye-lens distance. (c) Calculate the power lens needed to correct the patient's far point, again neglecting the eye-lens distance.

The distance between the eyepiece and the objective lens in a certain compound microscope is \(20.0 \mathrm{~cm}\). The focal length of the objective is \(0.500 \mathrm{~cm}\), and that of the eyepiece is \(1.70 \mathrm{~cm}\). Find the overall magnification of the microscope.
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