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Chapter 25: Problem 71

In bright light, the pupils of the eyes of a cat narrow to a vertical slit \(0.30 \mathrm{mm}\) across. Suppose that a cat is looking at two mice $18 \mathrm{m}$ away. What is the smallest distance between the mice for which the cat can tell that there are two mice rather than one using light of $560 \mathrm{nm} ?$ Assume the resolution is limited by diffraction only.

Short Answer

Expert verified
Answer: The smallest distance between the mice is \(0.0497\,\text{mm}\).

Step by step solution

01

Formula for Angular Resolution

For a slit aperture, such as the cat's eye, the angular resolution can be determined using the following formula: \[θ = \dfrac{1.22 \times λ}{a}\] Where θ is the angular resolution, λ is the wavelength of light, and a is the width of the slit aperture (pupil width in this case).
02

Plug in given values

We are given the wavelength of light, λ as \(560\,\text{nm}\), and the width of the cat's pupil, a as \(0.30\,\text{mm}\). However, we need to convert the units to meters. \[λ = 560\,\text{nm} = 560 \times 10^{-9} \text{m}\] \[a = 0.30 \,\text{mm} = 0.30 \times 10^{-3} \text{m}\] Now, we can plug these values into the angular resolution formula: \[θ = \dfrac{1.22 \times (560 \times 10^{-9} \text{m})}{0.30 \times 10^{-3} \text{m}}\]
03

Calculate Angular Resolution

Calculate the angular resolution, θ: \[θ = \dfrac{1.22 \times (560 \times 10^{-9})}{0.30 \times 10^{-3}} = 2.76 \times 10^{-6} \, \text{radians}\]
04

Calculate the smallest distance

We know that the cat is \(18\,\text{m}\) away from the mice. Using the angular resolution and the distance between the cat and the mice, we can determine the smallest distance, d, between the mice for which the cat can tell that there are two mice rather than one: \[d = θ \times D\] Where: \[d = \text{smallest distance between the mice}\] \[θ = \text{angular resolution}\] \[D = \text{distance between the cat and the mice}\] Plug in the values and calculate d: \[d = (2.76 \times 10^{-6}) \times 18 = 4.97 \times 10^{-5}\, \text{m}\] \[d = 4.97 \times 10^{-5} \text{m} = 0.0497\,\text{mm}\]
05

Final Answer

The smallest distance between the mice for which the cat can tell that there are two mice rather than one using light of \(560\, \text{nm}\) is \(0.0497\,\text{mm}\).

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Most popular questions from this chapter

You are given a slide with two slits cut into it and asked how far apart the slits are. You shine white light on the slide and notice the first-order color spectrum that is created on a screen \(3.40 \mathrm{m}\) away. On the screen, the red light with a wavelength of 700 nm is separated from the violet light with a wavelength of 400 nm by 7.00 mm. What is the separation of the two slits?
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