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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

A grating is made of exactly 8000 slits; the slit spacing is $1.50 \mu \mathrm{m} .\( Light of wavelength \)0.600 \mu \mathrm{m}$ is incident normally on the grating. (a) How many maxima are seen in the pattern on the screen? (b) Sketch the pattern that would appear on a screen \(3.0 \mathrm{m}\) from the grating. Label distances from the central maximum to the other maxima.
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