/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 71 In bright light, the pupils of t... [FREE SOLUTION] | 91影视

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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: \[胃 = \dfrac{1.22 \times 位}{a}\] Where 胃 is the angular resolution, 位 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, 位 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. \[位 = 560\,\text{nm} = 560 \times 10^{-9} \text{m}\] \[a = 0.30 \,\text{mm} = 0.30 \times 10^{-3} \text{m}\] Now, we can plug these values into the angular resolution formula: \[胃 = \dfrac{1.22 \times (560 \times 10^{-9} \text{m})}{0.30 \times 10^{-3} \text{m}}\]
03

Calculate Angular Resolution

Calculate the angular resolution, 胃: \[胃 = \dfrac{1.22 \times (560 \times 10^{-9})}{0.30 \times 10^{-3}} = 2.76 \times 10^{-6} \, \text{radians}\]
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: \[d = 胃 \times D\] Where: \[d = \text{smallest distance between the mice}\] \[胃 = \text{angular resolution}\] \[D = \text{distance between the cat and the mice}\] 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

Diffraction by a single Slit The central bright fringe in a single-slit diffraction pattern from light of wavelength 476 nm is \(2.0 \mathrm{cm}\) wide on a screen that is $1.05 \mathrm{m}$ from the slit. (a) How wide is the slit? (b) How wide are the first two bright fringes on either side of the central bright fringe? (Define the width of a bright fringe as the linear distance from minimum to minimum.)
A soap film has an index of refraction \(n=1.50 .\) The film is viewed in transmitted light. (a) At a spot where the film thickness is $910.0 \mathrm{nm},$ which wavelengths are weakest in the transmitted light? (b) Which wavelengths are strongest in transmitted light?
A pinhole camera doesn't have a lens; a small circular hole lets light into the camera, which then exposes the film. For the sharpest image, light from a distant point source makes as small a spot on the film as possible. What is the optimum size of the hole for a camera in which the film is $16.0 \mathrm{cm}$ from the pinhole? A hole smaller than the optimum makes a larger spot since it diffracts the light more. A larger hole also makes a larger spot because the spot cannot be smaller than the hole itself (think in terms of geometrical optics). Let the wavelength be \(560 \mathrm{nm}\).
A grating spectrometer is used to resolve wavelengths \(660.0 \mathrm{nm}\) and \(661.4 \mathrm{nm}\) in second order. (a) How many slits per centimeter must the grating have to produce both wavelengths in second order? (The answer is either a maximum or a minimum number of slits per centimeter.) (b) The minimum number of slits required to resolve two closely spaced lines is $N=\lambda /(m \Delta \lambda),\( where \)\lambda$ is the average of the two wavelengths, \(\Delta \lambda\) is the difference between the two wavelengths, and \(m\) is the order. What minimum number of slits must this grating have to resolve the lines in second order?
Light incident on a pair of slits produces an interference pattern on a screen \(2.50 \mathrm{m}\) from the slits. If the slit separation is $0.0150 \mathrm{cm}$ and the distance between adjacent bright fringes in the pattern is \(0.760 \mathrm{cm},\) what is the wavelength of the light? [Hint: Is the small-angle approximation justified?]
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