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Question
electromagnetic waves and human vision
the photoreceptors in the human eye, called rods and cones, have different sensitivities to different wavelengths of electromagnetic waves. (figure 1) (notice that the y axis in the figure is a logarithmic scale.)
the rods, which number over 100 million, can only be activated by a certain range of wavelengths, but they do not pass any color information to the brain. in other words, they note differences in shades of grey (from black to white) and are responsible for a persons ability to see in dim light. cones, which number around 6 million, give us
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part e
since rods are about 1000 times more sensitive than cones (at 470 nm), they should be able to detect smaller values of the electric field. assuming rods and cones are sensitive to the average energy density of an electromagnetic wave, which of the following statements is the most accurate?
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rods are able to detect electric fields 1000 times smaller than the fields detectable by cones
rods are able to detect electric fields 1000/3 = 333 times smaller than the fields detectable by cones
rods are able to detect electric fields \sqrt{1000}=32 times smaller than the fields detectable by cones
rods are able to detect electric fields 3 times smaller than the fields detectable by cones
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Step1: Recall the formula for average energy density
The average energy density of an electromagnetic wave is given by \(u_{avg}=\frac{1}{2}\epsilon_0E_{rms}^2\).
Step2: Relate sensitivity and electric field
If the sensitivity (related to energy density) of rods is \(1000\) times that of cones. Let \(u_{avg,rod} = 1000u_{avg,cone}\). Then \(\frac{1}{2}\epsilon_0E_{rms,rod}^2=1000\times\frac{1}{2}\epsilon_0E_{rms,cone}^2\). Canceling out \(\frac{1}{2}\epsilon_0\) from both sides, we get \(E_{rms,rod}^2 = 1000E_{rms,cone}^2\), or \(E_{rms,rod}=E_{rms,cone}\sqrt{1000}\). So \(E_{rms,rod}\approx32E_{rms,cone}\) (since \(\sqrt{1000}\approx32\)).
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Rods are able to detect electric fields \(\sqrt{1000} = 32\) times smaller than the fields detectable by cones.