Class Management |Help Homework 9 Begin Date: 6/12/2016 12:30:00 PM -- Due Date:
ID: 1416787 • Letter: C
Question
Class Management |Help Homework 9 Begin Date: 6/12/2016 12:30:00 PM -- Due Date: 6/15/2016 11:59:00 PM End Date: 6/24/2016 12:00:00 AM (800) Problem 3: The wavelength range of visible light is usually taken to be from 400 to 700 nm, although it can vary from person to person. 50% Part (a) Calculate the ratio of the highest to lowest frequency of electromagnetic waves that most us can see Grade Summary Deductions Potential 6% 94% tan() | HOME sin cos cotan) asin0 acos atanacotan sinh0 Submissions Attempts remaining::2 (2% per attempt) detailed view 4 56 cosh() coshnh cotanh 0 END 2% 2% 2% O Degrees O Radians BA PACEDEL CLEAR Submit Hint I give up! Hints: 1 for a 0% deduction. Hints remaining: 0 Feedback: 0% deduction per feedback. -How is frequency related to wavelength for electromagnetic waves? Make sure you have the correct correspondence between maximum and minimum for wavelength and frequency 50% Part (b) For comparison, calculate the ratio of the highest to lowest frequency the ear can hear. This range is usually taken to be from 20 kHz to 20 HzExplanation / Answer
(a)
400 nm to 700 nm
We know,
v = f*
3.0 * 10^8 = fmax * 400 * 10^-9
fmax = 7.5 * 10^14 Hz
3.0 * 10^8 = fmin * 700 * 10^-9
fmin = 4.285 * 10^14 Hz
fmax/fmin = (7.5 * 10^14 )/(4.285 * 10^14)
fmax/fmin = 1.75
(b)
Ratio = (20*10^3)/20
Ratio of freq ear can hear = 1000