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***10points***physics optics help?
please help me with these questions:
1.two vertical narrow slits sperated by 0.20mm are illuminated perpendicularly by 500nm plane waves. if the centers of the first black bands in the interference pattern are 1.00 mm to the right and left of the central axis, how far away is the screen on which the fringes appear?
2.a soap flim of indec 1.340 has a region where it is 550.0 nm thick. determine the vacuum wavelengths of the radiation that is not reflected when the film is illuminated from above with sunlight.
3.3. a) the mount palomar telescope has a 508cm diameter objective mirror. determine its angular limit of resolution at a wavelength of 550nm in radians, degrees, and seconds of the arc.
b.)how far apart must two objects be on the surface of the moon if they are to be resolvable by palomar telescope? the eart-moon distance is 3.844*10^8 take wavelength 550nm
2 Answers
- 8 years agoFavorite Answer
1. d=0.20mm or 2.00*10^(-4)m.
lambda = 500nm or 500*10^(-9)m
x = 1.00mm = 1.00 * 10^(-3)m
L = ?
Use the equation sin (theta) = m(lambda) / d
Then x ~ L(theta). Solve for L.
2. I will admit that at first I didn't understand how to do this, but after a bit of reading through the textbook and looking for the answer on the internet, I found out how to do it and understand why.
I suggest upon doing this problem to read the textbook. It'll help a LOT when it comes to the exam.
n = 1.340
t = 550.0 nm
Use 2ntcos(theta) =m(lambda) --> Minimum reflection occurs when cos(theta) = 1 or when theta = 0.
lambda = 2nt/m
solve for when m is equal to 2, 3... or Lambdas between the range of 380nm and 740nm (the wavelengths of light). These will be your answers.
3a) d= 508 cm or 5.08m
lambda = 550nm or 550*10^-9m
Use this equation for angular limit of resolution.
theta = 1.22(lambda)/d -> the answer is automatically in radians.
to find the answer in degrees, simply convert radians into degrees (degrees = radians (180/pi))
an arc second is easy to find. In 1 hour or a full revolution of (2pi radians or 360 deg) is
3600 seconds (60s * 60min). So divide the answer you found for degrees by 3600s
b) L = 3.844*10^8m
theta = answer found in part a) in radians
Use s = L(theta)
s should be ~ 29.1m
Part b is what I am not a 100% sure of, but it seems correct.
It's good to ask questions, but sometimes the textbook reveals all the steps you need to know to do the problems, barring Q2.
Source(s): Source for Q1: Page 696 of chapter 24 of the Giancoli Physics Textbook (6th ed) Source for Q2: textbook. Took a bit of logic to figure it out. Answer available on question #20: http://univphys.szu.edu.cn/UserFiles/File/22_20121... Source for Q3) page 737 to 739 (same textbook) - Anonymous8 years ago
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