One thing to note, I did a calculation with 2.44 & the distance from Earth to moon being 3.56E8m and got 795.35 which is very close to choice E. Do you guys think E could be a possibility?
[source for distance (surface to surface) ...
Right so when I use 1.22 and 3.84E8 for distance form Earth to moon, I get 429.44
When I use the same distance but use 2.44 I get 858.88.
Neither are possible answer. I'm pretty lost :)
Well would it give me this relationship: 2t = ( m + 0.5 ) λi / n where lambda i is wave length in air and therefore n must be n1? That way I relate air to n1. Since n1 < n2 perhaps it bounces off n2 but doesn't pass it, therefore I do not use it in the equation?
Is that what you meant? I'm...
It is to measure the height of a circular aperture (one circle only, not 2 merging ) . Theta is the angle between the top part of the circle and bottom if you are looking at it as a flat circular surface. Width is top of the circle to bottom so diameter of the circle.
Ohhh the first part makes more sense now, its like a unit circle and adding 2pi correct? if you take the sin of 60deg and sin of 60+360deg its the same value. Is that the correct way of thinking about m?
As for the second part I believe there is a phase change between air and n1 because n_air...
So I guess my problem is I don't really understand what m is. I know its an integer and thought is it the number of times light passes through a medium that causes part of the ray to reflect. That is why I thought it is 2.
As for destructive, I know what it means ( the waves, when added...
So I guess since it goes through a phase change m cannot be zero (maybe m = 2?)? Also would the n be n1 because it never enters n2? (assuming it has 1 phase change when entering n1 and then bounces off n2 without entering it)
Opps I meant 2.44 but wrote 2.33 on the post. Sadly I got no such info for the distance. I played around with the values though and think that might be the reason why...
Homework Statement
Monochromatic light of wavelength, λ is traveling in air. The light then strikes a thin film having an index of refraction n1 that is coating a material having an index of refraction n2. If n2 is larger than n1, what minimum film thickness will result in minimum reflection of...
Homework Statement
Suppose you wanted to be able to see astronauts on the moon. What is the smallest diameter of the objective lens required to resolve a 0.60 m object on the moon? Assme the wavelength of the light is near the middle of the visible spectrum: 550 nm yellow light.
(in m)
A...