Snell's Law & Index of Refraction - Wavelength, Frequency and Speed of Light
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- เผยแพร่เมื่อ 5 ส.ค. 2016
- This physics video tutorial discusses snell's law and the index of refraction of light. It discusses the difference between the law of reflection and refraction. It contains plenty of notes, equations / formulas, with examples and practice problems.
Law of Reflection - Geometric Optics:
• Law of Reflection - Ge...
The Mirror Equation:
• Spherical Mirrors & Th...
Refraction of Light:
• Refraction of Light
Snell's Law & Index of Refraction:
• Snell's Law & Index of...
Total Internal Reflection & Critical Angle:
• Total Internal Reflect...
Thin Lens Equation:
• Thin Lens Equation, Op...
_____________________________
Ray Diagrams:
• Ray Diagrams
The Huygens Principle:
• Huygens Principle - Ph...
Young's Double Slit Experiment:
• Young's Double Slit Ex...
Single Slit Diffraction:
• Single Slit Diffractio...
Diffraction Grating Problems:
• Diffraction Grating Pr...
Polarization of Light Problems:
• Polarization of light ...
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I believe you've made a mistake in the problem with light passing through air, glass and diamond (around 20:00). When calculating the final angle you used sin(35.26), whereas I believe it should be sin(57.74), which is the angle between the normal line and the light. You would then get an angle of 30.69 instead of 21.1. Please correct me if I'm wrong.
right
he misplaced the 35.26 which is supposed to be an incident angle. so it's still sin(35.25) to be used.
Yes, I also noticed this mistake.
No, actually, he has a point. The new angle of incidence is equal to the old angle of refraction, because these are two angles with parallel lines and a secant in the form of a reflected ray.
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please do a video on refraction through a glass prism.. thanks
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I was really hoping to find a formula which relates the index of refraction, or the change in the speed of light on the one hand to the wavelength or the frequency of the light on the other hand. Obviously, different frequencies/wavelengths of light diverge in a prism due to this principle, but what is the mathematical formula that allows you to predict the degree of bending or slowing by knowing the frequency or wavelength?
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At minute 20:15 isn't the angle of incidence as the light moves from glass to diamond suppose to be approximately 54.74° and not 35.26°?
look at the angle of the air and the location of the angle
If you used alternate angles, then you should get around 54.74° which is incorrect as it is not specified if it's a parallel line.
I'm wondering the same thing. I don't follow the reasoning of the replies you've received either.
His initial process involves relating angles between the ray of light and the "normal line". Then he switches for the second calculation to using the angle between the ray of light and the surface to find the third angle.
Just use some geometry or trig on the problem and you’ll answer you’re own question
using AIA, it actually is the angle. I thought what you thought at first
How do you calculate the angle for the total reflection that is caused when light is from air to water? For example if you are besides a lake, you can see the light reflected off from the water surface
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When we have a lot of materials with different index of refraction can we just use the first one and the last one independently of the order of them?
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@06:53 So close. The wavelength does not change. Different wavelengths refract at different angles. Refraction does not change colors, just the angle they are deflected. You cannot put a red laser into a prism and get a blue laser out (unless it's a KTP or BBO crystal, but that's another can of worms).
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To find Theta 3, dont you have to use the complementary angle of Theta 2, in order to use properly the Snell's law?
If you’re talking about the angle about the normal line of the refraction light then no you do not. Theta 2 (reflection) = theta 1(incident) and theta 3 can just be solved using the formula n1sin(theta1)=n3sin(theta3)
The numbers are interchangeable so it can be considered angle 1,2, or 3 as long as you know the difference in each and know how to apply them in each formula
Actually if you’re talking about 20:15 then yes you are right I think he made a mistake
Never mind I saw what he did , it’s because it’s the opposite angle that is equal
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for one of the last example i dont get why for n3 you used the n of air. isn't it contacting with the material again? so shouldn't n3 be 1.2?
20:01
But, It is not supposed to use the (90-35.26 or 54,72 angle), as we used on the first one!!!?
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Why didn't you use the angle of incidence for the second step of the second example problem? Shouldn't the final equation be 1.5sin(90-35.26)=2.4sin(theta)
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