How do I find this index of refraction?
I know how to find the angle of refraction, but I don't know how to find the index.
Here's the question;
"A ray of light has an angle of incidence of 36.0 degrees on a block of quartz and an angle of refraction of 20.0 degrees. What is the index of refraction for the block of quartz?"
I think I read somewhere that quarts refracts at 1.458, but I don't know how to work out the question.
Here's the question;
"A ray of light has an angle of incidence of 36.0 degrees on a block of quartz and an angle of refraction of 20.0 degrees. What is the index of refraction for the block of quartz?"
I think I read somewhere that quarts refracts at 1.458, but I don't know how to work out the question.
Answer:
refractive index = sine(incident angle)/sine(refractive angle)
=(sin 36)/(sin 20) =1.719
this does not match with experimental data because question may have been formed arbitrarily
=(sin 36)/(sin 20) =1.719
this does not match with experimental data because question may have been formed arbitrarily
How do you find the index of refraction of the material?
A beam of light of wavelength 553 nm travel-ing in air is incident on a slab of transparent
material.
Given: The incident beam makes an angle of 48.1 ◦ with the normal, and the refracted beam makes an angle of 26.3 ◦ with the nor-mal.
Find the index of refraction of the material.
material.
Given: The incident beam makes an angle of 48.1 ◦ with the normal, and the refracted beam makes an angle of 26.3 ◦ with the nor-mal.
Find the index of refraction of the material.
Answer:
Use Snell's Law with n1 = 1
so 1*sin(48.1) = n2*sin(26.3)
So n = sin(48.1)/sin(26.3) = 1.68
so 1*sin(48.1) = n2*sin(26.3)
So n = sin(48.1)/sin(26.3) = 1.68
How do you use refraction index to explain this experiment?
My teacher put a clear test tube in some refractory fluid, and you couldn't see it. All i know is that it has something to do with the refraction index. i don't know how to explain how the refraction index played a role in this. Can someone explain this to me, and why?
Answer:
Let's look at a similar case. If you have eyeglasses, the lenses are very very transparent, but they clearly bend light. We know the bending takes place at the interface between air and the lens material and the amount of bending depends on the different indices of refraction of the two materials (air and lens material). If you pop the lens out and hold it in your hand, you can still see it. But if you were to drop the lens into a liquid with the same index of refraction, there would be no bending (if you know snell's law, imagine that n1 and n2 are the same) of light. And the lens would 'vanish'. Optically, the lens is the same as the surrounding liquid.
How do you calculate the index of refraction of water?
The speed of light is 2.25 x 10^8 in water, how do I use this to find the index of refraction of water?
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