Re: Optics B/C
Posted: February 26th, 2017, 3:20 pm
Correct! Your turn/UTF-8 U+6211 U+662F wrote:If 700 nm is red and 450 nm is blue, then green?
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Re: Optics B/C
Posted: February 27th, 2017, 4:29 pm
A laser is pointed at mirror A with an angle of incidence of 50 degrees. Mirror A is connected to Mirror B at a 30 degree angle. What is the angle of reflection off of mirror B?
(The mirrors look something like this:
Mirror A
<
Mirror B
)
(The mirrors look something like this:
Mirror A
<
Mirror B
)
Re: Optics B/C
Posted: March 2nd, 2017, 6:23 pm
UTF-8 U+6211 U+662F wrote:A laser is pointed at mirror A with an angle of incidence of 50 degrees. Mirror A is connected to Mirror B at a 30 degree angle. What is the angle of reflection off of mirror B?
(The mirrors look something like this:
Mirror A
<
Mirror B
)
20°?
Re: Optics B/C
Posted: March 3rd, 2017, 6:10 pm
That's what I got, yep!
Re: Optics B/C
Posted: March 4th, 2017, 6:47 am
Cool.
A beam of light enters a prism with an index of refraction of 1.35. If each of the angles of the prism is 60°, what maximum angle must the angle of incidence be so that no light comes out the other side?
A beam of light enters a prism with an index of refraction of 1.35. If each of the angles of the prism is 60°, what maximum angle must the angle of incidence be so that no light comes out the other side?
Re: Optics B/C
Posted: March 5th, 2017, 10:03 am
Avogadro wrote:Cool.
A beam of light enters a prism with an index of refraction of 1.35. If each of the angles of the prism is 60°, what maximum angle must the angle of incidence be so that no light comes out the other side?
Assuming a surrounding index of refraction of 1, arcsin(1/1.35) = 47.8 deg, which requires an angle of refraction of 77.8 deg, so 1*sin(theta) = 1.35*sin(77.8 deg), theta = arcsin(1.35*sin(77.8 deg)), and theta = undefined. Therefore, light always comes out the other side?
Re: Optics B/C
Posted: March 7th, 2017, 7:58 am
Yeah, looks like I accidentally made a trick question lol. Good practice anyway!UTF-8 U+6211 U+662F wrote:Avogadro wrote:Cool.
A beam of light enters a prism with an index of refraction of 1.35. If each of the angles of the prism is 60°, what maximum angle must the angle of incidence be so that no light comes out the other side?Assuming a surrounding index of refraction of 1, arcsin(1/1.35) = 47.8 deg, which requires an angle of refraction of 77.8 deg, so 1*sin(theta) = 1.35*sin(77.8 deg), theta = arcsin(1.35*sin(77.8 deg)), and theta = undefined. Therefore, light always comes out the other side?
Re: Optics B/C
Posted: March 8th, 2017, 11:37 am
Avogadro wrote:Yeah, looks like I accidentally made a trick question lol. Good practice anyway!UTF-8 U+6211 U+662F wrote:Avogadro wrote:Cool.
A beam of light enters a prism with an index of refraction of 1.35. If each of the angles of the prism is 60°, what maximum angle must the angle of incidence be so that no light comes out the other side?Assuming a surrounding index of refraction of 1, arcsin(1/1.35) = 47.8 deg, which requires an angle of refraction of 77.8 deg, so 1*sin(theta) = 1.35*sin(77.8 deg), theta = arcsin(1.35*sin(77.8 deg)), and theta = undefined. Therefore, light always comes out the other side?
It looks like you just used snell's law and the critical angle and didn't use the 60 degree apex angle. Using geometry (which is kind of complicated to explain here), I'm getting the formula of theta=arcsin( n*sin( 60 - arcsin(1/n))). This gives be 16.58 degrees. If this angle was decreased (I think) any more, no light would escape.
Re: Optics B/C
Posted: March 8th, 2017, 12:54 pm
Tom_MS wrote:Avogadro wrote:Yeah, looks like I accidentally made a trick question lol. Good practice anyway!UTF-8 U+6211 U+662F wrote:Assuming a surrounding index of refraction of 1, arcsin(1/1.35) = 47.8 deg, which requires an angle of refraction of 77.8 deg, so 1*sin(theta) = 1.35*sin(77.8 deg), theta = arcsin(1.35*sin(77.8 deg)), and theta = undefined. Therefore, light always comes out the other side?It looks like you just used snell's law and the critical angle and didn't use the 60 degree apex angle. Using geometry (which is kind of complicated to explain here), I'm getting the formula of theta=arcsin( n*sin( 60 - arcsin(1/n))). This gives be 16.58 degrees. If this angle was decreased (I think) any more, no light would escape.
the angle of 77.8 deg
A Science Olympiad Optics supervisor holds up two white flashlights. The first has a magenta filter, and the second has a cyan filter. What color is the overlap of the two flashlights?
Re: Optics B/C
Posted: March 8th, 2017, 2:12 pm
UTF-8 U+6211 U+662F wrote:Tom_MS wrote:Avogadro wrote: Yeah, looks like I accidentally made a trick question lol. Good practice anyway!It looks like you just used snell's law and the critical angle and didn't use the 60 degree apex angle. Using geometry (which is kind of complicated to explain here), I'm getting the formula of theta=arcsin( n*sin( 60 - arcsin(1/n))). This gives be 16.58 degrees. If this angle was decreased (I think) any more, no light would escape.Anyways new question:the angle of 77.8 deg
A Science Olympiad Optics supervisor holds up two white flashlights. The first has a magenta filter, and the second has a cyan filter. What color is the overlap of the two flashlights?
I'm getting the same exact answer of 16.5836 degrees and I made sure it was right by doing it the long way and going through each step that leads to the condensed formula Tom described
I'm not 100% sure about this but is it light blue?