That's correct; your turn!You can answer I just wanted to give it a shot.What is the smallest time delay required between two waves of 400nm light to obtain complete destructive interference?
- I'm getting...
That's correct; your turn!You can answer I just wanted to give it a shot.What is the smallest time delay required between two waves of 400nm light to obtain complete destructive interference?
- I'm getting...
A certain lens has an index of refraction of 1.5, a front lens radius of 0.09 m, and a back surface lens of -0.04 m according to the cartesian sign convention. If it is 0.10 m thick, determine the front vertex power.That's correct; your turn!
Using the lensmakers formula and its derivation for thick lenses, I came up with an answer of 13.42 dioptersA certain lens has an index of refraction of 1.5, a front lens radius of 0.09 m, and a back surface lens of -0.04 m according to the cartesian sign convention. If it is 0.10 m thick, determine the front vertex power.That's correct; your turn!
Almost. To find the front vertex power you need to subtract the equivalent focal length by the distance from the front of the lens to the first principal plane.Using the lensmakers formula and its derivation for thick lenses, I came up with an answer of 13.42 dioptersA certain lens has an index of refraction of 1.5, a front lens radius of 0.09 m, and a back surface lens of -0.04 m according to the cartesian sign convention. If it is 0.10 m thick, determine the front vertex power.That's correct; your turn!
Is it 64 diopters then? Since the equivalent focal length is the inverse of power and then subtract .09 from thatAlmost. To find the front vertex power you need to subtract the equivalent focal length by the distance from the front of the lens to the first principal plane.Using the lensmakers formula and its derivation for thick lenses, I came up with an answer of 13.42 dioptersA certain lens has an index of refraction of 1.5, a front lens radius of 0.09 m, and a back surface lens of -0.04 m according to the cartesian sign convention. If it is 0.10 m thick, determine the front vertex power.
Keep in mind that the front surface radius will not be the same as the distance to the principal plane.Is it 64 diopters then? Since the equivalent focal length is the inverse of power and then subtract .09 from thatAlmost. To find the front vertex power you need to subtract the equivalent focal length by the distance from the front of the lens to the first principal plane.
Using the lensmakers formula and its derivation for thick lenses, I came up with an answer of 13.42 diopters
okay, so I think it would be 80.55 diopters since the power of lens 1 is 5.55 m^-1 and P2 is 12.5 m^-1 soKeep in mind that the front surface radius will not be the same as the distance to the principal plane.Is it 64 diopters then? Since the equivalent focal length is the inverse of power and then subtract .09 from thatAlmost. To find the front vertex power you need to subtract the equivalent focal length by the distance from the front of the lens to the first principal plane.
Nice! Your turnokay, so I think it would be 80.55 diopters since the power of lens 1 is 5.55 m^-1 and P2 is 12.5 m^-1 soKeep in mind that the front surface radius will not be the same as the distance to the principal plane.
Is it 64 diopters then? Since the equivalent focal length is the inverse of power and then subtract .09 from that
Using the principle that all light coming from a point gets focused to the same point (no matter where it reflects), it would make sense that the image would grow a bit dimmer whenever one part of it gets covered. When two parts are covered, it is dimmed more.Alright so here's a quick conceptual question. You have a converging lens and divide the face into 4 equally sized areas. You then try to project an image using the lens. What happens when you place a piece of paper over quadrant 1,2,3 and 4. How about just 1 and 2 , or 1 and 4
Correct! Your turnUsing the principle that all light coming from a point gets focused to the same point (no matter where it reflects), it would make sense that the image would grow a bit dimmer whenever one part of it gets covered. When two parts are covered, it is dimmed more.Alright so here's a quick conceptual question. You have a converging lens and divide the face into 4 equally sized areas. You then try to project an image using the lens. What happens when you place a piece of paper over quadrant 1,2,3 and 4. How about just 1 and 2 , or 1 and 4
Polarized light of intensity I strikes a rotating polarizing film whose angle is given by arcsin(1/t) where t is time in seconds. At one time does exactly 25% of the initial intensity of the polarized light get through the film?Correct! Your turn
Polarized light of intensity I strikes a rotating polarizing film whose angle is given by arcsin(1/t) where t is time in seconds. At one time does exactly 25% of the initial intensity of the polarized light get through the film?
Correct. Your turn.Polarized light of intensity I strikes a rotating polarizing film whose angle is given by arcsin(1/t) where t is time in seconds. At one time does exactly 25% of the initial intensity of the polarized light get through the film?
- No one's tried in a while, so let me.
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