Keep in mind that the front surface radius will not be the same as the distance to the principal plane.Sean_Sylvester1 wrote:Is it 64 diopters then? Since the equivalent focal length is the inverse of power and then subtract .09 from thatTom_MS wrote: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.Sean_Sylvester1 wrote:
Using the lensmakers formula and its derivation for thick lenses, I came up with an answer of 13.42 diopters
Optics B/C
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Re: Optics B/C
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Re: Optics B/C
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 soTom_MS wrote:Keep in mind that the front surface radius will not be the same as the distance to the principal plane.Sean_Sylvester1 wrote:Is it 64 diopters then? Since the equivalent focal length is the inverse of power and then subtract .09 from thatTom_MS wrote: 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.
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Re: Optics B/C
Nice! Your turnSean_Sylvester1 wrote: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 soTom_MS wrote:Keep in mind that the front surface radius will not be the same as the distance to the principal plane.Sean_Sylvester1 wrote:
Is it 64 diopters then? Since the equivalent focal length is the inverse of power and then subtract .09 from that
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Re: Optics B/C
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
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Re: Optics B/C
correction I meant to say concave mirror
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Re: Optics B/C
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.Sean_Sylvester1 wrote: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
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Re: Optics B/C
Correct! Your turnTom_MS wrote: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.Sean_Sylvester1 wrote: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
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Re: Optics B/C
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?Sean_Sylvester1 wrote:Correct! Your turn
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Re: Optics B/C
Tom_MS wrote: 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?
Malus' law states I = I0cos(x). In this case, 0.25=1*cos(arcsin(1/t))^2. Solving the expression for a positive t, we find t=1.1547 seconds.
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Re: Optics B/C
Correct. Your turn.jkang wrote:Tom_MS wrote: 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?Malus' law states I = I0cos(x). In this case, 0.25=1*cos(arcsin(1/t))^2. Solving the expression for a positive t, we find t=1.1547 seconds.
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