Optics B/C

[Adi1008]

What is the smallest time delay required between two waves of 400nm light to obtain complete destructive interference?

You can answer I just wanted to give it a shot.

I'm getting... (Click to Show Hidden Text)

6.67*10^-16 seconds. This is using the distance over time definition of the speed of light with a 200nm distance to create fully destructive interference.

That's correct; your turn!

[Tom_MS]

That's correct; your turn!

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.

[Sean_Sylvester1]

That's correct; your turn!

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.

Using the lensmakers formula and its derivation for thick lenses, I came up with an answer of 13.42 diopters

[Tom_MS]

That's correct; your turn!

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.

Using the lensmakers formula and its derivation for thick lenses, I came up with an answer of 13.42 diopters

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]

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.

Using the lensmakers formula and its derivation for thick lenses, I came up with an answer of 13.42 diopters

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.

Is it 64 diopters then? Since the equivalent focal length is the inverse of power and then subtract .09 from that

[Tom_MS]

Using the lensmakers formula and its derivation for thick lenses, I came up with an answer of 13.42 diopters

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.

Is it 64 diopters then? Since the equivalent focal length is the inverse of power and then subtract .09 from that

Keep in mind that the front surface radius will not be the same as the distance to the principal plane.

[Sean_Sylvester1]

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.

Is it 64 diopters then? Since the equivalent focal length is the inverse of power and then subtract .09 from that

Keep in mind that the front surface radius will not be the same as the distance to the principal plane.

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 so

[Tom_MS]

Is it 64 diopters then? Since the equivalent focal length is the inverse of power and then subtract .09 from that

Keep in mind that the front surface radius will not be the same as the distance to the principal plane.

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 so

Nice! Your turn

[Sean_Sylvester1]

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

[Sean_Sylvester1]

correction I meant to say concave mirror