Optics B/C

[kenniky]

Are you sure? (Click to Show Hidden Text)

Perfectly unpolarized light does lose half of its intensity when going through the first polarizer. According to Wikipedia, "A beam of unpolarized light can be thought of as containing a uniform mixture of linear polarizations at all possible angles. Since the average value of cos^2  is 1/2, the transmission coefficient becomes I/I_o=1/2"

Next question (Click to Show Hidden Text)

An upright image is reduced to 1/4 of the object’s height when the object is placed 26.9 cm from the lens. What is the focal length of the lens

Huh, I took a test that said differently... weird

Either way...

answer (Click to Show Hidden Text)

the distance of the image is 26.9/4 = 6.725 cm

Upright so it's virtual so distance is negative

1/f = 1/29.6 + 1/-6.725 = 1/29.6 - 4/29.6 = -3/29.6

So f = 29.6/-3 = - 9.86666... cm

[jonboyage]

Are you sure? (Click to Show Hidden Text)

Perfectly unpolarized light does lose half of its intensity when going through the first polarizer. According to Wikipedia, "A beam of unpolarized light can be thought of as containing a uniform mixture of linear polarizations at all possible angles. Since the average value of cos^2  is 1/2, the transmission coefficient becomes I/I_o=1/2"

Next question (Click to Show Hidden Text)

An upright image is reduced to 1/4 of the object’s height when the object is placed 26.9 cm from the lens. What is the focal length of the lens

Huh, I took a test that said differently... weird

Either way...

answer (Click to Show Hidden Text)

the distance of the image is 26.9/4 = 6.725 cm

Upright so it's virtual so distance is negative

1/f = 1/29.6 + 1/-6.725 = 1/29.6 - 4/29.6 = -3/29.6

So f = 29.6/-3 = - 9.86666... cm

Whoops! (Click to Show Hidden Text)

You accidentally turned 26.9 into 29.6 halfway through. The answer should be -8.966...cm. You did it the right way, though.

Your turn.

[kenniky]

A telescope has a focal length of 1.25 m and an eyepiece diameter of 5.00 mm, what is the magnification?

[Avogadro]

A telescope has a focal length of 1.25 m and an eyepiece diameter of 5.00 mm, what is the magnification?

Seems pretty simple (Click to Show Hidden Text)

250 times?

[CVMSAvalacheStudent]

[quote="kenniky"]A telescope has a focal length of 1.25 m and an eyepiece diameter of 5.00 mm, what is the magnification?[/quote]
[hide]Why/idk[/hide]

[kenniky]

A telescope has a focal length of 1.25 m and an eyepiece diameter of 5.00 mm, what is the magnification?

Seems pretty simple (Click to Show Hidden Text)

250 times?

Correct, your turn

[Tom_MS]

A telescope has a focal length of 1.25 m and an eyepiece diameter of 5.00 mm, what is the magnification?

Seems pretty simple (Click to Show Hidden Text)

250 times?

I'm used to the formula for magnification being the focal length of the objective divided by the focal length of the eyepiece. If you treat the eyepiece diameter as the focal length of the eyepiece, you get this result; however, the focal length should be independent of the diameter for the eyepiece. I looked at what this question may have originated from on wikipedia: https://en.wikipedia.org/wiki/Optical_t ... nification and I believe that the section where it has "(or diameter)" for the magnification formula means that it is the ratio of both diameters, not the focal length of one and the diameter of the other. I found the referenced site for that sentence on wikipedia, and I found this page:http://www.nexstarsite.com/_RAC/form.html. This page leads to this conclusion that I just explained. If you work out the math yourself, you will also see what I mean.

I saw this question on the MIT test, and I really don't think it makes sense.

[kenniky]

I'm used to the formula for magnification being the focal length of the objective divided by the focal length of the eyepiece. If you treat the eyepiece diameter as the focal length of the eyepiece, you get this result; however, the focal length should be independent of the diameter for the eyepiece. I looked at what this question may have originated from on wikipedia: https://en.wikipedia.org/wiki/Optical_t ... nification and I believe that the section where it has "(or diameter)" for the magnification formula means that it is the ratio of both diameters, not the focal length of one and the diameter of the other. I found the referenced site for that sentence on wikipedia, and I found this page:http://www.nexstarsite.com/_RAC/form.html. This page leads to this conclusion that I just explained. If you work out the math yourself, you will also see what I mean.

I saw this question on the MIT test, and I really don't think it makes sense.

Interesting... I will admit my question was blatantly stolen from the MIT Test with different numbers lol

This will be helpful to know at least

[Avogadro]

Here's your question:

The absorption spectra for chlorophyll a and chlorophyll b are shown above. How does this help to explain why leaves appear green?

[kenniky]

Here's your question:

The absorption spectra for chlorophyll a and chlorophyll b are shown above. How does this help to explain why leaves appear green?

ANSWER (Click to Show Hidden Text)

Almost none of the yellow and green light is absorbed. It is instead reflected and picked up by our eyes, so the leaves appear green.