Optics B/C

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Re: Optics B/C

Postby UTF-8 U+6211 U+662F » February 19th, 2018, 11:02 am

All right, my question killed the marathon, so I'll just post the answer.
Answer
If the refractive index of the lens > 1.33: The focal length increases. If the refractive index of the lens = 1.33: The lens becomes invisible. If the refractive index of the lens < 1.33: The lens becomes diverging.
1) Fill in the blank: Stars moving toward us appear more _ (color).
2) Why is this?
Blue; wavelength decreases due to Doppler shift
Good! Although I wanted you to explain what the Doppler effect is. Your turn anyway. :)

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Re: Optics B/C

Postby jonboyage » February 19th, 2018, 12:05 pm

1. What is the power, in diopters, of a lens with n=1.6, r1=3cm, r2=-10cm, D=1cm?

2. Is the lens: Bi-convex, bi-concave, plano-convex, plano-concave, convexo-concave, or concavo-convex?

3&4. Repeat questions 1 and 2 for a lens with n=1.4, r1=5cm, r2=infinity, D=.25cm.

5. Which of these lenses is more likely to experience more profound spherical of aberration? How do you know?
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Re: Optics B/C

Postby UTF-8 U+6211 U+662F » February 19th, 2018, 12:33 pm

1. What is the power, in diopters, of a lens with n=1.6, r1=3cm, r2=-10cm, D=1cm?

2. Is the lens: Bi-convex, bi-concave, plano-convex, plano-concave, convexo-concave, or concavo-convex?

3&4. Repeat questions 1 and 2 for a lens with n=1.4, r1=5cm, r2=infinity, D=.25cm.

5. Which of these lenses is more likely to experience more profound spherical of aberration? How do you know?
Oh boy...
I'm not exactly sure what n, r1, r2, and D mean, but I'm going to assume they're the refractive index, the radius of the surface closest to the light, the radius of the surface farthest from the light, and the thickness of the lens respectively. I'm also going to assume that you're using the Cartesian sign convention. Now that we have the necessary info, we can begin the problem: 1) [math](1.6-1)\left(\frac{1}{3 cm} - \frac{1}{-10 cm} + \frac{(1.6-1)(1 cm)}{1.6(3 cm)(-10 cm)}\right) = 0.2525 cm^{-1} * \frac{100 cm}{1 m} = 25.25 m^{-1} = 25.25 \textrm{ diopters or } 3 * 10^1 \textrm{ diopters.}[/math] 2) Bi-concave 3)[math](1.4-1)\left(\frac{1}{5 cm} - \frac{1}{\infty} + \frac{(1.4-1)(0.25 cm)}{1.4(5 cm)(\infty)}\right) = 0.08 cm^{-1} * \frac{100 cm}{1 m} = 8 m^{-1} = 8 \textrm{ diopters.}[/math] 4) Plano-concave 5) Lens 1: It has two spherical sides while lens 2 has one.

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Re: Optics B/C

Postby jonboyage » February 19th, 2018, 1:02 pm

1. What is the power, in diopters, of a lens with n=1.6, r1=3cm, r2=-10cm, D=1cm?

2. Is the lens: Bi-convex, bi-concave, plano-convex, plano-concave, convexo-concave, or concavo-convex?

3&4. Repeat questions 1 and 2 for a lens with n=1.4, r1=5cm, r2=infinity, D=.25cm.

5. Which of these lenses is more likely to experience more profound spherical of aberration? How do you know?
Oh boy...
I'm not exactly sure what n, r1, r2, and D mean, but I'm going to assume they're the refractive index, the radius of the surface closest to the light, the radius of the surface farthest from the light, and the thickness of the lens respectively. I'm also going to assume that you're using the Cartesian sign convention. Now that we have the necessary info, we can begin the problem: 1) [math](1.6-1)\left(\frac{1}{3 cm} - \frac{1}{-10 cm} + \frac{(1.6-1)(1 cm)}{1.6(3 cm)(-10 cm)}\right) = 0.2525 cm^{-1} * \frac{100 cm}{1 m} = 25.25 m^{-1} = 25.25 \textrm{ diopters or } 3 * 10^1 \textrm{ diopters.}[/math] 2) Bi-concave 3)[math](1.4-1)\left(\frac{1}{5 cm} - \frac{1}{\infty} + \frac{(1.4-1)(0.25 cm)}{1.4(5 cm)(\infty)}\right) = 0.08 cm^{-1} * \frac{100 cm}{1 m} = 8 m^{-1} = 8 \textrm{ diopters.}[/math] 4) Plano-concave 5) Lens 1: It has two spherical sides while lens 2 has one.
Almost...
1. Correct 2. It's actually convexo-concave since r1 is positive, aka "(" shape, and r2 is negative, meaning "(" shape again. 3. Correct 4. Oh so close. It's plano-convex because r1 is positive, aka "(" shape followed by "|" shape. 5. I was looking for an answer that incorporated the diopters. Since lens 1 has higher diopter values, it will experience more spherical aberration.
Your turn.
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Re: Optics B/C

Postby UTF-8 U+6211 U+662F » February 19th, 2018, 1:26 pm

1. What is the power, in diopters, of a lens with n=1.6, r1=3cm, r2=-10cm, D=1cm?

2. Is the lens: Bi-convex, bi-concave, plano-convex, plano-concave, convexo-concave, or concavo-convex?

3&4. Repeat questions 1 and 2 for a lens with n=1.4, r1=5cm, r2=infinity, D=.25cm.

5. Which of these lenses is more likely to experience more profound spherical of aberration? How do you know?
Oh boy...
I'm not exactly sure what n, r1, r2, and D mean, but I'm going to assume they're the refractive index, the radius of the surface closest to the light, the radius of the surface farthest from the light, and the thickness of the lens respectively. I'm also going to assume that you're using the Cartesian sign convention. Now that we have the necessary info, we can begin the problem: 1) [math](1.6-1)\left(\frac{1}{3 cm} - \frac{1}{-10 cm} + \frac{(1.6-1)(1 cm)}{1.6(3 cm)(-10 cm)}\right) = 0.2525 cm^{-1} * \frac{100 cm}{1 m} = 25.25 m^{-1} = 25.25 \textrm{ diopters or } 3 * 10^1 \textrm{ diopters.}[/math] 2) Bi-concave 3)[math](1.4-1)\left(\frac{1}{5 cm} - \frac{1}{\infty} + \frac{(1.4-1)(0.25 cm)}{1.4(5 cm)(\infty)}\right) = 0.08 cm^{-1} * \frac{100 cm}{1 m} = 8 m^{-1} = 8 \textrm{ diopters.}[/math] 4) Plano-concave 5) Lens 1: It has two spherical sides while lens 2 has one.
Almost...
1. Correct 2. It's actually convexo-concave since r1 is positive, aka "(" shape, and r2 is negative, meaning "(" shape again. 3. Correct 4. Oh so close. It's plano-convex because r1 is positive, aka "(" shape followed by "|" shape. 5. I was looking for an answer that incorporated the diopters. Since lens 1 has higher diopter values, it will experience more spherical aberration.
Your turn.
Actually, if we're using Cartesian sign convention
The surface for r1 is on the left, so if r1 is positive, the radius is to the right of the surface, making it convex. Inversely, the surface for r2 is on the right, so if r2 is negative, the radius is to the left of the surface, making it also convex. Thus, 2) Biconvex 4) Plano-convex (which can be confirmed with the answers for 1 and 3, as positive power -> converging lens)
Why do dry roads sometimes seem like they are wet to a passing motorist?

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Re: Optics B/C

Postby jonboyage » February 19th, 2018, 2:02 pm

Oh boy...
I'm not exactly sure what n, r1, r2, and D mean, but I'm going to assume they're the refractive index, the radius of the surface closest to the light, the radius of the surface farthest from the light, and the thickness of the lens respectively. I'm also going to assume that you're using the Cartesian sign convention. Now that we have the necessary info, we can begin the problem: 1) [math](1.6-1)\left(\frac{1}{3 cm} - \frac{1}{-10 cm} + \frac{(1.6-1)(1 cm)}{1.6(3 cm)(-10 cm)}\right) = 0.2525 cm^{-1} * \frac{100 cm}{1 m} = 25.25 m^{-1} = 25.25 \textrm{ diopters or } 3 * 10^1 \textrm{ diopters.}[/math] 2) Bi-concave 3)[math](1.4-1)\left(\frac{1}{5 cm} - \frac{1}{\infty} + \frac{(1.4-1)(0.25 cm)}{1.4(5 cm)(\infty)}\right) = 0.08 cm^{-1} * \frac{100 cm}{1 m} = 8 m^{-1} = 8 \textrm{ diopters.}[/math] 4) Plano-concave 5) Lens 1: It has two spherical sides while lens 2 has one.
Almost...
1. Correct 2. It's actually convexo-concave since r1 is positive, aka "(" shape, and r2 is negative, meaning "(" shape again. 3. Correct 4. Oh so close. It's plano-convex because r1 is positive, aka "(" shape followed by "|" shape. 5. I was looking for an answer that incorporated the diopters. Since lens 1 has higher diopter values, it will experience more spherical aberration.
Your turn.
Actually, if we're using Cartesian sign convention
The surface for r1 is on the left, so if r1 is positive, the radius is to the right of the surface, making it convex. Inversely, the surface for r2 is on the right, so if r2 is negative, the radius is to the left of the surface, making it also convex. Thus, 2) Biconvex 4) Plano-convex (which can be confirmed with the answers for 1 and 3, as positive power -> converging lens)
Why do dry roads sometimes seem like they are wet to a passing motorist?
My bad, you're right
I mixed up the negatives and what they mean. I guess that was good practice for me too so I won't mess it up again haha.
Next problem's answer: I think this is what you're referring to but I may be completely off
This is due to a mirage on a hot day. The light from the blue sky refracts upwards as it gets closer to the road because the hotter, less-dense air closer to the ground increases the index of refraction. This makes the road appear as though it is reflecting light from the sky, which is what water on the road would do.
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Re: Optics B/C

Postby UTF-8 U+6211 U+662F » February 19th, 2018, 6:48 pm

Why do dry roads sometimes seem like they are wet to a passing motorist?
Next problem's answer: I think this is what you're referring to but I may be completely off
This is due to a mirage on a hot day. The light from the blue sky refracts upwards as it gets closer to the road because the hotter, less-dense air closer to the ground increases the index of refraction. This makes the road appear as though it is reflecting light from the sky, which is what water on the road would do.
Yep, your turn!

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Re: Optics B/C

Postby jonboyage » February 20th, 2018, 2:40 pm

How do helium-neon lasers work? (Incorporate the explanation of population inversion in your answer)
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Re: Optics B/C

Postby c0c05w311y » February 26th, 2018, 2:24 pm

Answer:
A helium neon laser has a cavity filled with a mixture of helium and neon (85% He) with an anode and cathode at opposite ends plus a reflector at the back, a laser bore tube in the middle for the beam to form in, and an "output coupler" (aka a mirror with 1% transmittance) at the front. When the electricity is turned on, energetic electrons flow from the anode to the cathode and collide with He atoms. This excites the He, and since excited energy levels of He are very close to excited energy levels of Neon, collisions between He and Ne excite the Neon atoms. (Note that the excited He is considered metastable, so it doesn't emit light itself, but it can transfer energy through collision. (HeNe lasers use He because it makes it much easier, otherwise the Ne would mostly only be excited to lower states.)) This produces the population inversion (when more particles are excited than not) and allows for amplification. Light bounces back and forth (oscillates) between the mirrors on either side of the gas cavity, and waves gain energy with each pass: the laser light causes excited Ne atoms to emit more of the same laser light (collimated, coherent, same wavelength). Just to be clear, the light is coming from excited neon, and has a wavelength of 632.991 nm (632.816 in air). also sorry that its hard to tell but i promise its not written in LISP)))))))))))))))))))))))))))))))))))))))))))))))))))))

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Re: Optics B/C

Postby jonboyage » February 26th, 2018, 2:53 pm

Answer:
A helium neon laser has a cavity filled with a mixture of helium and neon (85% He) with an anode and cathode at opposite ends plus a reflector at the back, a laser bore tube in the middle for the beam to form in, and an "output coupler" (aka a mirror with 1% transmittance) at the front. When the electricity is turned on, energetic electrons flow from the anode to the cathode and collide with He atoms. This excites the He, and since excited energy levels of He are very close to excited energy levels of Neon, collisions between He and Ne excite the Neon atoms. (Note that the excited He is considered metastable, so it doesn't emit light itself, but it can transfer energy through collision. (HeNe lasers use He because it makes it much easier, otherwise the Ne would mostly only be excited to lower states.)) This produces the population inversion (when more particles are excited than not) and allows for amplification. Light bounces back and forth (oscillates) between the mirrors on either side of the gas cavity, and waves gain energy with each pass: the laser light causes excited Ne atoms to emit more of the same laser light (collimated, coherent, same wavelength). Just to be clear, the light is coming from excited neon, and has a wavelength of 632.991 nm (632.816 in air). also sorry that its hard to tell but i promise its not written in LISP)))))))))))))))))))))))))))))))))))))))))))))))))))))
That was a really nice, detailed explanation. Now just remember that for the competition :P
Your turn!
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Re: Optics B/C

Postby c0c05w311y » February 27th, 2018, 7:04 am

Consider an f/4 refracting telescope with an aperture diameter of 550mm, and an f/14 telescope.

a. which telescope would be better for taking pictures?
b. which telescope would be better for making precise observations, and why?
c. Name three typical abberations. What is a coma and how do you fix it?
d. If the eyepiece of the f/4 telescope has a focal length of 12mm, what is the magnification of the telescope? What is the angular resolution for IR light?


if i messed something up let me know

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Re: Optics B/C

Postby UTF-8 U+6211 U+662F » February 27th, 2018, 3:14 pm

Consider an f/4 refracting telescope with an aperture diameter of 550mm, and an f/14 telescope.

a. which telescope would be better for taking pictures?
b. which telescope would be better for making precise observations, and why?
c. Name three typical abberations. What is a coma and how do you fix it?
d. If the eyepiece of the f/4 telescope has a focal length of 12mm, what is the magnification of the telescope? What is the angular resolution for IR light?


if i messed something up let me know
Answer
a) If the eyepieces are the same size, the f/14 would have a bigger focal length and thus a bigger magnification and a smaller FOV, so [b]the f/4[/b]. b) [b]The f/14[/b] c) Comatic, spherical, chromatic; A coma is a comet-like tail on a star seen through a telescope. It can be minimized by using a spherical or bestform/aplanatic lens (although spherical and chromatic aberrations are more common in refracting telescopes?). d) 14 = focal length of objective / 550mm. M = 14 * 550mm / 12 mm = [b]642[/b] Angular resolution = 1.220 * 1 mm / 550 mm = [b]0.0022 radians[/b]

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Re: Optics B/C

Postby c0c05w311y » March 1st, 2018, 4:16 pm

Consider an f/4 refracting telescope with an aperture diameter of 550mm, and an f/14 telescope.

a. which telescope would be better for taking pictures?
b. which telescope would be better for making precise observations, and why?
c. Name three typical abberations. What is a coma and how do you fix it?
d. If the eyepiece of the f/4 telescope has a focal length of 12mm, what is the magnification of the telescope? What is the angular resolution for IR light?


if i messed something up let me know
Answer
a) If the eyepieces are the same size, the f/14 would have a bigger focal length and thus a bigger magnification and a smaller FOV, so [b]the f/4[/b]. b) [b]The f/14[/b] c) Comatic, spherical, chromatic; A coma is a comet-like tail on a star seen through a telescope. It can be minimized by using a spherical or bestform/aplanatic lens (although spherical and chromatic aberrations are more common in refracting telescopes?). d) 14 = focal length of objective / 550mm. M = 14 * 550mm / 12 mm = [b]642[/b] Angular resolution = 1.220 * 1 mm / 550 mm = [b]0.0022 radians[/b]

Looks pretty good! sorry ive taken so long to get back to you, I've been sick plus I had my regional competition on wednesday. The reason, according to wikipedia, that higher f numbers are better for precise measurements and such is that it is easier to reduce aberrations. Also, I wasn't really thinking about the magnification for part a but yes, if you assume the diameter is the same, there is more magnification. I was thinking about the fact that if you assume the focal length is the same, the diameter is bigger for a smaller f number, which means you get more light for pictures in a smaller amount of time.


Anyway, your turn!

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Re: Optics B/C

Postby UTF-8 U+6211 U+662F » March 1st, 2018, 7:25 pm

What is the primary difference between gamma and X rays? (Don't say wavelength/frequency/energy since there isn't a defined boundary between gamma and X rays there)

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Re: Optics B/C

Postby Tom_MS » March 2nd, 2018, 5:46 am

What is the primary difference between gamma and X rays? (Don't say wavelength/frequency/energy since there isn't a defined boundary between gamma and X rays there)
Gamma rays are produced by nuclear processes (like nuclear decay) and particle interactions (like annihilation of particle-antiparticle pairs). X-rays are produced by electrons outside of the nucleus.


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