Optics B/C

[John Richardsim]

I want a mirror that can give me an image that is upright and has a height that is less than the object's height. What type of mirror do I need and where must the object be placed?

Answer (Click to Show Hidden Text)

You need a diverging mirror. The object can be placed anywhere in front of the mirror

Ja boi, you got it. Your turn.

[jonboyage]

You have a sheet of glass that has n=1.6. You spray a thin coating (n=1.3) on the glass. A light containing photons with 265.837kJ/mol of energy is shined from a vacuum (n=1) onto the glass with the thin coating at 0 degrees to the normal. How thick does the coating need to be for there to be complete destructive interference with the light reflecting off of the glass and off of the coating?

[kenniky]

You have a sheet of glass that has n=1.6. You spray a thin coating (n=1.3) on the glass. A light containing photons with 265.837kJ/mol of energy is shined from a vacuum (n=1) onto the glass with the thin coating at 0 degrees to the normal. How thick does the coating need to be for there to be complete destructive interference with the light reflecting off of the glass and off of the coating?

answer (Click to Show Hidden Text)

265.387 kJ/mol = 4.407 x 10^-19 J/photon

thus wavelength = 4.516*10^-7 m = 451.6 nm

2*1.3*t + 451.6/2 = (1+1/2)451.6 ==> t = [b]173.7 nm[/b]?

[jonboyage]

You have a sheet of glass that has n=1.6. You spray a thin coating (n=1.3) on the glass. A light containing photons with 265.837kJ/mol of energy is shined from a vacuum (n=1) onto the glass with the thin coating at 0 degrees to the normal. How thick does the coating need to be for there to be complete destructive interference with the light reflecting off of the glass and off of the coating?

answer (Click to Show Hidden Text)

265.387 kJ/mol = 4.407 x 10^-19 J/photon

thus wavelength = 4.516*10^-7 m = 451.6 nm

2*1.3*t + 451.6/2 = (1+1/2)451.6 ==> t = [b]173.7 nm[/b]?

You're so close! (Click to Show Hidden Text)

173.7nm would be the total distance that the wave travels, in order to get lined up with the 1/2 phase of the other reflection (since complete destructive interference is caused by the two waves being exactly opposite each other, requiring them to be 1/2 of a phase apart from each other). Since the wavelength of the wave in the thin film is about 346.7nm, 1/2 of the phase would be your answer, 173.7nm (I got 173.4nm). Because the wave travels back and forth, to get the answer. you have to divide by 2 again, to get about [b]86.7nm[/b], which is the answer

Sorry this was so late
Your turn!

[kenniky]

darn, I need to figure out what all the equations in my binder actually mean

Question:

Light with intensity I is passed through five subsequent polarizing films with axes of polarization 0°, 15°, 60°, 85°, and 89° (in order). What is the intensity of the light after it passes through all of these films?

*assume the light is perfectly unpolarized

[jonboyage]

darn, I need to figure out what all the equations in my binder actually mean

Question:

Light with intensity I is passed through five subsequent polarizing films with axes of polarization 0°, 15°, 60°, 85°, and 89° (in order). What is the intensity of the light after it passes through all of these films?

*assume the light is perfectly unpolarized

Answer (Click to Show Hidden Text)

This is just doing Malus's law over and over again. 
Since the light is perfectly unpolarized from the start, the first polarizer at 0° will make it 0.5*I
2nd polarizer: 0.5I*cos^2(15°) = 0.4665I
3rd: 0.4665I*cos^2(60) = 0.1166I
4th: 0.1166I*cos^2(85) = 0.00088591I
5th: .00088591I*cos^2(89) = [b]0.0000002698I[/b]

[kenniky]

darn, I need to figure out what all the equations in my binder actually mean

Question:

Light with intensity I is passed through five subsequent polarizing films with axes of polarization 0°, 15°, 60°, 85°, and 89° (in order). What is the intensity of the light after it passes through all of these films?

*assume the light is perfectly unpolarized

Answer (Click to Show Hidden Text)

This is just doing Malus's law over and over again. 
Since the light is perfectly unpolarized from the start, the first polarizer at 0° will make it 0.5*I
2nd polarizer: 0.5I*cos^2(15°) = 0.4665I
3rd: 0.4665I*cos^2(60) = 0.1166I
4th: 0.1166I*cos^2(85) = 0.00088591I
5th: .00088591I*cos^2(89) = [b]0.0000002698I[/b]

hmm you may want to check that... not correct although you have the right idea

[jonboyage]

darn, I need to figure out what all the equations in my binder actually mean

Question:

Light with intensity I is passed through five subsequent polarizing films with axes of polarization 0°, 15°, 60°, 85°, and 89° (in order). What is the intensity of the light after it passes through all of these films?

*assume the light is perfectly unpolarized

Answer (Click to Show Hidden Text)

This is just doing Malus's law over and over again. 
Since the light is perfectly unpolarized from the start, the first polarizer at 0° will make it 0.5*I
2nd polarizer: 0.5I*cos^2(15°) = 0.4665I
3rd: 0.4665I*cos^2(60) = 0.1166I
4th: 0.1166I*cos^2(85) = 0.00088591I
5th: .00088591I*cos^2(89) = [b]0.0000002698I[/b]

hmm you may want to check that... not correct although you have the right idea

Whoops! (Click to Show Hidden Text)

I totally forgot to take into account that once it's polarized the first time I needed to change the angle. The answer didn't look right to me :/ Lets try again:
2nd: .5I*cos^2(15) = .4665I
3rd: .4665I*cos^2(45) = .2333I
4th: .2333I*cos^2(25) = .1916I
5th: .1916I*cos^2(4) = [b].1907I[/b]

Did I do it right this time?

[kenniky]

Whoops! (Click to Show Hidden Text)

I totally forgot to take into account that once it's polarized the first time I needed to change the angle. The answer didn't look right to me :/ Lets try again:
2nd: .5I*cos^2(15) = .4665I
3rd: .4665I*cos^2(45) = .2333I
4th: .2333I*cos^2(25) = .1916I
5th: .1916I*cos^2(4) = [b].1907I[/b]

Did I do it right this time?

Almost right (Click to Show Hidden Text)

You don't need to multiply it by anything the first time, unpolarized light doesn't lose any intensity when it becomes polarized. Only after it's polarized does it lose intensity after subsequent polarizations.

So the answer would be twice what you got (0.3815 I)

You got the idea though, so feel free to ask a question

[jonboyage]

Whoops! (Click to Show Hidden Text)

I totally forgot to take into account that once it's polarized the first time I needed to change the angle. The answer didn't look right to me :/ Lets try again:
2nd: .5I*cos^2(15) = .4665I
3rd: .4665I*cos^2(45) = .2333I
4th: .2333I*cos^2(25) = .1916I
5th: .1916I*cos^2(4) = [b].1907I[/b]

Did I do it right this time?

Almost right (Click to Show Hidden Text)

You don't need to multiply it by anything the first time, unpolarized light doesn't lose any intensity when it becomes polarized. Only after it's polarized does it lose intensity after subsequent polarizations.

So the answer would be twice what you got (0.3815 I)

You got the idea though, so feel free to ask a question

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