Thanks Scott! The prism analogy for lens shaping was a big revelation. I could never ground focal length of lenses to anything conceptual other than the abstractions of memorized formulas. This is taking off the aperture stop!
Answer 1: Taking the asumption that index of refraction is proportional to speed of bike on pavement divided by the speed of bike on the beach, the index of refraction for the beach is 1.5384. Then we can compute the angle based on the snell's law n1*sin(theta1) = n2*sin(theta 2), we know n1=1, theta1 is 30 degrees, n2=1.5384 thefore theta2 is 18.966 degrees. Answer 2: Basically we used the equation D=A*(n-1)= 18.6degrees*(2-1)= 18.6 degrees.
Scott. If you shine a HeNe laser into a pool, illuminating the side of the pool, you see a red spot. What if you dunk your head under the water and looked at the same spot? What color would it be? If color was a function of wavelength, then you would see a blue spot instead of red. But if you do this experiment, you still see a red spot, right? So color is not wavelength dependent. It’s frequency dependent. Both wavelength and frequency get smaller by the same amount when light enters a denser medium. You know this, but I had to figure it out. So many people associate color with wavelength, but that’s wrong. Color is a function of frequency. P
Assume the index of the pavement is 1.0. The index of sand is a ratio of the velocities. Recall, Index >1.0. Then use Snell's law to compute the angle (hint: angle will be smaller than on the pavement or
Thanks for all of the videos for geometrical optics and applications with zemax from your site! Words cannot express how useful this material. I was wondering if there was a place on the Optics Realm site that has pdf versions of the slides you presented for your video series. I couldn't seem to find this on the site, but this would be great to have so that I can have a quick reference to these very useful notes.
Loving the tutorial, just one minor thing so far, is the frequency of light not around 600 Terra Hertz not Giga Hertz, I'm concerned that my new Mac proccessor might start to light up if gets close to 600Ghz ;0)
Scott thanks for uploading this, it's very helpful. would you explain why the frequency of photon or light doesnt change in different matter (so the wavelength should change to maintain the speed)?
Wavelength is not equal to 1/f. (Remember wavelength is a measure of distance not time) But Wavelength = c/f, so when the speed reduces, the wave length reduces proportionately. In other words wavelength is the distance traveled by the wave in one cycle. So when the speed of the wave reduces, it's wavelength reduces.
Here's a tip; proofread your punctuation, before uploading tutorials to the internet or else your English teachers will break out the red pens, haha! FYI - The Chandra space telescope uses several nested, cylindrical mirrors to focus X-rays onto a detector for NASA. The technical information and the images are fascinating! Thanks for the waves.
I understand that this video is a bit old *cough* 8+ years old *cough*, but I was wondering what the answer to the second question was? Is it 9.176694816770244206698807138084° ? If it is wrong, can you please elaborate? (I am not that good with math, it took me almost a minute to figure out that this video was 8+ years old 🤣)
Thanks Scott! The prism analogy for lens shaping was a big revelation. I could never ground focal length of lenses to anything conceptual other than the abstractions of memorized formulas. This is taking off the aperture stop!
Answer 1: Taking the asumption that index of refraction is proportional to speed of bike on pavement divided by the speed of bike on the beach, the index of refraction for the beach is 1.5384. Then we can compute the angle based on the snell's law n1*sin(theta1) = n2*sin(theta 2), we know n1=1, theta1 is 30 degrees, n2=1.5384 thefore theta2 is 18.966 degrees.
Answer 2: Basically we used the equation D=A*(n-1)= 18.6degrees*(2-1)= 18.6 degrees.
Perfect
It's nice when two things you enjoy, physics and photography, complement each other so well.
Thanks so much Scott, I really appreciate this great tutorial
I learned a lot! Now, I'll go on to part 2.
Great and unique video.
Very useful to amateur photographers.
I love how I understand this all
....... the frequency of 0.5um visible light is 600THz, not GHz!!!!!!!!
Cool so far, exited for part2. (orig, reason I found your vid but deiced to binge lol)
Scott. If you shine a HeNe laser into a pool, illuminating the side of the pool, you see a red spot. What if you dunk your head under the water and looked at the same spot? What color would it be?
If color was a function of wavelength, then you would see a blue spot instead of red. But if you do this experiment, you still see a red spot, right?
So color is not wavelength dependent. It’s frequency dependent. Both wavelength and frequency get smaller by the same amount when light enters a denser medium.
You know this, but I had to figure it out. So many people associate color with wavelength, but that’s wrong. Color is a function of frequency.
P
Thank you for this great tutorial! You're good teacher!
Well organised lectures
Assume the index of the pavement is 1.0. The index of sand is a ratio of the velocities. Recall, Index >1.0. Then use Snell's law to compute the angle (hint: angle will be smaller than on the pavement or
Very cool videos. Thanks dude
Good stuff. Your window beam displacement at 7:30 is in the wrong direction though.
This is very helpful! Thanks!
Thanks for all of the videos for geometrical optics and applications with zemax from your site! Words cannot express how useful this material. I was wondering if there was a place on the Optics Realm site that has pdf versions of the slides you presented for your video series. I couldn't seem to find this on the site, but this would be great to have so that I can have a quick reference to these very useful notes.
Loving the tutorial, just one minor thing so far, is the frequency of light not around 600 Terra Hertz not Giga Hertz, I'm concerned that my new Mac proccessor might start to light up if gets close to 600Ghz ;0)
Perfect
Nice explanations.. minor error in the slide... visible light is 100s of THz.. Not GHz
How do you find the index of refraction with the information given in question #1?
Thank you...........
thanks sir
Scott thanks for uploading this, it's very helpful. would you explain why the frequency of photon or light doesnt change in different matter (so the wavelength should change to maintain the speed)?
I don't understand this either. If you have lower wavelength then you should have a greater frequency (i.e. WL = 1/f) -- Right?
Wavelength is not equal to 1/f. (Remember wavelength is a measure of distance not time)
But Wavelength = c/f, so when the speed reduces, the wave length reduces proportionately.
In other words wavelength is the distance traveled by the wave in one cycle. So when the speed of the wave reduces, it's wavelength reduces.
I am not and engineer but I fint this very interesting.
Here's a tip; proofread your punctuation, before uploading tutorials to the internet or else your English teachers will break out the red pens, haha! FYI - The Chandra space telescope uses several nested, cylindrical mirrors to focus X-rays onto a detector for NASA. The technical information and the images are fascinating! Thanks for the waves.
I understand that this video is a bit old *cough* 8+ years old *cough*, but I was wondering what the answer to the second question was?
Is it 9.176694816770244206698807138084° ?
If it is wrong, can you please elaborate? (I am not that good with math, it took me almost a minute to figure out that this video was 8+ years old 🤣)
I thought it was 18.6, because D = A(n-1) -> D = 18.6(2-1) -> D = 18.6(1). I don't know the answer to the first question though