Recent content by rinalai

Thanks for pointing out the mistake. Yes, by using the formula Spectral width= (800x10^-9 m)^2 x (4.41x10^12 sec)/3x10^8(m/s) = 9.4nm could finally be derived. Thank you so much for patiently guiding through this question. It had been a difficult one for me :oldshy:

Thanks, Steve4Physics, for your kind comments. Here is my answers and the professor's feedback(in italic): 1.6x10^14(Hz) corresponds to 9.4 nm g. Peak power is pulse energy per pulse duration: 12.5 nJ/100 fs = 125 kW h. Focus is the same as in h. So we want 125 kW from the sun. Since...

Here is the full question: The circled ones are those I am still trying to figure out :oldeyes:

0.3µm/fs is the speed of light c=3x10^8 m/s= 0.3µm/fs 100fs x 0.3µm/fs = 30 µm (sorry that I mistyped) Is this the spectral width of the pulse? Besides, I think the pulse-shape was a Gaussian pulse. Should I go for FWHM in this case?

Thank you for pointing out the rounding error. Recalculated cycles of pulse is 37.

There are a few more questions on the same laser: mean power: 1 W pulse duration: 100 fs repetition rate: 80 MHz center wavelength: 800 nm What is the (1)spectral width (2)peak power of a pulse ? For spectral width, I tried calculation as 100fs x 0.3µm/fs = 100 µm but it seemed incorrect. I...

Thanks, Steve4Physics,Tom.G, jbrggs, SammyS, for your comments. I tried again: The wavelength of an electromagnetic wave is 800nm Q1. For interest, in what part of the EM spectrum is the wave? A. infrared Q2. Calculate the wave’s frequency in Hz. A. f=c/λ = 3x10^8 m/s ÷ (800nm = 8x10^-7 m) =...

Q1. For interest, in what part of the EM spectrum is the wave? A. electrical part Q2. Calculate the wave’s frequency in Hz. A. frequency f = repetition rate: 80 MHz= 8x10^7 Hz Q3. Calculate the wave’s period (i.e. the time for 1 optical cycle) in seconds. A. period T=1/f = 1/8x10^7 =...

I utilized an Ultrashort Pulse Calculator on a website (https://www.rp-photonics.com/ultrashort_pulses.html) and got the bandwidth of 9.39nm with the specifications listed above. Then divided the wavelength by the bandwidth, and it came out 85(optical cycles). However, I am not sure if it is...

Here is my answer to this question: Stokes shift is the difference in wavelength between positions of the band maxima of the excitation and emission spectra of the same electronic transition. When Stokes shift is large, it means there is more energy loss, which is not favorable regarding...

Thank you so much for helping out! I had a discussion with my professor, and here is the explanation from him: "Even with the highest possible resolution of optical microscopes, the size of the spheres cannot be resolved; both are sub-resolution spheres. Can they be seen? Yes, since they emit...

Thank you so much for helping out :smile:

The power of the microscope was not given by the professor... It seems that there's some requirements missing in the question.

I wonder if this question is about two-point resolution or magnification... Would someone here be so kind as to offer some guidance on this question? Thank you!