Physics 552 Optical Spectroscopy Problem Set 1, Assignments of Optics

Download Physics 552 Optical Spectroscopy Problem Set 1 and more Assignments Optics in PDF only on Docsity! Problem set 1 of Physics 552 optical spectroscopy 2008 1) The usual way to define the width of a spectrum is the full width at half height. A) Calculate the natural width (in frequency, cycles per second) of the emission line of two different electron oscillators that emit at 300 nm and at 500 nm. B) Change these frequency widths to wavelengths. C) Compare the widths of the spectra to the width of a hydrogen atom in its ground state (the Bohr radius). D) Which oscillator has the smallest width on the wavelength scale? E) Look up on the web the “classical width of the electron”. Is the width of a spectrum in the visible larger or smaller that the classical with of an electron. F) What kind of light would have a wavelength on this scale? 2) When we calculated the rate of emission induced by an oscillator that is continuously driven, st emW , we called this “induced emission”. We also calculated the rate of absorption (induced absorption), stabsW . We then found the ratio of the two. Calculate this ratio (show the rates of the two, and then the ratio) to show that it must be <1 (you cannot emit from a forced oscillator more rapidly than the rate at which the oscillator absorbs energy from the forcing function; that is, from the oscillating electric field). Explain why it is less than one. 3) In the lecture (see the notes) we defined the Number of charges (electron oscillators) per volume to be N (we will assume this is the number of charges per cm3). We defined the complex index of refraction to be ( )'n n iκ= + . Thus, κ is the absorption coefficient of the electric field per wavelength. A) What is the absorption coefficient of the intensity per wavelength? Also give your reason. B) What is therefore the usual absorption coefficient of a solution per cm, in terms of κ ? C) Now calculate the actual absorption coefficient of a solution with 1 molar concentration of electron oscillators. Assume that the oscillator strength is 0.3. 4) The following parts of this question refer to the lecture notes on the home page. A. Calculate the value of μ for an electron oscillator corresponding to a frequency of 600 nm light (this would be 0ω ). B. Calculate the value of Q (we called this the “quality factor of the oscillator). C. Remembering what 'ω is, give an interpretation of Q. D. If n waves pass by an observer in time τ , what is the frequency ν of the wave train? E. If you are counting the waves, you cannot make any better precision than 1± wave. This sets the error in your estimate of the number of waves passing by. For the decaying electronic oscillator of the lecture, the decay time of the emitted light pulse is the time it takes for the oscillator to decay to 1 e of its original amplitude. Using this estimate for the time τ of the emitted light pulse, and using the results you calculated above, what is the approximate spread in frequency νΔ of this emitted light pulse? This represents the uncertainty of the emitted light frequency (remember the free oscillator – no damping - is oscillating exactly at the frequency corresponding to 600 nm light – that is, no uncertainty). F. Calculate the spread of the emitted spectrum in nanometers. G. Extra credit: Using the resolution of the BEST spectrograph you can find (such as the gratings you used in the lab), calculate whether you could resolve the width of this emitted spectrum. H. Say you have a grating spectrometer in your lab that can resolve 0.01 nm. Using the expression ( ) ' 2 em η μ ω + = , what does the ratio of η μ have to be, in order for you to be able to resolve the width of the emitted spectrum? (we have not discussed η very much, but it is a nonradiative contribution to the damping of the oscillator – for instance, collisions.) I. Now, go to the notes where we calculated the width of the spectrum by using the Fourier transforms. Does this estimate of the width of the spectrum agree with what you calculated above? J. Does the value of the oscillator strength affect the width of the emitted spectrum? Why or why not? 5) Homework question on page 4 of the Scattering lecture notes. 6) Homework question on page 5 of the Scattering lecture notes. 7) The cross section of a molecule for absorption is defined as the opaque area that the molecule presents to the incident light beam. For N molecules per ml, each with a cross section of σ , a layer of thickness dx would present a total black-out area of N dxσ . So, the beam of light would decrease in intensity by dI I N dxσ= − . Calculate the cross section of the Cy7 dye at 743 nm has -1 1250,000 M cmε −= .