Fabry Perot Interferometer

[ayesha203]

Design a Fabry perot interferometer, that provides a transmission of T =1 for light with λ1 = 499nm under orthogonal incidence and a transmission for λ2 = 500nm of T≤0.01 (without any further maxima in between). How big has to be the finesse?

Draw the transmission curve of several Fabry - Perot interferometer with this particular geometry (i.e. the same optical path length) for different finesse - values of F-1, 10 and 100 in the range from 498 nm to 501nm.

 

[BvU]

hello Ayesha,

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[ayesha203]

I have tried to solve the above question in the following way:

T = 1 for λ1 = 499nm

T ≤ 0.01 for λ2 = 500nm

for orthogonal incidence : Δs = 2nd, n = 1

Δs = m (λm)

m=1; n=1

d= λ1/2 = 249.5 nm

and i couldn't solve further...

Please help :)

 

[BvU]

It will be difficult to build such an instrument. 249.5 nm is too small, and a deviation of +0.5 nm would already let the 500 nm through with the same T. A hair is 50 μm !

You need an expression for the transmission of such a thing as a function of the wavelength. Nicely to be accommodated under 2): relevant equations...

 

[HalfLostAndSearching]

The Fabry-Perot interferometer is a powerful tool used in optics to measure the properties of light, such as wavelength and intensity. It consists of two parallel, partially reflective mirrors separated by a fixed distance, forming a resonant cavity. When light is incident on the mirrors, it undergoes multiple reflections and interference, resulting in a characteristic transmission spectrum.

To design a Fabry-Perot interferometer that meets the given specifications, we need to calculate the required finesse. Finesse is a measure of the spectral resolution of the interferometer and is defined as the ratio of the free spectral range (FSR) to the full width at half maximum (FWHM) of the transmission peak. In this case, we want the transmission to be 1 for a specific wavelength (499nm) and less than 0.01 for a neighboring wavelength (500nm).

The FSR of a Fabry-Perot interferometer is given by FSR = c/2nd, where c is the speed of light, n is the refractive index of the medium between the mirrors, and d is the distance between the mirrors. For a transmission of 1 at 499nm, we can calculate the required value of d to be approximately 249.5nm. This will result in a FSR of 299.8nm for light with a wavelength of 499nm.

To achieve a transmission of less than 0.01 at 500nm, we need to calculate the required FWHM. This can be done using the formula FWHM = λ2^2/FSR, where λ2 is the wavelength at which the transmission needs to be less than 0.01. Plugging in the values, we get a FWHM of 0.167nm for light with a wavelength of 500nm.

The finesse can then be calculated as F = FSR/FWHM, which in this case gives a value of approximately 1800. This means that the interferometer must have a finesse of at least 1800 to meet the given specifications.

To visualize the effect of varying finesse on the transmission spectrum, we can draw the transmission curves for different values of finesse (F-1, 10, and 100) in the range of 498nm to 501nm. As shown in the attached diagram, as the finesse increases, the transmission peaks become narrower and taller, resulting in a