Learning Forum

[uzmayaqoob1960]

In Franhaufer diffraction theory in your paid courses i read this statement" the pupil has a hard edge, its Fourier transform cannot have a hard
edge, and it ripples out to infinity".plz explain it what does it mean?

[Akhil Dutt Vijayakumar]

Hi,

Thank you for posting in the forum,

I am giving clarification to the few terms used in the statement.

Pupil with a hard edge: The “pupil” refers to the opening through which light passes in an optical system, like the iris in your eye or the aperture in a camera lens.

“Hard edge” means that the pupil has a clear, abrupt transition from open to closed area. Unlike a blurry or gradual transition, a hard edge has a sharp distinction between where light can pass and where it cannot.

When light passes through an aperture with a hard edge, it diffracts, meaning it bends and spreads out slightly beyond the geometric edges of the aperture. This diffraction creates a characteristic pattern of light and dark areas, called the diffraction pattern.

The “hard edge” of the pupil cannot directly translate into a “hard edge” in its diffraction pattern. Because of the inherent wave nature of light, diffraction creates a smooth, gradual transition between light and dark regions in the pattern, even if the aperture itself has a sharp edge. This transition forms the “ripples” that extend towards infinity.

Hope that Helps.
Akhil

[uzmayaqoob1960]

Many Thanks akhil .
Can you plz let me know one more thing as i posted earlier but couldn't get a satisfactory answer.I am working on zemax spectroscopy tutorial as provided in this link i couldn't get why -1 diffractionorder whats the science behind selecting it not zero diffraction order ?.If you could clarify i will be really obliged.

Link from where the following pic is copied https://support.zemax.com/hc/en-us/articles/1500005578762-How-to-build-a-spectrometer-theory

[Akhil Dutt Vijayakumar]

Hi,

 

Thank you for posting in the forum,

 

The 0th diffraction order carries the most intense light, usually appearing as a peak in the center of the diffraction pattern. However, for precise analysis of a specific wavelength, this order is often not ideal.

The 0th order contains the desired wavelength information along with light that passed straight through (undiffracted), this inclusion of other wavelengths dilutes the spectral purity of the target wavelength. This makes it difficult to accurately measure its intensity and characteristics. 

Therefore, most spectrometers prioritize the information in the higher diffraction orders, primarily the -1st order, for its concentrated peak and purer spectral content.

 

Hope that helps.

Akhil

[uzmayaqoob1960]

many thanks akhil your efforts are much appreciated.You respond with good clarity.WE can even go for +1 diffraction order? or why he stick with -ve diffraction order?

[Akhil Dutt Vijayakumar]

 

Hi,

Thanks for the follow-up,

I can add some more clarification on this aspect.

The spectrometer design shown here uses a transmission grating and this configuration is called Littrow spectrometer. Here the grating is already placed at an angle, also the incident light angle is equal to diffraction angle (α = β = 33.367°). 

In this specific setup, the angled grating (33.367°) allows the diffracted beam (-1st order) to retrace its path back through the same collimating and focusing lens. This simplifies alignment and reduces the number of optical components needed.

In the Littrow configuration the -1st order typically has the highest intensity compared to other orders, maximizing signal strength at the detector. Also, the -1st order is further away from the undiffracted beam and scattered light, minimizing their interference with the desired signal.

In general, the -1 order is preferred in spectrometers demanding high intensity and stray light reduction in the diffraction pattern, but depending on the configuration and type of spectrometer(reflection or transmission) other diffraction orders (+1 and higher orders) are also used.

Hope you understand clearly.

Akhil

 

 

[uzmayaqoob1960]

yes many thanks Akhil.I really appreciate your efforts.