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## Derive an expression for intensity distribution due to Fraunhofer diffraction at a single slit.

Answer:

**Fraunhofer Diffraction by Single Slit**

There are some assumptions that must be made for this description of the * single slit diffraction pattern*:

- The slit size is small, relative to the wavelength of light; The screen is very far.; Cylindrical waves can be represented in 2D diagrams as circular waves.; The intensity at any point on the screen is independent of the angle made between the ray to the screen and the normal line between the slit and the screen (this angle is called T below). This is possible because the slit is narrow. (Fraunhofer
*Diffraction*by Single Slit}

Let a monochromatic ray incident normally on a single slit UV and width of the slit is w. According to Huygen’s principle the wave front divide into two components of equal amplitude α. The inclined angle is q with the direction of incident beam are focused at another point. Now we want to measure the resultant intensity at point S, we draw a perpendicular UD; it is known that the optical paths of the waves travelled after the plane UD to the point S are equal.

From figure 2 the intensity and resultant amplitude at point S obtain by vector method. Let β be the phase difference then in triangle KBP