# In the double slit experiment with light, we know that the spacing of the interference fringes...

## Question:

In the double slit experiment with light, we know that the spacing of the interference fringes (which is what we call the bright spots on the screen) depends on the wavelength of the light. How? Specifically, if you make the wavelength of the light bigger, how does the spacing of the fringes change?

## The Young's Double Slit Experiment

In Young's double-slit experiment, light with the same frequency emerging from two slits is allowed to undergo coherent addition. In other words, monochromatic light is allowed to superpose with a fixed phase difference. If a screen is inserted in the region of interference then at points where the path difference from the slits is an integral multiple of the wavelength we have constructive interference. These points are bright. At points where the path difference is an odd multiple of half-wavelength, there is destructive interference and these points are dark. Thus an alternately dark and bright fringe pattern is observed on the screen. The separation between two consecutive bright (or dark) fringes is called the fringe width.

If the separation between the slit plate and the screen is {eq}\displaystyle {D} {/eq}, the separation between the slits is {eq}\displaystyle {d} {/eq} and the wavelength is {eq}\displaystyle {\lambda} {/eq} then the fringewidth or the separation between the fringes is given by,

{eq}\displaystyle { \beta=\frac{\lambda D}{d}} {/eq}.

Clearly, if the wavelength is made bigger then the fringe width too will increase linearly. The fringe width may be enhanced also by increasing D or by decreasing d. Effectively the arrangement functions as an amplifier of the wavelength. The amplification factor is {eq}\displaystyle {\frac{D}{d}} {/eq}. 