# The equation of diffraction

I have learned the equation of diffraction, which is: $$\sin \theta=\frac{\lambda}{d},$$ where $\lambda$ is the wavelength and $d$ is the gap width.

So, when the gap width is closer to the wavelength, wave diffracts the most. But what if the gap width is less than wavelength, so how is the equation possible, as the value of the sine is greater than 1?

The intensity of the light in the diffraction pattern from a double slit is given by:

$$I(\theta) = \cos^2\left(\frac{\pi d \sin\theta}{\lambda}\right) \tag{1}$$

We get maxima when the $\cos$ on the right side of the equation is $\pm 1$, and this happens when:

$$\frac{\pi d \sin\theta}{\lambda} = \pi n$$

for the integer $n = 0, \pm 1, \pm 2$ and so on. A quick rearrangement gives the positions of the maxima as:

$$\sin\theta = n \frac{\lambda}{d}$$

Which is where we get your expression for the spacing of the fringes. To see what happens as we decrease the slit spacing $d$ towards $\lambda$ let's use equation (1) to draw a graph of the intensity for various values of $d$:

For this graph I've set $\lambda$ to 500nm and I've graphed the intensity for four values of $d$ from 10$\mu$m to 500nm. The graph shows that as we decrease $d$ towards $\lambda$ the spacing gets bigger and bigger until at $d = \lambda$ the spacing becomes so big that we only get one fringe. Remember that $\theta$ can't be less than $-\pi/2$ or greater than $\pi/2$ i.e. the maximum fringe spacing possible is $\pi$.

And that's why your equation does't have any solutions for $d < \lambda$. It's because the spacing betwen the fringes gets greater than $\pi$ and there is only one fringe.

This is one of the classic examples where the electromagnetic theory of light fails to answer. When the slitwidth is less than the wavelength the above equation is no more valid. Actually when the slitwidth is so small the probability of light to pass through the slit becomes vanishingly zero.

• On the contrary, the EM theory of light does answer this problem just not in the quasi-optical limit. To answer the problem properly you have to take into account the full EM theory of wave scattering from arbitrary sized object. This is done routinely by numerical EM simulation programs such as HFSS, Ansys, etc. Sep 10 '15 at 12:21