Problem

# A collimated light beam is incident normally on three very narrow, identical slits. At the...

A collimated light beam is incident normally on three very narrow, identical slits. At the center of the pattern projected on a screen, the irradiance is Imax.

a. If the irradiance IP at some point P on the screen is zero, what is the phase difference between light arriving at P from neighboring slits?

b. If the phase difference between light waves arriving at P from neighboring slits is π, determine the ratio IP/Imax.

c. What is IP/Imax at the first principal maximum?

d. If the average irradiance on the entire screen is Iav, what is the ratio IP/Iav at the central maximum?

#### Step-by-Step Solution

Solution 1

(a)

Given that light is incident normally on three very narrow identical slits.

When there are $$N$$ identical slits, then the amplitude phasors form a closed polygon with equal sides and the phase difference between the adjacent amplitudes is $$\frac{2 \pi}{N}$$.

Therefore here for three slits the amplitude phasors makes polygon of 3 sides. Therefore the polygon formed is an equilateral triangle.

Then the phase difference between the adjacent phasors is $$\varphi=\frac{2 \pi}{3}=120^{\circ}$$

(b)

Now the phase difference between light waves arriving at $$\mathrm{P}$$ from neighboring slits is $$\pi$$. Therefore the alternate phasors are opposite to each other. Therefore for the resultant amplitude, out of three amplitudes two will cancel each other and only one will remain.

Therefore the amplitude at point $$\mathrm{P}$$ will be $$E=\left(E_{0}\right)+\left(-E_{0}\right)+\left(E_{0}\right)=E_{0}$$

Therefore the irradiance at point $$P$$ will be proportional to $$E^{2}$$

Therefore $$I_{P}=E_{0}^{2}$$

And the amplitudes reaching the central spot are in phase with each other.

Therefore the amplitude at center will be $$E_{\max }=E_{0}+E_{0}+E_{0}=3 E_{0}$$

Then the irradiance at the center will be proportional to square of $$E_{\max }$$

Therefore,

\begin{aligned} I_{\max } &=\left(3 E_{0}\right)^{2} \\ &=9 E_{0}^{2} \end{aligned}

Then $$\frac{I_{p}}{I_{\max }}=\frac{E_{0}^{2}}{9 E_{0}^{2}}$$

$$I_{p}=\frac{1}{9} I_{\max }$$

(c)

The principal maximum takes place for $$p=0, \pm N, \pm 2 N, \ldots$$

For the first principal maximum $$p=\pm N$$

Then the first principal maximum $$I_{p}=N^{2} I_{0}$$

But we have $$I_{\max }=N^{2} I_{0}$$

Then for principal maximum,

$$\begin{gathered} \frac{I_{P}}{I_{\max }}=\frac{N^{2} I_{0}}{N^{2} I_{0}} \\ I_{p}=I_{\max } \end{gathered}$$

(d)

If $$I_{0}$$ is the irradiance coming from each slit on to the screen, then the average irradiance on the screen will be $$I_{\sigma r}=I_{0}+I_{0}+I_{0}=3 I_{0}$$ (since there are three slits)

But the irradiance at principal maximum $$I_{p}=N^{2} I_{0}$$

\begin{aligned} &I_{P}=(3)^{2} I_{0} \\ &I_{P}=9 I_{0} \\ &\frac{I_{P}}{I_{a v}}=\frac{9 I_{0}}{3 I_{0}} \\ &\frac{I_{p}}{I_{\alpha v}}=3 \\ & I_{p}=3 I_{\text {av }} \end{aligned}