Question

A triple-slit experiment consists of three narrow slits, equally spaced by distance $d$and illuminated by light of wavelength $\lambda$. Each slit alone produces intensity $I_1$on the viewing screen at distance$L$.
$\left(a\right)$Consider a point on the distant viewing screen such that the path-length difference between any two adjacent slits is$\lambda$. What is the intensity at this point?
$\left(b\right)$What is the intensity at a point where the path-length difference between any two adjacent slits is$\raisebox{1ex}{$\lambda $}\!\left/ \!\raisebox{-1ex}{$2$}\right.$?

Short answer

Part $\left(a\right)$

$\left(a\right)$At this time, the intensity is$I_A=9I_1.$

Part $\left(b\right)$

$\left(b\right)$When the path-length difference of two adjacent slits at the intensity is$I_B=I_1.$

Step-by-step solution

Step: 1 Intensity of slit:

The light casts a shadow when the widths of the slits are greater than the wavelength of the light. Light diffraction occurs when the slit widths are small, and the light waves overlap on the screen. As a result, the light intensity rises as the slit width rises.

Step: 2 Finding intensity: (part a)

If the direction difference between any two adjacent sources is $\lambda$, all three sources' light will be in phase. As a result, the intensity will be

$$\begin{gathered} I_A=3^2{I}_1 \\ {I}_A=9I_1. \end{gathered}$$

Step: 3 Finding path-length at intensity: (part b)

If any two neighbouring slits have a path difference of $\raisebox{1ex}{$\lambda $}\!\left/ \!\raisebox{-1ex}{$2$}\right.$, two of them will cancel each other out, resulting in the intensity emanating from a single slit, which is,

$I_B=I_1.$

So, the values are equal.