Question

(a) Derive an expression for path difference in Young’s double slit experiment and obtain the conditions for constructive and destructive interferences at a point on the screen.

(b) The intensity at the central maxima in Young's double slit experiment is $I_0$. Find out the intensity at a point where the path difference is $\dfrac{\lambda}{6},\; \dfrac{\lambda}{4},\; \dfrac{\lambda}{3}$.

Answer 1

We can find the conditions for constructive and destructive interference by considering two coherent sources of light passing through the two slits and forming fringes on a screen placed at some distance from the slits. We will find the path difference of the two wavelengths of light and will then find that at what intervals from the central maxima, the constructive or destructive interference will occur. Later in the second part, we can use the intensity formula i.e, $I=I_0 cos^2(\phi /2)$ to determine the intensity value for given path differences by finding their phase differences.

Formula used: Intensity, $I=I_0 cos^2(\phi /2)$

**Complete step by step solution:
**

(a) Let us consider ${S}_{1}$ and ${S}_{2}$ be two narrow slits placed perpendicular to the plane of a paper and a screen is placed on the perpendicular bisector of ${S}_{1}{S}_{2}$, which is illuminated by monochromatic light.

The slits have been placed $d$ distance apart and a screen at a distance D from the slits.

Now, we have a point P on the screen at $x$ distance away from point O on the screen.

Now, the path difference at P between the waves reaching from ${S}_{1}$ and ${S}_{2}$ is $\delta x={S}_{2}P-{S}_{1}P$.

Now, we will draw a perpendicular ${S}_{1}N$ on ${S}_{2}P$

Therefore, $\delta x={S}_{2}P-{S}_{1}P={S}_{2}P-NP={S}_{2}N$

Now, from the triangle ${S}_{1}{S}_{2}N$, we can write that $sin\theta=\dfrac{{S}_{2}N}{{S}_{2}{S}_{1}}$

Therefore, $\delta x={S}_{2}N={S}_{2}{S}_{1}sin\theta =dsin\theta$

Since, here $\theta$ is so small, we can write $sin\theta \simeq \theta =tan\theta =\dfrac{x}{D}$

Thus, path difference $\delta x=\dfrac{xd}{D}$

Now, we can find the condition for constructive interference as

$\dfrac{xd}{D}=n\lambda ;n=0,1,2.....$

Thus, the position of $n{th}$ bright fringe, ${X}_{n}=\dfrac{nD\lambda}{d}$, where n=0 and ${X}_{n}=0$, central bright fringe will be formed at O.

The condition for destructive interference can be given by

$\dfrac{xd}{D}=(2n + 1)\dfrac{\lambda}{2}$, for n = 1, 2…..

(b) The intensity in Young’s double slit experiment is calculated using the formula,

$I=I_0 cos^2 (\dfrac{\phi}{2})$, where $\phi$ is the phase difference.

Phase difference from a certain path difference can be calculated using the formula

$\phi = \dfrac{2\pi}{\lambda}\times \Delta x$, where $\Delta x$ is the path difference.

So, for path difference, $\dfrac{\lambda}{6}$, phase difference $\phi_1 =\dfrac{2\pi}{\lambda}\times \dfrac{\lambda}{6}=\dfrac{\pi}{3}$

Thus, intensity $I_1 = I_0 cos^2\dfrac{\pi}{6}=\dfrac{3I_0}{4}$

Similarly, for path difference, $\dfrac{\lambda}{4}$, phase difference $\phi_2 =\dfrac{2\pi}{\lambda}\times \dfrac{\lambda}{4}=\dfrac{\pi}{2}$

Thus, intensity $I_2 = I_0 cos^2\dfrac{\pi}{2}=0$

Again, for path difference, $\dfrac{\lambda}{3}$, phase difference $\phi_3 =\dfrac{2\pi}{\lambda}\times \dfrac{\lambda}{3}=\dfrac{2\pi}{3}$

Thus, intensity $I_1 = I_0 cos^2\dfrac{\pi}{3}=\dfrac{I_0}{4}$

Note: One more method, we can derive the conditions for constructive and destructive interferences by using the wave equations for the two light sources and finding the phase difference or the path difference. In the second part, conversion of path difference into phase difference is a must and may become a point of mistake. And the zero intensity indicates the existence of a dark fringe.