1
$\begingroup$

I studied that in Young's Double Slit Experiment the variation of intensity ($I$) of the fringes on the screen with respect to the phase difference ($Φ$) is given by :

$I = 4I_{0} \cos^{2}\frac{Φ}{2}$

$I_{0}$ is the intensity of light coming from each slit. At maximas or constructive interference, $Φ = nλ$, where $n$ is any whole number and hence we get $I = 4I_{0}$ Below I have given the image of an interference pattern from a laser beam passing through double slit. As you can see as we move away from the central maxima, the intensity decreases and eventually it becomes zero. But how is this possible? According to our equation, the intensity of the centre all the bright fringes should be $4I_{0}$ and hence we should get equal brightness in all the maximas. But why does the intensity decrease and become zero at some point? Shouldn't the interference pattern extend upto infinity and there should be equal brightness at all the maximas? Please explain. I am so confused. Interference Pattern

$\endgroup$
1

3 Answers 3

2
$\begingroup$

In your formula, $I_0$ is the intensity of wave from either one of the sources, at the point of consideration. Now, as we move further from Central Bright Fringe, $I_0$ decreases too, varying as $I \propto \frac 1r$, if we consider line source (and hence cylindrical wavefront). Hence bright fringes become dimmer.

$\endgroup$
2
  • $\begingroup$ But isn't $I \propto \frac{1}{r^2}$ instead of $I \propto \frac{1}{r}$? Doesn't intensity follow the inverse square law? $\endgroup$ Commented May 16, 2021 at 16:10
  • $\begingroup$ No, that is for point source. In YDSE usually line source is considered. $\endgroup$ Commented May 16, 2021 at 16:46
0
$\begingroup$

Why do you imagine that the intensity would be the same all the way to infinity? That would require an infinite amount of energy to illuminate an infinitely wide screen.

The best way to think about the observed phenomenon is to imagine what you would see if you had only one slit. You would find that the incident light from a single slit would not be constant everywhere on the screen, stretching in infinity in either direction, but would vary, having a maximum directly opposite the single slit and quickly fading away either side of it.

The mistake you have made in your thinking is to assume that the intensity of the incident light is constant everywhere after it passes through the slit.

$\endgroup$
4
  • $\begingroup$ I do understand this but what is basically Io to be plugged in the equation? The way the question is written confuses me as well. $\endgroup$
    – Alchimista
    Commented May 14, 2021 at 10:50
  • $\begingroup$ But Marco Ocram look at the equation I wrote. The equation is basically telling us that intensity is constant at the maximas regardless of the distance from the central maxima as there is no 'distance factor' in the equation. Yes, Alchimista I am facing the same problem. $\endgroup$ Commented May 15, 2021 at 2:04
  • $\begingroup$ @RIPANBARUAH well my doubt was transitory because of the question. Io is that emerging from the slit(s). $\endgroup$
    – Alchimista
    Commented May 15, 2021 at 13:02
  • $\begingroup$ Yes Alchemista, $I_{0}$ is the intensity of light source coming from each slit. $\endgroup$ Commented May 16, 2021 at 16:07
0
$\begingroup$

If you shine a spherical lightwave on the wall, you will observe that away from the line that connects the source with the wall (assuming that the direction of the lightwave is perpendicular to the wall) the intensity of the light will diminish. In the double-slit experiment, you use basically two spherical light waves. The intensities of both waves will drop upon hitting the screen.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.