# Double-Slit Experiment: Effect of Intensity Reduction on Fringes

Monochromatic light passes through a double-slit arrangement. The intensity of the monochromatic light passing through one of the slits of the double-slit arrangement is reduced. State, and explain, the effect of this change on the appearance of the bright fringes and of the dark fringes.

It makes sense that the brighter fringes will reduce in brightness due to the fact that the sum of the amplitudes will not be as high a number as before.

However, I'm a little confused as for the effect on the appearance of darker fringes. I thought that because darker fringes means 0 light intensity, I thought that even after the reduction in intensity of the monochromatic light, destructive interference will still take place at some point, therefore giving rise to no change in appearance to the dark fringes.

However, the answer to the above question says "dark fringes will be brighter / less dark because summing amplitudes no longer gives zero"

Could someone please explain the flaw in my conceptual understanding?

• The maximum/minimum intensity of the fringe pattern will be $I_1\pm I_2$. Where $I_{1,2}$ is intensity on slit 1 and slit2. If both intensity are equal then only minima is having 0 intensity. Jun 15, 2016 at 5:02
• In a nutshell: If you decrease the intensity of the incident light by x percent, then the intensity at every point on the screen will be decreased by x percent. May 3, 2017 at 18:14

Suppose the amplitude of the wave from slit $1$ arriving at a point is $A_1$ and the amplitude of the wave from slit $2$ arriving at the same point is $A_2$ and let $A_2>A_1$.

The relative phase between the waves from the two slits arriving at that point depends on the path difference between the slits and the point and not on the amplitude of the waves.

So if the path difference is an integer number of wavelengths then the two waves will arrive exactly in phase with one another and the resultant amplitude will be $A_2+A_1$ which means that the intensity will be proportional to $(A_2+A_1)^2$ at a maximum.

If the path difference is such that the waves arriving at a point are exactly out of phase with one another the resulting amplitude is $A_2-A_!$ which means that the intensity is proportional to $(A_2-A_1)^2$ at a minimum.
The minimum intensity is only zero if the amplitudes of the two waves is the same $(A_2=A_1)$.

So starting from equal intensity (amplitude) from each slit by decreasing the intensity from one slit you decrease the maximum intensities of the interference pattern but increase the minimum intensities of the fringe pattern - the contrast between the bright and dark fringes decreases.

I'm not sure if my understanding is correct, but i hope this would help!

When it comes to destructive interference, usually we would be given a situation where two coherent waves move in opposite direction. However, the resultant displacement only becomes zero when both have the same amplitude.

As an illustration, a wave of amplitude 2a, moving in positive direction, interfere with another wave of amplitude 2a, moving in negative direction. Thus, the resultant displacement due to the destructive interference is zero.

So, for this case where the dark fringes becomes brighter, we need to know this.

When the intensity of ONE of the fringes is reduced, the amplitude of ONE of the fringes is also reduced (I = A²) .

So, using the same example as above, assuming one of the amplitude of the waves is reduced, the wave of amplitude 2a, moving in positive direction, interfere with the other wave of amplitude 1a (I assume the amplitude of this wave is reduced), moving in negative direction, the resultant displacement would be ( 2a - 1a) 1a.

As you can see, the resultant displacement increases from zero (from first example) to 1a (from second example). Since resultant displacement of wave indicates the brightness of light wave, increase in resultant displacement of the waves indicates that the dark fringes becomes brighter.

Thank you!

• All you have written in your answer is correct but the two coherent waves need not necessarily move in opposite direction for destructive interference. It would be correct to say that two waves will destructively interfere and the resultant would come 0 if they of off-phase. Apr 30, 2017 at 2:49
• Oh my, I'm sorry for the mistake. I'll keep in mind about this. Thanks for correcting me!
– Myra
Apr 30, 2017 at 11:15

The bright fringes will get darker as the intensity of the bright fringes is when the two beams interact constructively and the intensity becomes proportional to the sum of the two beam intensities. In the destructive interference case the light beam intensity becomes proportional to the difference of the two beam intensities which if equal will lead to absolutely dark fringe otherwise the fringes for destructive interference will have a non-zero intensity.

$I = I_1 + I_2 + 2 \sqrt{(I_1~ I_2)} \cos\theta$ , thus $I$ is zero only when $I_1 = I_2$ and $\theta = (2n+1)\pi\;.$

Suppose intensity of one light is $I_1=2$ and another light $I_2=2$, when they intefere destructively, the resulatant is zero and hence dark fringe is formed.

But when intensity of $i_1$ becomes 1, their destructive inteference results in (2-1)=1, so what was once a dark fringe has now the intensity of 1.

NOTE: Although amplitudes cancel each other out instead of intensity, it is easier to understand the concept with intensity.

They become less dark because as you know the intensity in one slit decreases and so the amplitude decreases in that slit. Complete destructive interference occurs when the amplitudes from both slits are the same as they completely cancel out. However, if one of the amplitudes out of the two sources is reduced while the other is kept the same , they do not completely cancel out, hence the resultant wave at that point still has some amplitude, even after destructive interference, unlike before when the amplitudes completely cancelled out. This amplitude and hence resultant displacement makes the dark fringes from before less dark.