enter image description here

$p$ = Path difference

$r$ = Distance travelled by the rays

$x$ = Perpendicular distance between interference of the rays to the medium point of the incident rays

$a$ = Vertical distance between the incident rays

$D$ = Horizontal distance

I knew that

$r_2-r_1 = p \tag{1} $

$r_1^2 = D^2+(x-a/2)^2 \tag{2}$

$r_2^2 = D^2+(x+a/2)^2 \tag{3}$

$(3)-(2) = r_2^2-r_1^2 = (r_2-r_1)(r_2+r_1)=2ax \tag{4}$

$(4) = (r_2-r_1)(r_2+r_1) = 2pD \tag{5}$

$$(4) = (5) = 2pD = 2ax$$

Hence path difference $p =ax/D$

The question is why $(r_2+r_1) = 2D$ ?

  • $\begingroup$ That's a condition for a max or a min where the path difference is either $N\lambda$ or $ (N+\frac{1}{2})\lambda$ . If you do a little trigonometry you should be able to see where the similar triangles are. BTW "derivative" has a different meaning, which is why I changed the title. $\endgroup$ Commented Oct 6, 2015 at 12:12
  • 1
    $\begingroup$ Who says that $r_1+r_2=2D$? In general this is false (though it's true to leading order in $x$). Are you following a specific reference that claims this? Linking to it would help clarify the question. $\endgroup$ Commented Oct 6, 2015 at 12:17
  • $\begingroup$ @EmilioPisanty The reference came out from my school, sorry. $\endgroup$
    – ytan11
    Commented Oct 6, 2015 at 12:57

2 Answers 2


Here's how I go about it. Let's call the middle point of the 2 slits point $A$. Also, let's call the angle between the line joining $A$ to the point of the central maxima ( on the screen, horizontally infront of $A$) and $A$ to the point on the screen under observation $\theta$.


Now, since the slits and the distance between $a$ them are very very small as compared to the distance $D$ between the slits and the screen, the 2 light paths $r_1$ and $r_2$ may be considered almost parallel.

path difference

In that case the path difference becomes $a sin\theta$ as is evident from the image. On the other hand, from the bigger triangle, $sin\theta \approx tan\theta = \frac{x}{D}$ because $\theta$ is very small.

So you get path difference $p = \frac{ax}{D}$.

There are a lot of approximations, but they work well under suitable conditions. (When $\theta$ is very small, that is, $x \ll D$ and $a \lll D$.)


Here we should have to relate the slit width w,distance from the screen D ,wave length lambda Where the wave length is the path difference w/wave length=x/D

  • $\begingroup$ Using a trigonometric approximation is very important, we can approximate for example $\endgroup$ Commented Jan 18, 2017 at 4:20

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