# Derivation of equation of path difference in double slit

$$p$$ = 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 $$(r_2+r_1) = 2D$$ ?

• 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. Oct 6, 2015 at 12:12
• 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. Oct 6, 2015 at 12:17
• @EmilioPisanty The reference came out from my school, sorry. Oct 6, 2015 at 12:57

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.

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

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