# 10.3.2 Derivation of the dsinθ formula

For the double-slit, the path difference δ can be calculated using this $\delta =d\sin \theta$ formula. The detailed derivation of this formula is arguably not required by the H2 syllabus. But the geometrical trick involved is so fascinating I have to share it with you.

Remember that in O-level optics, you were taught that rays originating from a point at infinity can be treated as parallel rays when they arrive at the lens. But why?

• Consider the circular sector PAB with sector angle β.
• If we make β really small, the sector would be so squeezed it looks like a thin line.
• If we were to zoom into the arc AB, we would see that the radii AP and AB appear as parallel lines, and the arc AB a straight line.

Basically, APB has become an isosceles triangle with angles 90°, 90° and 0°. Amazing right? This “parallel ray approximation” is at the centre of the $\delta =d\sin \theta$ derivation.

• Consider light waves departing from slits A and B, arriving at destination P on the screen. Note that PAB’ is a circular section, which means PA and PB’ are radii and AB’ is a circular arc. However, this diagram is not reflective of the double-slit dimensions. The distance between the slits, called the slit separation d, is usually less than 1 mm. The distance between the slits and the screen, called the screen distance L, is seldom less than 30 cm.
• If drawn to scale, AOB would look like a single point. And AP, OP and BP all make the same angle θ with the centre line.
• If we zoom into the double-slits, we would see that the two radii PA and PB’ are practically parallel lines, and the arc AB’ is practically a straight line at right angle to PA and PB’. (APB’ is an isosceles triangle with angles 90°, 90° and 0°.)

From the right angled triangle ABB’, we obtain $\text{BB }\!\!'\!\!\text{ }=d\sin \theta$.  But $\text{BB}'=\left| {\text{AP}-\text{BP}} \right|$ is actually the path difference! Hence $\delta =d\sin \theta$!

In general, to calculate the path difference $\displaystyle \delta =|\text{AP}-\text{BP}|$ , we must first calculate AP and BP, which can be quite tedious. However, for double-slits, the path difference can be calculated directly using $\delta =d\sin \theta$ (as long as $L\gg d$ so the parallel ray approximation holds). Hence

Bright fringes are formed at angles θ where $\displaystyle d\sin \theta =n\lambda$ , $n=0,1,2...$

Dark fringes are formed at angles θ where $\displaystyle d\sin \theta =(n+\frac{1}{2})\lambda$  , $n=0,1,2...$

Video Explanation

Why is the path difference equal to dsinθ?