For the double-slit, the path difference *δ* can be calculated using this 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 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 . But is actually the path difference! Hence !

In general, to calculate the path difference , we must first calculate AP and BP, which can be quite tedious. However, for double-slits, the path difference can be calculated directly using (as long as so the parallel ray approximation holds). Hence

Bright fringes are formed at angles *θ* where ,

Dark fringes are formed at angles *θ* where ,

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**Video Explanation**

Why is the path difference equal to *d*sin*θ*?

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