Why do two different methods to calculate the path difference yield different results?

HCV Concepts(Physics) 17.26

Note that $$\Delta$$ means triangle

Consider a point source of light S, which sends two rays of light of wavelength $$\lambda$$ (order of magnitude $$10^{-7}$$ m) - one to point O (directly in front of it) and another to point P, which is at a distance $$a$$ (order of magnitude $$10^{-4}$$ m) from O. Points O and P lie on a screen and the ray SO is perpendicular (normal) to the screen, while ray SP makes an angle $$\theta$$ with ray SO. SO = $$D$$ (order of magnitude $$10^{0}$$ m), such that $$D$$ >> $$a$$. A diagram of this is given below: It is given in the question that the path difference between the two rays is $$\frac{\lambda}{4}$$ (i.e. SP $$= D+ \frac{\lambda}{4}$$ $$\implies$$ SP - SO = $$\frac{\lambda}{4})$$ and we are told to express $$a$$ in terms of $$D$$ and $$\lambda$$. Using two different approaches, I get two different results.

Approach 1: Because $$\Delta$$SOP is a right angled triangle, and SP is the hypotenuse,

SP = $$\sqrt{D^2+a^2}$$ $$\implies \frac{\lambda}{4} = \sqrt{D^2+a^2} - D \\ \frac{\lambda}{4} = D\sqrt{1+\frac{a^2}{D^2}} - D \\ \frac{\lambda}{4} = D(1+\frac{a^2}{D^2})^{\frac{1}{2}} - D$$

Because $$a <, $$\frac{a^2}{D^2}$$ is very small, meaning that $$(1+\frac{a^2}{D^2})^{\frac{1} {2}} \approx 1+\frac{a^2}{D^2} \times \frac{1} {2} = 1+\frac{a^2}{2D^2}$$

Using this, $$\frac{\lambda}{4} = D(1+\frac{a^2}{2D^2}) - D \\ \frac{\lambda}{4} = D+ D\frac{a^2}{2D^2} - D \\ \frac{\lambda}{4} = \frac{a^2}{2D} \\ a^2 = \frac{ \lambda \times 2D} {4} \\ a^2 = \frac{ D \lambda} {2} \\ a = \sqrt{ \frac{ D \lambda} {2}}$$

This ($$\sqrt{ \frac{ D \lambda} {2}}$$) is also the answer given in the textbook.

But, using an approach often used in diffraction and interference questions, I got a different answer...

Approach 2: $$\tan \theta = \frac{a}{D}$$ Because $$a< is very small. Because $$\tan \theta \$$ is very small, $$\tan \theta \approx \theta$$

Now, another right triangle can be constructed by drawing a line segment (OQ) which is perpendicular to SP and passes through O: From the diagram, it can be seen that $$\angle$$ POQ is also equal to $$\theta$$. As $$\theta$$ is so small, $$\angle$$ SOQ is nearly $$\frac{\pi}{2}$$ ($$90^\circ$$), making $$\Delta$$ SOQ nearly an isosceles triangle and making SQ $$\approx$$ SO = $$D$$. Given SQ = SO, the path difference, SP-SO, is equal to QP, i.e. QP = $$\frac{\lambda}{4}$$

In $$\Delta$$ POQ $$\implies \sin \theta = \frac{\left(\frac{\lambda}{4} \right)}{a} \\ \sin \theta = \frac{\lambda}{4a}$$

Now, because $$\theta$$ is small, $$\sin \theta \approx \theta$$ (I could have just done $$\sin \theta \approx \tan \theta$$ instead of $$\tan \theta \approx \theta$$ and $$\sin \theta \approx \theta$$)

$$\implies \frac{\lambda}{4a} = \frac{a}{D} \\ \frac{\lambda \times D}{4} = a^2 \\ \therefore a = \sqrt{ \frac{D \lambda}{4} }$$

As I have already mentioned, this is not the answer which is given in my textbook. Although I make a few more approximations in approach 2 than in approach 1, in which I only make 1 assumption, it is worth noting that many of the formulae for single slit and double slit interference have been derived using similar methods in my textbook.

As the two answers vary quite significantly - by a factor of $$\sqrt 2$$, both answers cannot be simultaneously correct, and I don't think that simply approximating more times in approach would bring about such an error. I don't know for sure which of the two approaches is correct even though the textbook's answer matches that of approach 1. I would greatly appreciate an answer which clarifies which answer is correct, and why the approach that lead to the other answer is wrong. Thank you for your time.

• HCV Concepts(Physics) is my textbook, HC Verma's "Concepts of Physics". I slightly modified question 26 a) of the exercises for chapter 17 (light waves) of the same book. A simple and accurate answer would be greatly appreciated. Oct 13 '20 at 16:21

The answers are different because you have, in the course of two successive approximations in your second method, unknowingly assumed that θ/2 = θ.

It can be shown that the exact path difference is a tan(θ/2).

In approach one, the path difference is a(tanθ)/2.

While that in that 2nd one is a sinθ, which is way off compared to the first.

Clearly the first is a much better approximation than the second, as θ approaches 0.

Thus, while both reach the same limit while θ approaches 0.00, it is simply that θ/2, and hence the 1st method, is a better approximation than θ.

You can correct the second approach by accounting for the fact that if Δ SOQ is isosceles, then angle QOP is θ/2 (exact) rather than θ, and then calculate. You will now find that the error in actual path difference and your answer is negligible even in terms of θ.

Have a nice day. :)