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: Figure 1, the initial situation

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 $a <<D$, $\frac{a^2}{D^2}$ 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<<D \implies \tan \theta \ $ 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:

Here, a point Q is dropped onto SP such that OQ is perpendicular to SP

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.

  • $\begingroup$ 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. $\endgroup$
    – Meripadhai
    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. :)


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