# Is the formula, for finding the distance between two slits, my books gives correct?

In my book'a section on Young's double-slit experiment, the formula, $d = m \lambda \sin\theta$, is given. In this equation $d$ is the distance between two slits, $\lambda$ is the wavelength of light coming through the slits, and $\theta$ is the angle between the central reference to the brightest maximum on the screen opposite the slits.

I am assuming this formula's derivation involves some degree of approximation, because another formula in the same section assumes the distance between the slit and the screen is similar in length to the hypotenuse the picture above.

Using the same approximation, I got something similar but not the same: $$d = \frac{m\lambda}{\sin\theta} \, .$$

Which result is correct??

• Sounds like most of this question is actually not relevant to what you want to know. I think the only question is why the angles in the green and red lines are the same. Could you remove the parts that are not needed for that specific question? We try to make questions as clear as possible by removing irrelevant material. – DanielSank Aug 2 '15 at 23:40
• @DanielSank, I can make another question on mathSE. Asking that. – most venerable sir Aug 2 '15 at 23:42
• Actually one physicsSE since it has a geometry tag, – most venerable sir Aug 2 '15 at 23:57
• What is the question? Are you asking which formula is correct? Are you asking how to derive the correct formula? Be specific. – DanielSank Aug 3 '15 at 0:23

If the slits are on top of each other, then the light travelling through each slit goes the same distance and therefore has the same phase. In this case, the distance between the fringes is infinite. On the other hand, if the slits are very far apart, then even a small angle incurs a large path difference, so the fringes are very close together. Thus we have reasoned that the distance between the fringes goes down as $d$ goes up.

Consider the formula written by OP:

$$d = \frac{m \lambda}{\sin \theta} \, .$$

The first intensity maximum occurs when $m=1$ giving

$$\sin \theta_1 = \frac{\lambda}{d} \, .$$

Expanding the $\sin$ to lowest order we get

$$\theta_1 = \frac{\lambda}{d}$$

which says that increasing $d$ makes the angle of the first maximum smaller, as we predicted above. From this reasoning, we see that OP's formula is probably correct and that putting the $\sin$ in the numerator would give the wrong behavior.

Note that I didn't actually derive the correct answer, I just showed that moving the $\sin$ function from denominator to numerator would probably be incorrect. That said, given the definitions in the question, the correct formula for the maxima of the two slit interference is in fact

$$d = \frac{m \lambda}{\sin \theta}$$

as written by OP.

• So my textbook is wrong! – most venerable sir Aug 3 '15 at 15:17
• @user132522 looks that way. It would be nice if someone else could confirm this. – DanielSank Aug 3 '15 at 15:55
• @DanielSank: Okay, sir, could you tell what you meant by even a small angle incurs a large path difference, so the fringes are very close together? Which angle are you talking of? And also, how by a large path difference, the fringe-width becomes smaller? – user36790 Sep 2 '15 at 8:53
• @user36790 Draw a picture. If the slits are very far apart, then at small $\theta$ (see the picture in the question) the paths from either slit to a point on the screen have very different length. – DanielSank Sep 2 '15 at 14:57
• Okay, if increasing the distance between two slits make the paths larger, how does it lead to decreasing the fringe-width? – user36790 Sep 2 '15 at 15:04