Chapter 47: Problem 5

Question: Light of wavelength $\lambda~=~600~{\rm nm}$ is incident normally on a diffraction grating. Two adjacent principle maximia occur at $\sin\theta~=~0.20$and $\sin\theta~=~0.30$. The fourth order is missing.

1.

What is the seperation between adjacent slits?

2.

What is the smallest possible slit width?

3.

Name all the orders actually appearing on the screen with the values derived in parts (a) and (b).

Solution: Firstly, we can figure out the order of the intereference fringes from $d\sin\theta_k~=~k\lambda$, and taking the ratio of two adjacent maxima gives

$${\sin\theta_k\over\sin\theta_{k+1}}~=~{k\over k+1}~=~{0.2\over 0.3}$$

from which we conclude k = 2. Given, this we then can determine the seperation between adjacent slits to be

$$d~=~{2\lambda\over\sin\theta_2}~=~6~{\rm\mu m}$$

We can determine the angular position of the fourth order fronge to be

$${\sin\theta_4\over\sin\theta_2}~=~{4\over 2}$$

and hence $\sin\theta_4~=~2\sin\theta_2~=~0.4$. Now since the fourth order interference fringe is missing, we know that we have a diffraction minima at that angular position. Therefore, the minimum slit width that could give rise to a missing fourth interference fringe is when the first diffraction minimum is at that location, $a\sin\theta_1~=~\lambda$, and hence

$$a~=~{1\lambda\over\sin\theta_4}~=~1.5~{\rm\mu m}$$

The highest order interference fringe occurs at $\sin\theta_{\rm max}~=~1$, and thus $d~=~k_{\rm max} \lambda$, from which we now can determine $k_{\rm max}~=~10$. Given that the 4th fringe is absent, therefore, there is also a diffraction minimum located precisely at the 8th fringe also. Hence, present are interference fringes 1,2,3,5,6,7,9,10.