Question 49: Wave Optics

Question

In a Young's double slit experiment, the distance between the two identical slits is 6.1 times larger than the slit width. Then the number of intensity  maxima observed within the central maximum of the single slit diffraction pattern is :

Accepted Answer

In a double slit experiment, the bright spots (maxima) from interference are found where:
Ã¢ÂÂÃ¢ÂÂ$d\,\sin 	heta = m\lambda$,Ã¢ÂÂwhere $m$ can be 0, ÃÂ±1, ÃÂ±2, and so on.
The dark spots (minima) from single slit diffraction appear at:
Ã¢ÂÂÃ¢ÂÂ$a\,\sin 	heta = \pm\lambda$.
For a bright interference fringe to be inside the central bright area of the single slit (central maximum), it must fall between these two first diffraction minima.
This means we need:
Ã¢ÂÂ$|m\lambda/d| \leq \lambda/a$,  or  $|m| \leq d/a$.
The problem says the distance between the slits ($d$) is 6.1 times the width of each slit ($a$), so $d/a = 6.1$.
The biggest whole number for $m$ is 6, so there are maxima for $m = \pm1, \pm2, ..., \pm6$.
If we count only the bright spots on the sides (not the middle one), that's 6 on one side and 6 on the other, making 12 bright maxima.
If the central bright spot ($m=0$) is counted too, there would be 13, but usually Ã¢ÂÂwithin the central maximumÃ¢ÂÂ means only the fringes between the two minima, not the central line.
So, you will see 12 bright interference fringes inside the central maximum of the single slit envelope.
Answer = 12. (Option C)

Answer

3

Answer

6

Answer

12

Answer

24

Wave Optics

In a Young's double slit experiment, the distance between the two identical slits is 6.1 times larger than the slit width. Then the number of intensity maxima observed within the central maximum of the single slit diffraction pattern is :

Choose the correct answer: