Question

In the diffraction from a single slit of width $2.5\lambda$, the total number of minimas and secondary maximum (maxima) on either side of the central maximum are :

• A. $4$ minimas, $2$ secondary maximas
• B. $2$ minimas, $2$ secondary maximas
• C. $4$ minimas, $4$ secondary maximas
• D. $4$ minimas, $3$ secondary maximas

Answer 1

Answer: (A)

$4$ minimas, $2$ secondary maximas

Given, width of single slit, $b=2.5\lambda$
We know that, for minima
$b\sin \theta =n\lambda$
$\sin \theta =\frac{n\lambda }{b}$
$=\frac{n\lambda }{2.5\lambda }$
$\sin \theta =\frac{n}{2.5}$
Since maximum value of $\sin \theta$ is $1$.
So, only $n=1,2$
Thus only $4$ minima can be obtained on the both sides of the screen.
For secondary maxima
$b\sin \theta =(2n+1)\frac{\lambda }{2}$
$\sin \theta =\frac{(2n+1)\lambda }{2b}$
$=\frac{2n+1}{2\times 2.5}$
$\sin \theta =\frac{2n+1}{5}$
Since maximum value of $\sin \theta$ is $1$.
So, only $n=1$, thus only $2$ secondary maxima can be obtained on the both sides of the screen.