Question

Young's double slit experiment apparatus shown in figure , $D>>d$ and $d=6\lambda$. Find total number of maxima on the screen.

• A. $11$
• B. $10$
• C. $5$
• D. $6$

Answer 1

The correct option is A $11$

Since $D>>d$, the path difference is,
$\Delta x=d\sin \theta$
$(\Delta x{)}_{min}=0$ for $\theta =0^{\circ }$
$(\Delta x{)}_{max}=d$ for $\theta ={90}^{\circ }$

Since, $d=6\lambda$
$\therefore (\Delta x{)}_{max}=6\lambda$

So, maximas are formed at,
$\Delta x=0,\pm \lambda ,\pm 2\lambda ,\pm 3\lambda ,\pm 4\lambda ,\pm 5\lambda$

$\Delta x=\pm 6\lambda$ is not possible because angle $\theta ={90}^{\circ }$cannot be obtained on the screen.
$\therefore$ There will be five maximas each, on both sides of the central maxima.

So, including central maxima, total number of maxima$=11$.

Hence, $\left(A\right)$ is the correct answer.

Why this question? To learn to find the extent of interference pattern on the screen.