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The maximum number of possible interference maxima for slit-separation equal to twice the wavelength in Young's double-slit experiment, is
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five
For possible interference maxima on the screen, the condition is
$d \sin \theta=n \lambda\ldots(i)$
Given : $d=$ slit - width $=2 \lambda$
$\begin{aligned}
& \therefore & 2 \lambda \sin \theta & =n \lambda \\
\Rightarrow & & 2 \sin \theta & =n
\end{aligned}$
The maximum value of $\sin \theta$ is 1 , hence,
$n=2 \times 1=2$
Thus, Eq. (i) must be satisfied by 5 integer values $i e,-2,-1,0,1,2$. Hence, the maximum number of possible interference maxima is 5 .
$d \sin \theta=n \lambda\ldots(i)$
Given : $d=$ slit - width $=2 \lambda$
$\begin{aligned}
& \therefore & 2 \lambda \sin \theta & =n \lambda \\
\Rightarrow & & 2 \sin \theta & =n
\end{aligned}$
The maximum value of $\sin \theta$ is 1 , hence,
$n=2 \times 1=2$
Thus, Eq. (i) must be satisfied by 5 integer values $i e,-2,-1,0,1,2$. Hence, the maximum number of possible interference maxima is 5 .
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