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The maximum number of possible interference maxima when slit separation is equal to 4 times the wavelength of light used in a double slit experiment is
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Verified Answer
The correct answer is:
4
In double slit experiment, for possible interference maxima on the screen is given as
$$
\begin{aligned}
d \sin \theta &=n \lambda \\
\text { Given, slit width, } d &=4 \lambda \\
\therefore \quad 4 \lambda \sin \theta &=n \lambda \\
\Rightarrow \quad 4 \sin \theta &=n
\end{aligned}
$$
Since, maximum value of $\sin \theta=1$
$\therefore$ For maximum number of possible interference maxima,
$$
\begin{aligned}
& & 4 \cdot 1 &=n \\
\Rightarrow & & n &=4 \\
\text { i.e., } & & n &=0,1,2,3
\end{aligned}
$$
$\therefore$ Maximum number of possible maxima will be 7 i.e. $0, \pm 1, \pm 2, \pm 3$
Hence, no option is correct.
$$
\begin{aligned}
d \sin \theta &=n \lambda \\
\text { Given, slit width, } d &=4 \lambda \\
\therefore \quad 4 \lambda \sin \theta &=n \lambda \\
\Rightarrow \quad 4 \sin \theta &=n
\end{aligned}
$$
Since, maximum value of $\sin \theta=1$
$\therefore$ For maximum number of possible interference maxima,
$$
\begin{aligned}
& & 4 \cdot 1 &=n \\
\Rightarrow & & n &=4 \\
\text { i.e., } & & n &=0,1,2,3
\end{aligned}
$$
$\therefore$ Maximum number of possible maxima will be 7 i.e. $0, \pm 1, \pm 2, \pm 3$
Hence, no option is correct.
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