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The maximum number of possible interference maxima for slit separation equal to $1.8 \lambda$, where $\lambda$ is the wavelength of light used, in a Young's double slit experiment is
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Verified Answer
The correct answer is:
3
3
As $\sin \theta=\frac{n \lambda}{d}$ and $\sin \theta$ cannot be $\ngtr 1$
$$
\therefore 1=\frac{n \lambda}{1.8 \lambda}
$$
or $n=1.8$
Hence maximum number of possible interference maximas, $0, \pm 1$ i.e. 3
$$
\therefore 1=\frac{n \lambda}{1.8 \lambda}
$$
or $n=1.8$
Hence maximum number of possible interference maximas, $0, \pm 1$ i.e. 3
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