Q. 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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A

zero

23%

B

3

29%

C

infinite

29%

D

5

19%

Solution:

As $sin\,\theta =\frac{n\lambda}{d}$ and $sin \theta$ cannot be $≯ 1$
$\therefore \, 1 = \frac{n\lambda}{1.8\lambda}$
or $n= 1.8$
Hence maximum number of possible
interference maximas, $0, \pm 1\, i.e.\, 3$

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