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Question

The maximum number of possible interference maxima for slit-separation equal to twice the wavelength in Young's double-slit experiment, is; 


A
infinite
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B
five
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C
three
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D
zero
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Solution

The correct option is B five
The condition for possible interference maxima is
$$dsin\theta =n\lambda $$ and 
$$d=2\lambda $$
$$\therefore 2\lambda sin\theta =n\lambda $$
$$\Rightarrow n=2sin\theta $$
Maximum value of $$sin\theta $$ is 1
$$\therefore n=2\times 1=2$$
So, possible values at which interference maxima occurs are
n = -2, -1, 0, +1, +2
$$\therefore 5 values$$

Physics

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