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Physics

Optics

Easy

Question

In a Young’s double slit experiment the intensity, at a point where the path difference is $fraction numerator lambda over denominator 6 end fraction$ (λ being the wavelength of the light used), $blank i s blank I.$ If $I subscript 0 end subscript$ denotes the maximum intensity, $I divided by I subscript 0 end subscript$ is equal to

$\frac{1}{\sqrt{2}}$
$\frac{\sqrt{3}}{2}$
$\frac{1}{2}$
$\frac{3}{4}$

The correct answer is: $fraction numerator 3 over denominator 4 end fraction$

Phase difference $equals fraction numerator 2 pi over denominator lambda end fraction cross times$ path difference
ie, $ϕ equals fraction numerator 2 pi over denominator lambda end fraction blank cross times fraction numerator lambda over denominator 6 end fraction equals fraction numerator pi over denominator 3 end fraction$
As, $I equals I subscript m a x end subscript cos to the power of 2 end exponent invisible function application open parentheses fraction numerator ϕ over denominator 2 end fraction close parentheses$
Or $fraction numerator I over denominator I subscript m a x end subscript end fraction equals cos to the power of 2 end exponent invisible function application open parentheses fraction numerator ϕ over denominator 2 end fraction close parentheses$
Or $fraction numerator I over denominator I subscript 0 end subscript end fraction equals cos to the power of 2 end exponent invisible function application open parentheses fraction numerator pi over denominator 6 end fraction close parentheses equals fraction numerator 3 over denominator 4 end fraction$

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