In Young’s double slit experiment the intensity of light at a point on the screen where the path difference λ is K…..

Ans: (c)

Sol: $\displaystyle I_R = I_1 + I_2 + 2 \sqrt{I_1 \times I_2 } cos\phi $

$\displaystyle I_R = I + I + 2 \sqrt{I \times I } cos\phi $

$\displaystyle I_R = 2I(1 + cos\phi) $

$\displaystyle I_R = 2I(2 cos^2 \frac{\phi}{2}) $

$\displaystyle I_R = 4 I cos^2 \frac{\phi}{2} $

According to question , $\displaystyle K = 4 I cos^2 \pi $ (As path diff. is λ hence phase diff. is 2π)

$\displaystyle K = 4 I $ ….(i)

In second case ,

$\displaystyle I_R’ = 4 I cos^2 \frac{\phi’}{2} $

$\displaystyle I_R’ = 4 I cos^2 \frac{\pi}{6} $ (As path diff. is λ/6 hence phase diff. is π/3)

$\displaystyle I_R’ = 4 I \times \frac{3}{4}= 3I $

$\displaystyle I_R’ = 3I = \frac{3 K}{4} $