QJEEM20Pheaf6: JEE Main 2020 Physics - Class 12 JEEM20Pheaf6 Solution: In the figure below, P and Q are two equally intense coherent…

Let the distace of each point from midpoint = d

For point A :

x_P – x_Q = (d + 2.5) – (d – 2.5) = 5 m

$\Delta$ $\phi$ due to path difference = ${{2\pi } \over \lambda }\left( {\Delta x} \right)$ = ${{2\pi } \over {20}}\left( 5 \right)$ = ${\pi \over 2}$

At A, Q is ahead of P by path, as wave emitted by Q reaches before wave emitted by P.

Total phase difference at A

= ${\pi \over 2} - {\pi \over 2}$ (due to P being ahead of Q by 90º) = 0

$\therefore$ I_A = I₁ + I₂ + $2\sqrt {{I_1}} \sqrt {{I_2}}$ cos $\Delta$ $\phi$

= I + I + $2\sqrt I \sqrt I \cos \left( 0 \right)$

= 4I

For point C :

x_Q – x_P = (d + 2.5) – (d – 2.5) = 5 m

$\Delta$ $\phi$ due to path difference = ${{2\pi } \over \lambda }\left( {\Delta x} \right)$ = ${{2\pi } \over {20}}\left( 5 \right)$ = ${\pi \over 2}$

Total phase difference at C = ${\pi \over 2} + {\pi \over 2}$ = $\pi$

$\therefore$ I_C = I₁ + I₂ + $2\sqrt {{I_1}} \sqrt {{I_2}}$ cos $\Delta$ $\phi$

= I + I + $2\sqrt I \sqrt I \cos \left( \pi \right)$

= 0

For point B :

x_P – x_Q = 0

$\Delta$ $\phi$ due to path difference = 0

$\Delta$ $\phi$ = ${\pi \over 2}$ (Due to P being ahead of Q by 90º)

$\therefore$ I_B = I₁ + I₂ + $2\sqrt {{I_1}} \sqrt {{I_2}}$ cos $\Delta$ $\phi$

= I + I + $2\sqrt I \sqrt I \cos \left( {{\pi \over 2}} \right)$

= 2I

I_A : I_B : I_C = 4I : 2I : 0

= 2 : 1 : 0