JEE Main 2020 — Wave Optics Question with Solution

Let the distace of each point from midpoint = d

For point A :

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

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

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

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

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

= 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

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

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

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

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

= 0

For point B :

x_P – x_Q = 0

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

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

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

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

= 2I

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

= 2 : 1 : 0