Question

In Young's double slit interference experiment, using two coherent waves of different amplitudes, the intensities ratio between bright and dark fringes is 3 Then, the value of the wave amplitudes ratio that arrive there is

Answer 1

Answer:

Given : $$\dfrac{I_{max}}{I_{min}} = 3$$

Let the amplitude of the waves be A

1

​

and A

2

​

respectively.

Maximum intensity $$I_{max} = (\sqrt{I_1} + \sqrt{I_2})^2$$ where $$I_i \propto A^2_i$$

⟹ $$I_{max} = (A_1 + A_2)^2$$ ............(1)

Similarly, minimum intensity $$I_{min} = (A_1 - A_2)^2$$

∴ $$\dfrac{(A_1 + A_2)^2}{(A_1 - A_2)^2} = \dfrac{I_{max}}{I_{min}} = 3$$

OR $$\dfrac{A_1 + A_2}{A_1 - A_2} = \sqrt{3}$$ $$\implies A_1 = \bigg[ \dfrac{\sqrt{3} +1}{\sqrt{3} - 1}\bigg] A_2$$

⟹ $$\dfrac{A_1}{A_2} = \bigg[ \dfrac{\sqrt{3} + 1}{\sqrt{3} -1}\bigg]$$ or, $$\dfrac{A_2}{A_1} = \bigg[ \dfrac{\sqrt{3} - 1}{\sqrt{3} +1}\bigg]$$

Explanation: