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(D) In Young's double-slit experiment,the intensity of the interference pattern is given by $I = I_1 + I_2 + 2\sqrt{I_1 I_2} \cos \phi$.If the amplitu

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The contrast between the fringes will decrease.

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(D) In Young's double-slit experiment,the intensity of the interference pattern is given by $I = I_1 + I_2 + 2\sqrt{I_1 I_2} \cos \phi$.If the amplitudes $a_1$ and $a_2$ are not equal,the maximum intensity is $I_{max} = (a_1 + a_2)^2$ and the minimum intensity is $I_{min} = (a_1 - a_2)^2$.Since $a_1 
eq a_2$,the minimum intensity $I_{min}$ is not zero $(I_{min} > 0)$.This means the dark fringes are not perfectly dark,leading to a reduction in the contrast (visibility) of the interference fringes.

In Young's double-slit experiment,if the amplitudes of the interfering waves are not equal,then . . . . .

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