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(D) Let the amplitude of light from the first slit be $a_1 = a$ and from the second slit be $a_2 = a$. The intensity $I$ is proportional to the square

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the intensities of both the maxima and the minima increase.

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(D) Let the amplitude of light from the first slit be $a_1 = a$ and from the second slit be $a_2 = a$. The intensity $I$ is proportional to the square of the amplitude $(I \propto A^2)$.Initially,for equal widths,the amplitudes are equal $(a_1 = a_2 = a)$. The minimum intensity is $I_{\min} \propto (a - a)^2 = 0$,and the maximum intensity is $I_{\max} \propto (a + a)^2 = 4a^2$.When one slit is made twice as wide,the amplitude of light from that slit becomes $a_1 = 2a$,while the other remains $a_2 = a$.The new minimum intensity is $I_{\min}' \propto (2a - a)^2 = a^2$. Since $a^2 > 0$,the intensity of the minima increases.The new maximum intensity is $I_{\max}' \propto (2a + a)^2 = 9a^2$. Since $9a^2 > 4a^2$,the intensity of the maxima also increases.Therefore,both the intensity of the maxima and the minima increase.

In a double-slit experiment,instead of taking slits of equal widths,one slit is made twice as wide as the other. Then,in the interference pattern:

Similar Questions

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