In Young's double slit experiment,if the slit | vedclass

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(C) The ratio of slit widths is equal to the ratio of intensities of the light waves emerging from them,i.e.,$\frac{I_1}{I_2} = \frac{w_1}{w_2} = 1:9$

Vedclass · Accepted Answer

$1:4$

Vedclass · Answer

(C) The ratio of slit widths is equal to the ratio of intensities of the light waves emerging from them,i.e.,$\frac{I_1}{I_2} = \frac{w_1}{w_2} = 1:9$.Since intensity $I \propto a^2$,where $a$ is the amplitude,we have $\frac{a_1}{a_2} = \sqrt{\frac{I_1}{I_2}} = \sqrt{\frac{1}{9}} = \frac{1}{3}$.Let $a_1 = a$ and $a_2 = 3a$.The intensity at maxima is $I_{\max} = (a_1 + a_2)^2 = (a + 3a)^2 = (4a)^2 = 16a^2$.The intensity at minima is $I_{\min} = (a_1 - a_2)^2 = (a - 3a)^2 = (-2a)^2 = 4a^2$.The ratio of intensity at minima to that at maxima is $\frac{I_{\min}}{I_{\max}} = \frac{4a^2}{16a^2} = \frac{1}{4}$.

In Young's double slit experiment,if the slit widths are in the ratio $1 : 9$,then the ratio of the intensity at minima to that at maxima will be

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