Wave Optics

JEE Main level, single correct, medium

99 of 191

Question 99 of 191

In a Young's double-slit experiment, the ratio of the slit's width is 4:14:1 . The ratio of the intensity of maxima to minima, close to the central fringe on the screen, will be

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Answer: (B) 9:19:1

IntensityIntensity∝\proptowidthofslitwidthofslit
I1I2=41\dfrac{{I}_{1}}{{I}_{2}}=\dfrac{4}{1}.
ImaxImin=(I1+I2I1−I2)2=(I1I2+1I1I2−1)2\dfrac{{I}_{max}}{{I}_{min}}={\left(\dfrac{\sqrt{{I}_{1}}+\sqrt{{I}_{2}}}{\sqrt{{I}_{1}}-\sqrt{{I}_{2}}}\right)}^{2}={\left(\dfrac{\sqrt{\dfrac{{I}_{1}}{{I}_{2}}}+1}{\sqrt{\dfrac{{I}_{1}}{{I}_{2}}}-1}\right)}^{2}
=(4+14−1)2=(31)2=91={\left(\dfrac{\sqrt{4}+1}{\sqrt{4}-1}\right)}^{2}={\left(\dfrac{3}{1}\right)}^{2}=\dfrac{9}{1}
Imax:Imin=9:1{I}_{max}:{I}_{min}=9:1

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