Wave Optics

JEE Main level, single correct, medium

43 of 191

Question 43 of 191

Width of one of the two slits in a Young's double slit interference experiment is half of the other slit. The ratio of the maximum to the minimum intensity in the interference pattern is :

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Answer: (B) (3+22):(3−22)(3+2 \sqrt{2}):(3-2 \sqrt{2})

I∝\mathrm{I} \propto width Imax⁡=(I1+I2)2\quad\quad \mathrm{I}_{\max }=\left(\sqrt{\mathrm{I}_1}+\sqrt{\mathrm{I}_2}\right)^2
∴I1=I0,I2=2I0Imin⁡=(I1−I2)2\therefore \mathrm{I}_1=\mathrm{I}_0, \mathrm{I}_2=2 \mathrm{I}_0 \quad\quad \mathrm{I}_{\min }=\left(\sqrt{\mathrm{I}_1}-\sqrt{\mathrm{I}_2}\right)^2
Imax⁡Imin⁡=(2+1)2(2−1)2⇒3+223−22\frac{I_{\max }}{I_{\min }}=\frac{(\sqrt{2}+1)^2}{(\sqrt{2}-1)^2} \Rightarrow \frac{3+2 \sqrt{2}}{3-2 \sqrt{2}}

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