Wave Optics

JEE Main level, single correct, medium

112 of 191

Question 112 of 191

In Young's double slit interference experiment, the slit widths are in the ratio 1:251: 25. Then the ratio of intensity at the maxima and minima in the interference pattern is

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Answer: (C) 9:49: 4

We know that,
Imax⁡Imin⁡=(ω1ω2+1)2(ω1ω2−1)2\frac{I_{\max }}{I_{\min }}=\frac{\left(\sqrt{\frac{\omega_1}{\omega_2}+1}\right)^2}{\left(\sqrt{\frac{\omega_1}{\omega_2}-1}\right)^2}
Imax⁡I_{\max } and Imin⁡I_{\min } are maximum and minimum intensity
ω1\omega_1 and ω2\omega_2 are widths of two slits

∴Imax⁡Imin⁡=(125+1)2(125−1)2(ω1ω2=125 given )\therefore \frac{I_{\max }}{I_{\min }}=\frac{\left(\sqrt{\frac{1}{25}}+1\right)^2}{\left(\sqrt{\frac{1}{25}}-1\right)^2}\left(\frac{\omega_1}{\omega_2}=\frac{1}{25} \text { given }\right)
On solving we get,
Imax⁡Imin⁡=36251625=94=9:4\frac{I_{\max }}{I_{\min }}=\frac{\frac{36}{25}}{\frac{16}{25}}=\frac{9}{4}=9: 4

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