Wave Optics

JEE Main level, single correct, medium

21 of 191

Question 21 of 191

In an interference experiment the ratio of amplitudes of coherent waves is a1a2=13\dfrac{{a}_{1}}{{a}_{2}}=\dfrac{1}{3} . The ratio of maximum and minimum intensities of fringes will be:

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Answer: (A) 44

ImaxImin=(a1+a2a1−a2)2=(a1/a2+1a1/a2−1)2=(13+113−1)2\dfrac{{I}_{max}}{{I}_{min}}={\left(\dfrac{{a}_{1}+{a}_{2}}{{a}_{1}-{a}_{2}}\right)}^{2}={\left(\dfrac{{a}_{1}/{a}_{2}+1}{{a}_{1}/{a}_{2}-1}\right)}^{2}={\left(\dfrac{\dfrac{1}{3}+1}{\dfrac{1}{3}-1}\right)}^{2}
Imax:Imin=4:1{I}_{max} : {I}_{min}=4:1

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