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Home  >>  JEEMAIN and AIPMT  >>  Physics  >>  Class12  >>  Wave Optics
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Two light waves whose equation are given respectively as $y_1 = 6\sin wt; y_2 =8 \sin (wt+ 8)$ super impose to form a interference pattern. The ratio of maximum to minimum intensities of the super imposed wave is

$(a)\;49:1 \\ (b)\;1:49 \\ (c)\;1:7 \\ (d)\;7:1 $

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From the two given equation of wave
$a_1 =6 \;a_2= 8$
$\large\frac{I_{max}}{I_{min}}=\large\frac{\bigg[\Large\frac{a_1}{a_2}+1\bigg]^2}{\bigg[\Large\frac{a_1}{a_2} -1\bigg]^2}$
$\qquad=\large\frac{\bigg[\Large\frac{6}{8}+1\bigg]^2}{\bigg[\Large\frac{6}{8} -1\bigg]^2}$
$\qquad= \large\frac{49}{1}$
Hence a is the correct answer.
answered Jan 21, 2014 by meena.p

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