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Q.

The ratio of intensities of consecutive maxima in the diffraction pattern due to a single slit is

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a

1 : 4 : 9

b

1 : 2 : 3

c

1 :49 π2:425 π2

d

1:1π2:9π2

answer is C.

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Detailed Solution

I=I0sinαα2,  where  α=φ2For  nth secondary maxima  dsinθ=2n+12λ⇒α=φ2=πλdsinθ=2n+12π∴   I=I0sin2n+12π2n+12π2So  I0:I1:I2=I0:49π2I0:425π2I0=1:49π2:425π2
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