Questions
The ratio of intensities of consecutive maxima in the diffraction pattern due to a single slit is
a
1 : 4 : 9
b
1 : 2 : 3
c
1 :49 π2:425 π2
d
1:1π2:9π2
detailed solution
Correct option is C
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=1:49π2:425π2Talk to our academic expert!
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In a single slit diffraction experiment first minimum for red light (660 nm) coincides with first maximum of some other wavelength '. The value of ' is
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