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In a YDSE with $D= 1m,d=1mm$ light of wave length $500 \;nm$ is incident at an angle of $0.57^{\circ}$ with respect to axis of symmetry of experimental set up. If centre of symmetry of screen is 0. Then the intensity at point 0 in terms of intensity of central maxima $I_0$ is

$(a)\;2\;I_0 \\ (b)\;3\;I_0 \\ (c)\;I_0 \\ (d)\;None $

1 Answer

For point $0, \theta =0$
Hence $\Delta P = d \sin \theta _0$
for small angles$ \sin \theta _0= \theta _0$
$0.57^{\circ}= (10^{-2} \;rad)$
$d \theta_0= 1mm \times (10^{-2} \;rad)$
$\qquad= 10,000 nm$
$\qquad= 20 \times (500 nm)$
=> $\Delta P= 20 \lambda$
Hence point 0 corresponds to $20th $ maxima
=>Intensity at $0=I_0$
Hence c is the correct answer


answered Feb 13, 2014 by meena.p
edited Jul 22, 2014 by thagee.vedartham

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