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Standard XII

Physics

Young's Modulus of Elasticity

If intensity ...

Question

If intensity of each wave in the observed interface pattern is Young's double slit experiment is $I_{0}$ . then for some point $P$ where the phase difference is $ϕ$ intensity $I$ will be

I = I_{0} cos ϕ

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I = I_{0} {cos}^{2} ϕ

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I = I_{0} (1 + cos ϕ)

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I = 2 I_{0} (1 + cos ϕ)

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Solution

The correct option is D $I = 2 I_{0} (1 + cos ϕ)$
$I = I_{1} + I_{2} + 2 \sqrt{I_{1} I_{2}}$ as $ϕ$
put $I_{1} = I_{2} = I_{0}$
we get $I = I_{0} + I_{0} + 2 \sqrt{I_{0}^{2}}$ as $ϕ$
$\Rightarrow I = Z I_{0} + Z I_{0} + Z I_{0} c o s ϕ$
$\Rightarrow I = Z I_{0} (1 + c o s ϕ)$

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If intensity of each wave in the observed interface pattern is Young's double slit experiment is I0. then for some point P where the phase difference is ϕ intensity I will be

The correct option is D I=2I0(1+cosϕ)I=I1+I2+2√I1I2 as ϕput I1=I2=I0we get I=I0+I0+2√I20 as ϕ⇒I=ZI0+ZI0+ZI0cosϕ⇒I=ZI0(1+cosϕ)

If intensity of each wave in the observed interface pattern is Young's double slit experiment is $I_{0}$ . then for some point $P$ where the phase difference is $ϕ$ intensity $I$ will be

The correct option is D $I = 2 I_{0} (1 + cos ϕ)$
$I = I_{1} + I_{2} + 2 \sqrt{I_{1} I_{2}}$ as $ϕ$
put $I_{1} = I_{2} = I_{0}$
we get $I = I_{0} + I_{0} + 2 \sqrt{I_{0}^{2}}$ as $ϕ$
$\Rightarrow I = Z I_{0} + Z I_{0} + Z I_{0} c o s ϕ$
$\Rightarrow I = Z I_{0} (1 + c o s ϕ)$