If S2P - S1P - -6- find intensity at point P-

Path diff. =S_2P S_1P Phase diff. = 2π/λ . ( λ/6) i.e. ϕ = π/3 = 60^o I_net = 4I_o cos^2 (ϕ/2) = 4I_o cos^2 (30^o) = 4I_o ×3/4 = 3I_o ...

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Byju's Answer

Standard XII

Physics

Intuition of Wave Speed

If S2P - S1P ...

Question

If $S_{2} P - S_{1} P = \frac{λ}{6},$ find intensity at point P.

4 I_{o}

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3 I_{o}

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2 I_{o}

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I_{o}

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Solution

The correct option is B $3 I_{o}$
Path diff. = $S_{2} P - S_{1} P$
Phase diff. = $\frac{2 π}{λ} . (\frac{λ}{6})$
i.e. $ϕ = \frac{π}{3} = 60^{o}$
$I_{n e t} = 4 I_{o} c o s^{2} (\frac{ϕ}{2})$
$= 4 I_{o} c o s^{2} (30^{o})$
$= 4 I_{o} \times \frac{3}{4} = 3 I_{o}$

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If S2P−S1P=λ6, find intensity at point P.

The correct option is B 3IoPath diff. =S2P−S1P Phase diff. = 2πλ.(λ6) i.e. ϕ=π3=60o Inet=4Iocos2(ϕ2) =4Io cos2(30o) =4Io×34=3Io

If $S_{2} P - S_{1} P = \frac{λ}{6},$ find intensity at point P.

The correct option is B $3 I_{o}$
Path diff. = $S_{2} P - S_{1} P$
Phase diff. = $\frac{2 π}{λ} . (\frac{λ}{6})$
i.e. $ϕ = \frac{π}{3} = 60^{o}$
$I_{n e t} = 4 I_{o} c o s^{2} (\frac{ϕ}{2})$
$= 4 I_{o} c o s^{2} (30^{o})$
$= 4 I_{o} \times \frac{3}{4} = 3 I_{o}$