Two beams of light having intensities I and 4I interfere to produce a fringe pattern on a screen- The phase difference between the beams is -2 at point A and - at point B- Then- the difference between the resultant intensities at A and B is

I_A = I + 4I + 2√(I)√(4I) cos 90 I_A = 5I . . . . . (1) I_B = I + 4I + 2√(I)√(4I) cos 180 = 5I 4I I_B = I . . . . .(2) Δ I = I_A I_B = 5I – I = ...

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Byju's Answer

Standard VIII

Physics

Luminous Intensity

Two beams of ...

Question

Two beams of light having intensities I and 4I interfere to produce a fringe pattern on a screen. The phase difference between the beams is $\frac{π}{2}$ at point A and $π$ at point B. Then, the difference between the resultant intensities at A and B is

2 I

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4 I

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5 I

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7 I

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Solution

The correct option is B 4 I
$I_{A} = I + 4 I + 2 \sqrt{I} \sqrt{4 I} c o s 90$
$I_{A} = 5 I$ . . . . . (1)
$I_{B} = I + 4 I + 2 \sqrt{I} \sqrt{4 I} c o s 180$
$= 5 I - 4 I$
$I_{B} = I$ . . . . .(2)
$Δ I = I_{A} - I_{B}$
$= 5 I - I$
$= 4 I$

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Two beams of light having intensities I and 4I interfere to produce a fringe pattern on a screen. The phase difference between the beams is π2 at point A and π at point B. Then, the difference between the resultant intensities at A and B is

The correct option is B 4 IIA=I+4I+2√I√4Icos90 IA=5I . . . . . (1) IB=I+4I+2√I√4Icos180 =5I−4I IB=I . . . . .(2) ΔI=IA−IB =5I–I =4I

Two beams of light having intensities I and 4I interfere to produce a fringe pattern on a screen. The phase difference between the beams is $\frac{π}{2}$ at point A and $π$ at point B. Then, the difference between the resultant intensities at A and B is

The correct option is B 4 I
$I_{A} = I + 4 I + 2 \sqrt{I} \sqrt{4 I} c o s 90$
$I_{A} = 5 I$ . . . . . (1)
$I_{B} = I + 4 I + 2 \sqrt{I} \sqrt{4 I} c o s 180$
$= 5 I - 4 I$
$I_{B} = I$ . . . . .(2)
$Δ I = I_{A} - I_{B}$
$= 5 I - I$
$= 4 I$