SATHEE: Optics 6 Question 23

Optics 6 Question 23

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

(2001, 2M)

(a) $2 I$

(b) $4 I$

(d) 7 I

Show Answer

Solution:

$I (φ) = I_{1} + I_{2} + 2 \sqrt{I_{1} I_{2}} \cos φ$

Here, $I_{1} = I$ and $I_{2} = 4 I$

At point $A, φ = \frac{π}{2}$

$∴ I_{A} = I + 4 I = 5 I$

At point $B, φ = π$

$\begin{aligned} ∴ I_{B} = I + 4 I - 4 I = I \\ ∴ I_{A} - I_{B} = 4 I \end{aligned}$

NOTE Eq. (i) for resultant intensity can be applied only when the sources are coherent. In the question it is given that the rays interfere. Interference takes place only when the sources are coherent. That is why we applied equation number (i). When the sources are incoherent, the resultant intensity is given by $I = I_{1} + I_{2}$

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Optics 6 Question 23

25. 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 π/2 at point A and π at point B. Then the difference between resultant intensities at A and B is

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