1. 2sqrt n/n+1 2. sqrt n/(n+1)² 3. 2sqrt n/(n+1)² 4. sqrt n/n+1

Answer: 1. 2sqrt n/n+1 With intensities I₁:I₂=n (take I₁=nI₂ ), I_max=(√(I₁)+√(I₂))² and I_min=(√(I₁)-√(I₂))² . fracI_max-I_minI_max+I_min=4√(I₁I₂)/2(I₁+I₂)=2√(n)/n+1 Option (1).

Topic: Interference and Young's double-slit experiment· Optics

The interference pattern is obtained with two coherent light sources of intensity ratio n. In the interference pattern, the ratio I_max-I_minI_max+I_min will be

1. $\frac{2\sqrt n}{n+1}$

2. $\frac{\sqrt n}{(n+1)^2}$

3. $\frac{2\sqrt n}{(n+1)^2}$

4. $\frac{\sqrt n}{n+1}$

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Answer: 1. $\frac{2\sqrt n}{n+1}$

With intensities $I_1:I_2=n$ (take $I_1=nI_2$), $I_{max}=(\sqrt{I_1}+\sqrt{I_2})^2$ and $I_{min}=(\sqrt{I_1}-\sqrt{I_2})^2$.

$$\frac{I_{max}-I_{min}}{I_{max}+I_{min}}=\frac{4\sqrt{I_1I_2}}{2(I_1+I_2)}=\frac{2\sqrt{n}}{n+1}$$

Option (1).

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The interference pattern is obtained with two coherent light sources of intensity ratio n. In the interference pattern, the ratio I_max-I_minI_max+I_min will be

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