Question

The intensity of two waves is 2 and 3 units then the average intensity of light in the overlapping region will have the value
a. 2.5 b. 6 c. 5 d. 13

Answer 1

Answer:

5

Explanation:

I1+ I2+2√I1√I2+

I1+ I2-2√I1√I2÷2=5

(5+2√6+5-2√6)

Answer 2

Answer:

The average intensity of light in the overlapping region will have the value of 5.

Explanation:

Given:
Intensity of wave 1(1₁)=2

Intensity of wave 2(I₂)=3

To find: Intensity of light in the overlapping region

Solution:

The maximum intensity is calculated using,

$Imax=I_{1}+I_{2}+2\sqrt{I_{1}}\sqrt{I_{2}}$

The minimum intensity is equals to

$Imin=I_{1}+I_{2}-2\sqrt{I_{1}}\sqrt{I_{2}}$

On taking average we get,

$Iavg= \frac{I_{1}+I_{2}+I_{1}+I_{2}}{2}$

=2+3+2+3/2=10/2=5

Therefore, the average intensity of light in the overlapping region will have the value of 5.