Answer :
Answer:
The intensity of laser 2 is 4 times of the intensity of laser 1.
Explanation:
The intensity in terms of electric field is given by :
[tex]U=\dfrac{1}{2}\epsilon_o E^2[/tex]
E is electric field
It means, [tex]U\propto E^2[/tex]
In this problem, lasers 1 and 2 emit light of the same color, and the electric field in the beam of laser 1 is twice as strong as the e-field of laser 2.
Let E is electric field in the beam of laser 1 and E' is the electric field in the beam of laser 2. So,
[tex]\dfrac{U}{U'}=(\dfrac{E}{E'})^2[/tex]
We have,
E'=2E
So,
[tex]\dfrac{U}{U'}=\dfrac{E^2}{(2E)^2}\\\\\dfrac{U}{U'}=\dfrac{E^2}{4E^2}\\\\\dfrac{U}{U'}=\dfrac{1}{4}\\\\U'=4\times U[/tex]
So, the intensity of laser 2 is 4 times of the intensity of laser 1.