Question

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. how do the intensities of the two lasers compare?

Answer 1

The intensity of laser 2 is 4 times of the intensity of laser 1.

Step-by-step explanation:

The intensity in terms of electric field is given by :

`U=(1)/(2)\epsilon_o E^2`

E is electric field

It means,
`U\propto E^2`

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,

`(U)/(U')=((E)/(E'))^2`

We have,

E'=2E

So,

`(U)/(U')=(E^2)/((2E)^2)\\\\(U)/(U')=(E^2)/(4E^2)\\\\(U)/(U')=(1)/(4)\\\\U'=4* U`

So, the intensity of laser 2 is 4 times of the intensity of laser 1.