Explanation:
It is given that,
Distance, r = 3.5 m
Electric field due to an infinite wall of charges, E = 125 N/C
We need to find the electric field 1.5 meters from the wall, r' = 1.5 m. Let it is equal to E'. For an infinite wall of charge the electric field is given by :

It is clear that the electric field is inversely proportional to the distance. So,


E' = 291.67 N/C
So, the magnitude of the electric field 1.5 meters from the wall is 291.67 N/C. Hence, this is the required solution.
Answer:
0.5
Explanation:
Data provided in the question:
The angle between their transmission axes, θ = 60°
Now,
We have the relation,
I₁ = I₀cos²θ
where,
I₁ is the intensity of the transmitted light
I₀ is the intensity of the incident light
on rearranging, we get
=cos²60°
or
=0.5