When light moves from a medium with higher refractive index to a medium with lower refractive index, the critical angle is the angle above which there is no refracted ray, and it is given by:

(2)
where

is the refractive index of the second medium and

is the refractive index of the first medium.
We can find the ratio

by using Snell's law:

(1)
where

is the angle of incidence

is the angle of refraction
By using the data of the problem and re-arranging (1), we find

and if we use eq.(2) we can now find the value of the critical angle:
The answer is A.
Explanation:
We know that the average acceleration a for an interval of time Δt is expressed as:
a = Δv
Δt
where Δv is the change in velocity that occurs during Δt.
e formula for the instantaneous acceleration a is almost the same, except that we need to indicate that we're interested in knowing what the ratio of Δv to Δt approaches as Δt approaches zero.
We can indicate that by using the limit notation.
So, the formula for the instantaneous acceleration is:
a = lim Δv
Δt→0 Δt
Answer:
The force per unit length (N/m) on the top wire is 16.842 N/m
Explanation:
Given;
distance between the two parallel wire, d = 38 cm = 0.38 m
current in the first wire, I₁ = 4.0 kA
current in the second wire, I₂ = 8.0 kA
Force per unit length, between two parallel wires is given as;

where;
μ₀ is constant = 4π x 10⁻⁷ T.m/A
Substitute the given values in the above equation and calculate the force per unit length

Therefore, the force per unit length (N/m) on the top wire is 16.842 N/m
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Answer:
where is the graph I can't see it how can I solve the problem if I don't see the graph can you show the graph please