A ray of light traveling in air strikes the surface of a liquid.
If the angle of incidence is 33.5◦
and the
angle of refraction is 19.8◦, find the critical
angle for light traveling from the liquid back
into the air.
Answer in units of◦
sinr1/sinr2 = n2/n1,
n1 = 1.003 (for air)
n2 = n1•( sinr1/sinr2) = 1.6244
sinr(o) = 1/n2 = 0.6156
r(o) – arcsin(0.6156) = 38 degr.
find n.
sin33.5=n*sin19.8
then, critical angle= arcsin n
do I put my calculator in degrees?
yes, I would.
I keep getting error
To find the critical angle, we need to use Snell's Law. Snell's Law relates the angle of incidence (θ1) and the angle of refraction (θ2) to the indices of refraction of the two mediums involved.
The formula for Snell's Law is:
n1 * sin(θ1) = n2 * sin(θ2)
Here, n1 is the index of refraction of the first medium (air) and n2 is the index of refraction of the second medium (liquid).
Given that the angle of incidence (θ1) is 33.5° and the angle of refraction (θ2) is 19.8°, we can rewrite Snell's Law as:
sin(33.5°) / sin(19.8°) = n2 / n1
Since we are looking for the critical angle when light travels from the liquid back into the air, the critical angle occurs when the angle of refraction is 90°. Substituting this into Snell's Law, we get:
sin(33.5°) / sin(90°) = n2 / n1
Now we can solve for the critical angle:
sin(33.5°) / 1 = n2 / n1
To get the critical angle, we need to find the inverse sine of the ratio n2 / n1:
critical angle = arcsin(n2 / n1)
However, we don't have the values for n1 and n2. The index of refraction for air is very close to 1, whereas the index of refraction for the liquid is typically given in the problem or can be looked up.
Once you have the value for the index of refraction of the liquid (n2), you can substitute it into the equation to find the critical angle.