What is the refractive index of a
material which has a critical angle of 61 degrees for a light beam traveling from it
To begin, let’s recall the formula
for determining the critical angle 𝜃 𝑐. The sin of 𝜃 𝑐 equals 𝑛 two
divided by 𝑛 one, where 𝑛 one is the refractive index of the first medium that the
incident ray is initially in and 𝑛 two is the refractive index of the second medium
on the other side of the medium boundary. To make these quantities clearer,
we can draw a diagram like this. Here, we’re considering a ray of
light traveling from an unknown medium to air. So 𝑛 two is the refractive index
of air, and 𝑛 one is the quantity we want to solve for to answer this question.
Let’s go ahead and rearrange this
formula to make 𝑛 one the subject. To do this, we can multiply both
sides by 𝑛 one over the sin of 𝜃 𝑐. This way, 𝑛 one cancels out of the
right-hand side and the sin of 𝜃 𝑐 cancels out of the left-hand side, leaving 𝑛
one by itself. The equation now reads 𝑛 one
equals 𝑛 two divided by the sine of the critical angle.
We’ve been told that the critical
angle equals 61 degrees. So this is the value of 𝜃 𝑐. Remember too that 𝑛 two is the
refractive index of air, which is simply equal to one. Since we have values for both of
the variables on the right-hand side of this equation, we’re ready to substitute
them in to reach the final answer. Doing so, we have that 𝑛 one
equals one over the sin of 61 degrees. Plugging this expression into a
calculator gives a result of 1.1434 and so on. Rounding to two decimal places,
this becomes 1.14, and so we have our final answer.
Thus, we’ve found that the
refractive index of this unknown material is 1.14.