Question

Given lambda = 470.2[nm] and theta = 40.4^degree, then what is the index of refraction to the nearest hundreth I know we already asked this type of question. This is just to make sure you know how to do this. Given lambda = 536.8[nm] and 0 = 38.3^degree: then what is the index of refraction to the nearest hundredth Given a blue marking of 220 degrees and a straight-on marking of 255 degrees, how much has the light bent degrees.

Answer 1

(1)To calculate the index of refraction, you need to have both the incident angle and resultant angle. The incident angle is the angle the light strikes the surface. The resultant angle is the angle the light takes inside the optic.
we will use θ1 and θ2. θ1 is the incident angle, θ2 is the resultant angle.

To find the index of refraction, you take n1/n2 = sin θ2/sin θ1.

Presume:
n1 = 1
θ1 = Φ=60°
θ2 = 40.4°

Now, solve for n2.

n2 = n1(sin θ1/sin θ2).
= 1 (sin 60 / sin 40.4)
= 0.866 / 0.648
= 1.336

(2)Same as (1)
n1 = 1
θ1 = Φ=60°
θ2 =38.3°

Now, solve for n2.

n2 = n1(sin θ1/sin θ2).
= 1 (sin 60 / sin 38.3)
= 0.866 / 0.6

(c)If light goes straight to the medium it does not bend.

magine a car drivin towards a sand pit at an angle, when the first wheel hit it, it would carry on turning but not move as fast so the others wheels, still on tarmac, would carry on going at the same speed as before and so the car would turn slightly until all the wheels were in the sand pit.

n1/n2=Sin220 /sin225

n1/n2=Sin220 /sin225
= -0.642/-0.707
=0.908

it bend very slightly negregable.