Question

Light attempts to move from cubic zirconia (n=1.61) to air. In a nearby experiment, light is attempting to move from cubic zirconia to water. Finally, in a third trial, light is attempting to move from cubic zirconia to a material with an index of refraction of 1.62. For the angles of refraction (θcz->air , θcz->water , θcz->material ) use greater than symbols (>) to rank these three angles. (e.g. θ2 > θ2' > θ2" ) [Remember, the greater the refraction, the smaller the refracted angle. So the angle indicating the greatest refraction (i.e. the smallest angle) will be on the right.]

Answer 1

Part A. for light traveling from zirconia to air:

Using snell's law:

n1*sin i = n2*sin r1

n1 = refractive index of zirconia = 1.61

n2 = refractive index of air = 1.00

i = angle of incidence

r1 = angle of refraction in air = ? ()

r1 = arcsin ((n1/n2)*sin i)

r1 = arcsin ((1.61/1.00)*sin i)

r1 = arcsin (1.61*sin i)

Part B. for light traveling from zirconia to water:

Using snell's law:

n1*sin i = n2*sin r2

n1 = refractive index of zirconia = 1.61

n2 = refractive index of air = 1.333

i = angle of incidence

r2 = angle of refraction in air = ?

r2 = arcsin ((n1/n2)*sin i)

r2 = arcsin ((1.61/1.333)*sin i)

r2 = arcsin (1.21*sin i)

Part C. for light traveling from zirconia to material:

Using snell's law:

n1*sin i = n2*sin r3

n1 = refractive index of zirconia = 1.61

n2 = refractive index of air = 1.62

i = angle of incidence

r3 = angle of refraction in air = ?

r3 = arcsin ((n1/n2)*sin i)

r3 = arcsin ((1.61/1.62)*sin i)

r3 = arcsin (0.994*sin i)

Step 2: Now we know that value of sine function increase from 0 deg to 90 deg, and since incidence angle is same in all three cases, So

1.61*sin i > 1.21*sin i > 0.994*sin i

arcsin (1.61*sin i) > arcsin (1.21*sin i) > arcsin (0.994*sin i)

r1 > r2 > r3

_cz->air > _cz->water > _cz->material

Let me know if you've any query.