A ray of light travels from air into a liquid, The ray enters the liquid at an angle of 30.0°. The angle of refraction is 22.0°. Using Snell’s law, calculate the:
index of refraction of the liquid. What might the liquid be?

Question

Benjamin Asiedu · Accepted Answer

Here #"angle of incidence" (i)=30^@# #"angle of refraction" (r)=22^@# Then refractive index  #mu=sini/sinr=sin30/sin22=1.33# So the liquid is water

Nora Arzate · Answer

The liquid is water #(H_2O)#.

Snell's law states that: #n_1sini=n_2sinr# where: #i# is the angle of incidence #r# is the angle of refraction #n_1# is the refractive index of substance #1# #n_2# is the refractive index of substance #2# For air, the refractive index is #1#, and so: #:.sin30=n_2*sin22# #n_2=(sin30)/(sin22)# #~~1.33# From here, water seems to have a refractive index of #1.33#, and so this liquid is water.

Leo Abeyta · Answer

undefined Using Snell's Law:

n1 * sin(angle of incidence) = n2 * sin(angle of refraction)

Given:
- Angle of incidence (θ1) = 30.0°
- Angle of refraction (θ2) = 22.0°

Let n1 be the index of refraction of air, which is approximately 1.

Therefore,

1 * sin(30.0°) = n2 * sin(22.0°)

n2 = (sin(30.0°) / sin(22.0°))

n2 ≈ (0.5 / 0.374)
n2 ≈ 1.34

The index of refraction of the liquid is approximately 1.34. Based on the common indices of refraction for liquids, this value suggests that the liquid might be water.

A ray of light travels from air into a liquid, The ray enters the liquid at an angle of 30.0°. The angle of refraction is 22.0°. Using Snell’s law, calculate the: index of refraction of the liquid. What might the liquid be?