Question

A ray in glass is incident onto a water-glass interface, at an angle of incidence equal to half the critical angle for that interface. The indices of refraction for water and the glass are 1.33 and 1.83, respectively. The angle that the refracted ray in the water makes with the normal is closest to:

Answer 1

38°

Step-by-step explanation:

A

the critical angle can be calculated using expresion below

Sin(θ)critical=nr/ni -------------eqn(1)

θ)critical= critical angle

ni= refractive index of the incident medium= 1.83

nr= refractive index of the refractive medium= 1.33

Substitute the values into eqn (1)

Sin(θ)critical=1.33/1.83

θ)critical=sin-(0.726776)

(θ)critical=46.6168°

From the question, we are told that angle of incidence equals half of the critical angle. This can be expressed as

Angle of incidence = 46.6168°/ 2 = 22.308°

From Snell's law which can be expressed mathematically as

ni × sin(θi = nr × (θr

(θ)i = ngle of incidence 22.308°

(θ)r=the angle of refraction

nr = refractive index of the refraction medium of that of water,1.33

nr is the refractive index of the incidence medium of that of glass, 1.43

Then if we substitute we have

1.83 × sin(23.304)= 1.33 × sin (θr

Sin (θr=[ 1.83×sin(23.304)]/1.33

sinθr=0.54434

Sin-1(0.54434)

θr=32.98°

Hence, The angle that the refracted ray in the water makes with the normal is closest to is 32.98°