A ray of light is incident on a glass—water interface at the critical angle c as

Question

A ray of light is incident on a glass—water interface at the critical angle c as the figure illustrates. The reflected light then passes through a liquid (immiscible with water) and into the air. The indices of refraction for the four substances are given in the drawing. (Assume that nGlass = 1.54 and nLiquid = 1.54.)

Concepts:
(i) What determines the critical angle when the ray strikes the glass—water interface? (Do not substitute numerical values; use variables only. Use the following as necessary: nGlass and  nWater.)

c=

(ii) When the light is incident at the glass—water interface at the critical angle, what is the angle of refraction?


How is angle 1 related to the critical angle? (Do not substitute numerical values; use variables only. Use the following as necessary: c. Note that the c is lowercase.)

1 = 180 + c

1 = c

1 = 90 c

1 = 90 + c

1 = c

(iii) When the reflected ray strikes the glass—liquid interface, how is the angle of refraction

3

related to the angle of incidence

2?

Note that the two materials have the same indices of refraction.

3 = 2

3 = 90 + 2

3 = 180 + 2

3 = 2

3 = 90 2

(iv) When the ray passes from the liquid into the air, is the ray refracted?

YesNo    

Explain.

The ray travels from a material with a higher refractive index to a material with a lower refractive index.The ray is only refracted when the incident angle is equal to the critical angle.    The ray is only refracted when the incident angle is greater than the critical angle.The ray travels from a material with a lower refractive index to a material with a higher refractive index.

Calculation:
Determine the angle of refraction

5

for the ray as it passes into the air.

Explanation / Answer

i)

using snell's law at glass-water interface

nglass Sinc = nair Sin90

so Sinc = nair/nglass

c = Sin-1(nair/nglass)

ii)

nglass = 1.54

so c = Sin-1(nair/nglass) = Sin-1(1/1.54) = 33.02

1 = c

iii)

2 = 90 - 1  

using snell';s law at glass-liquid interface

nglass Sin2 = nliquid Sin3

since  nglass = nliquid

2 = 3

iv)

yes

since the index of refraction of liquid and air are different

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.