Refraction of Light, Total Internal Reflection, and Prisms Practice Questions — JAMB UTME

Question 1Question →

Which of the following conditions is necessary for total internal reflection to occur when light travels between two media?

Light must travel from an optically denser medium to an optically less dense medium, and the angle of incidence must be greater than the critical angle.

Light must travel from an optically less dense medium to an optically denser medium, and the angle of incidence must be greater than the critical angle.

Light must travel from an optically denser medium to an optically less dense medium, and the angle of incidence must be less than the critical angle.

Light must travel from an optically less dense medium to an optically denser medium, and the angle of incidence must equal zero degrees.

Show answer & explanation

Answer: Light must travel from an optically denser medium to an optically less dense medium, and the angle of incidence must be greater than the critical angle.

Answer

Total internal reflection requires light to travel from an optically denser medium to an optically less dense medium with an angle of incidence greater than the critical angle.

For total internal reflection to happen, two strict conditions must be satisfied: (1) light must travel from a medium with a higher refractive index to one with a lower refractive index, and (2) the angle of incidence must be strictly greater than the critical angle for that boundary.

Step-by-Step Solution

1

Identify the optical density requirement

Light must attempt to pass from a medium of higher refractive index (denser) into a medium of lower refractive index (less dense) so that the refracted ray bends away from the normal.

Bending away from the normal allows the angle of refraction to reach

90^\circ

at the critical angle.

2

Identify the angle of incidence requirement

The angle of incidence in the denser medium must exceed the critical angle (

i > C

).

When the incidence angle exceeds the critical angle, no refraction is possible and all energy is reflected back into the denser medium.

Key Concept

Conditions for Total Internal Reflection

Question 2Question →

The speed of light in medium $A$ is $1.5 \times 10^8\text{ m/s}$ and in medium $B$ is $3.0 \times 10^8\text{ m/s}$ . What is the critical angle for a light ray traveling from medium $A$ towards medium $B$ ?

30^\circ

45^\circ

60^\circ

90^\circ

Show answer & explanation

Answer: $30^\circ$

Answer

The critical angle for light traveling from medium

A

to medium

B

is

30^\circ

.

The critical angle

C

is the angle of incidence in the denser medium for which the angle of refraction in the rarer medium is

90^\circ

. Using the wave speed form of Snell's law,

\sin C = \frac{v_A}{v_B} = \frac{1.5 \times 10^8}{3.0 \times 10^8} = 0.5

. Taking the inverse sine yields

C = 30^\circ

.

Step-by-Step Solution

1

Determine the relationship between critical angle and wave speeds in the two media.

\sin C = \frac{n_B}{n_A} = \frac{v_A}{v_B}

By Snell's Law, critical angle occurs when the angle of refraction is

90^\circ

, giving

\sin C = \frac{n_2}{n_1}

. Since refractive index

n = \frac{c}{v}

, the ratio

\frac{n_B}{n_A}

simplifies to

\frac{v_A}{v_B}

.

2

Substitute the given wave speeds into the equation.

\sin C = \frac{1.5 \times 10^8\text{ m/s}}{3.0 \times 10^8\text{ m/s}} = 0.5

Light is traveling from the optically denser medium

A

(lower speed) to the less dense medium

B

(higher speed).

3

Calculate the inverse sine to find the critical angle

C

.

C = \arcsin(0.5) = 30^\circ

The angle whose sine is

0.5

is

30^\circ

.

Key Concept

Critical Angle and Total Internal Reflection in terms of Wave Speed

Question 3Question →

A water tank has a real depth of $16.0\text{ cm}$ . If the refractive index of water relative to air is $\frac{4}{3}$ , what is the apparent depth of the tank in centimetres when viewed normally from above?

Show answer & explanation

Answer: 12

Answer

The apparent depth of the water tank is

12.0\text{ cm}

.

When light passes from water to air, refraction causes the apparent depth to be smaller than the real depth by a factor equal to the refractive index

n

. Substituting a real depth of

16.0\text{ cm}

and

n = \frac{4}{3}

into

\text{Apparent Depth} = \frac{\text{Real Depth}}{n}

yields

12.0\text{ cm}

.

Step-by-Step Solution

1

Identify the relationship between refractive index, real depth, and apparent depth.

n = \frac{\text{Real Depth}}{\text{Apparent Depth}}

Refraction at a plane boundary causes an object immersed in a denser medium to appear closer to the surface when viewed from a rarer medium.

2

Rearrange the equation to isolate Apparent Depth.

\text{Apparent Depth} = \frac{\text{Real Depth}}{n}

The unknown quantity requested by the question is the apparent depth.

3

Substitute the known numerical values into the equation.

\text{Apparent Depth} = \frac{16.0}{\frac{4}{3}} = 16.0 \times \frac{3}{4} = 12.0\text{ cm}

Dividing

16.0

by

\frac{4}{3}

gives

12.0

.

Key Concept

Real and Apparent Depth in Light Refraction

Question 4Question →

A light ray strikes the first face of a glass prism of refracting angle $30^\circ$ at normal incidence ( $i = 0^\circ$ ). If the refractive index of the glass is $1.50$ , calculate the angle of emergence, in degrees, as the ray leaves the second face into air. (Take $\arcsin(0.75) = 48.6^\circ$ )

Show answer & explanation

Answer: 48.6

Answer

The angle of emergence of the light ray as it exits the prism is

48.6^\circ

.

Because the ray is incident normally at the first surface, it continues undeviated into the glass (

r_1 = 0^\circ

). By prism geometry, the angle of incidence at the second face is equal to the apex angle of the prism (

r_2 = A = 30^\circ

). Applying Snell's Law at the glass-air boundary gives

1.50 \sin(30^\circ) = 1.00 \sin(e)

, leading to

\sin(e) = 0.75

, which evaluates to an emergent angle of

48.6^\circ

.

Step-by-Step Solution

1

Determine the angle of refraction at the first surface

r_1 = 0^\circ

Light entering a surface normally (

i_1 = 0^\circ

) passes straight through without bending.

2

Find the angle of incidence at the second surface inside the prism using prism geometry

r_2 = 30^\circ

For any triangular prism, the refracting angle

A = r_1 + r_2

. Since

r_1 = 0^\circ

,

r_2 = A = 30^\circ

.

3

Apply Snell's Law at the second interface (glass to air)

\sin(e) = 0.75

n_{\text{glass}} \sin(r_2) = n_{\text{air}} \sin(e) \implies 1.50 \times \sin(30^\circ) = 1.00 \times \sin(e)

.

4

Calculate the emergent angle

e

e = 48.6^\circ

Taking the inverse sine of

0.75

yields

e = \arcsin(0.75) = 48.6^\circ

.

Key Concept

Prism Geometry and Snell's Law Refraction

Question 5Question →

A ray of light traveling inside a glass prism of refractive index $1.60$ strikes the boundary with an adjacent oil medium of refractive index $1.20$ . What is the sine of the critical angle for total internal reflection at this glass-oil interface?

0.75

1.33

0.625

0.80

Show answer & explanation

Answer: $0.75$

Answer

The sine of the critical angle at the glass-oil interface is

0.75

.

For light traveling from a medium of refractive index

n_1

into a medium of refractive index

n_2

(where

n_1 > n_2

), the critical angle

C

satisfies

\sin C = \frac{n_2}{n_1}

. Substituting

n_1 = 1.60

and

n_2 = 1.20

gives

\sin C = \frac{1.20}{1.60} = 0.75

.

Step-by-Step Solution

1

Identify the refractive indices of the two media.

Denser medium (glass):

n_1 = 1.60

; less dense medium (oil):

n_2 = 1.20

.

Total internal reflection occurs when light originates in the optically denser medium and strikes the boundary with a less dense medium.

2

Apply Snell's law at the critical angle

C

.

n_1 \sin C = n_2 \sin 90^\circ \implies \sin C = \frac{n_2}{n_1}

.

At the critical angle of incidence, the angle of refraction in the second medium is

90^\circ

, so

\sin 90^\circ = 1

.

3

Substitute the given numerical values to compute

\sin C

.

\sin C = \frac{1.20}{1.60} = \frac{3}{4} = 0.75

.

Dividing the refractive index of the oil by that of the glass gives the exact sine of the critical angle.

Key Concept

Critical Angle between Two Media

Question 6Question →

A ray of light passes symmetrically through an equilateral glass prism of refractive index $\sqrt{2}$ . What is the angle of minimum deviation of the ray, in degrees?

Show answer & explanation

Answer: 30

Answer

The angle of minimum deviation of the ray is

30^\circ

.

For an equilateral triangular prism, the apex angle

A

is

60^\circ

. At minimum deviation, light travels symmetrically through the prism and obeys the exact relation

n = \frac{\sin\left(\frac{A + D_m}{2}\right)}{\sin\left(\frac{A}{2}\right)}

. Substituting

n = \sqrt{2}

and

A = 60^\circ

gives

\sin\left(\frac{60^\circ + D_m}{2}\right) = \frac{\sqrt{2}}{2} = \sin(45^\circ)

. Equating arguments yields

\frac{60^\circ + D_m}{2} = 45^\circ

, leading directly to

D_m = 30^\circ

.

Step-by-Step Solution

1

Determine the refracting angle of the prism

A = 60^\circ

An equilateral prism has interior angles of

60^\circ

each, so the apex angle

A = 60^\circ

.

2

Set up the prism formula for minimum deviation

n = \frac{\sin\left(\frac{A + D_m}{2}\right)}{\sin\left(\frac{A}{2}\right)}

When light passes symmetrically through a prism, the deviation is at its minimum value

D_m

.

3

Substitute known values into the equation

\sqrt{2} = \frac{\sin\left(\frac{60^\circ + D_m}{2}\right)}{\sin(30^\circ)}

Given

n = \sqrt{2}

and

A = 60^\circ

, with

\sin(30^\circ) = 0.5

.

4

Calculate the sine of the half-angle

\sin\left(\frac{60^\circ + D_m}{2}\right) = \sqrt{2} \times 0.5 = \frac{\sqrt{2}}{2}

Multiplying both sides by

\sin(30^\circ) = 0.5

.

5

Solve for the minimum deviation angle

D_m

D_m = 30^\circ

Since

\arcsin\left(\frac{\sqrt{2}}{2}\right) = 45^\circ

, we set

\frac{60^\circ + D_m}{2} = 45^\circ \implies 60^\circ + D_m = 90^\circ \implies D_m = 30^\circ

.

Key Concept

Refraction of light through a prism at the angle of minimum deviation

Question 7Question →

A coin lies at the bottom of a vessel filled with a liquid to a depth of $14.0\text{ cm}$ . If the refractive index of the liquid relative to air is $1.40$ , calculate the apparent upward displacement of the coin in centimeters when viewed vertically from directly above.

Show answer & explanation

Answer: 4

Answer

The apparent upward displacement of the coin is

4.0\text{ cm}

.

Refraction at the liquid-air boundary makes an object at real depth

h = 14.0\text{ cm}

appear at an apparent depth

h' = \frac{h}{n} = \frac{14.0}{1.40} = 10.0\text{ cm}

. The apparent upward displacement is the difference between real depth and apparent depth:

d = 14.0\text{ cm} - 10.0\text{ cm} = 4.0\text{ cm}

.

Step-by-Step Solution

1

Identify the given values and formula for refractive index in terms of depth.

Real depth

h = 14.0\text{ cm}

, refractive index

n = 1.40

. Formula:

n = \frac{\text{Real depth}}{\text{Apparent depth}} = \frac{h}{h'}

.

Light rays bending away from the normal upon leaving the denser liquid cause the coin to appear closer to the surface.

2

Calculate the apparent depth (

h'

).

h' = \frac{14.0\text{ cm}}{1.40} = 10.0\text{ cm}

.

Rearranging the refractive index formula gives

h' = \frac{h}{n}

.

3

Calculate the apparent upward displacement (

d

).

d = h - h' = 14.0\text{ cm} - 10.0\text{ cm} = 4.0\text{ cm}

.

The displacement is the distance between the actual position at the bottom and the virtual image position.

Key Concept

Real depth, apparent depth, and apparent displacement

Question 8Question →

A ray of light traveling in air strikes the flat surface of a transparent glass slab at an angle of incidence of $60^\circ$ . If the refractive index of the glass slab relative to air is $\sqrt{3}$ , what is the angle of refraction inside the glass slab in degrees?

Show answer & explanation

Answer: 30

Answer

The angle of refraction inside the glass slab is

30^\circ

.

According to Snell's Law,

n = \frac{\sin i}{\sin r}

. Substituting

n = \sqrt{3}

and

i = 60^\circ

gives

\sqrt{3} = \frac{\sin 60^\circ}{\sin r}

. Since

\sin 60^\circ = \frac{\sqrt{3}}{2}

, rearranging gives

\sin r = \frac{\sqrt{3}/2}{\sqrt{3}} = 0.5

. Taking the inverse sine of

0.5

gives

r = 30^\circ

.

Step-by-Step Solution

1

Identify Snell's law formula relating the angle of incidence and angle of refraction.

n = \frac{\sin i}{\sin r}

Snell's law describes how light bends when crossing the boundary between two optical media.

2

Substitute the given values

i = 60^\circ

and

n = \sqrt{3}

into the equation.

\sqrt{3} = \frac{\sin 60^\circ}{\sin r}

Plugging in the known parameters allows us to isolate the unknown sine of the angle of refraction.

3

Substitute

\sin 60^\circ = \frac{\sqrt{3}}{2}

and rearrange for

\sin r

.

\sin r = \frac{\sqrt{3}/2}{\sqrt{3}} = 0.5

Canceling

\sqrt{3}

from both sides yields a simple numerical value for

\sin r

.

4

Take the inverse sine of

0.5

to find

r

.

r = \arcsin(0.5) = 30^\circ

The angle whose sine is

0.5

is

30^\circ

.

Key Concept

Snell's Law of Refraction

Question 9Question →

A rectangular glass block of thickness $6.0\text{ cm}$ has a refractive index of $1.50$ . Calculate the apparent depth, in centimeters, of a mark placed at the bottom of the block when viewed normally from above.

Show answer & explanation

Answer: 4

Answer

The apparent depth of the mark is

4.0\text{ cm}

.

The refractive index of a medium relative to air is given by the ratio of real depth to apparent depth (

n = \frac{\text{Real Depth}}{\text{Apparent Depth}}

). Substituting the given values gives

\text{Apparent Depth} = \frac{6.0\text{ cm}}{1.50} = 4.0\text{ cm}

.

Step-by-Step Solution

1

Identify the relationship between refractive index, real depth, and apparent depth for normal view

n = \frac{\text{Real Depth}}{\text{Apparent Depth}}

By definition of optical refraction when looking normally from an optically less dense medium (air) into a denser medium (glass).

2

Substitute the known values (

n = 1.50

,

\text{Real Depth} = 6.0\text{ cm}

) and solve for the apparent depth

\text{Apparent Depth} = \frac{6.0\text{ cm}}{1.50} = 4.0\text{ cm}

Dividing the real thickness by the refractive index yields the perceived (apparent) depth.

Key Concept

Real and Apparent Depth in Refraction

Question 10Question →

Which of the following conditions must be satisfied for light to undergo total internal reflection at the boundary between two transparent media?

The light ray must travel from an optically denser medium into an optically less dense medium, and the angle of incidence must exceed the critical angle.

The light ray must travel from an optically less dense medium into an optically denser medium, and the angle of incidence must exceed the critical angle.

The light ray must travel from an optically denser medium into an optically less dense medium, and the angle of incidence must be less than the critical angle.

The light ray must travel from an optically less dense medium into an optically denser medium at an angle of incidence equal to

0^\circ

.

Show answer & explanation

Answer: The light ray must travel from an optically denser medium into an optically less dense medium, and the angle of incidence must exceed the critical angle.

Answer

The light ray must travel from an optically denser medium into an optically less dense medium, and the angle of incidence must exceed the critical angle.

Total internal reflection occurs only when light passes from a medium of higher optical density into a medium of lower optical density, and the angle of incidence exceeds the critical angle for that boundary.

Step-by-Step Solution

1

Identify the boundary requirement for wave propagation direction in total internal reflection.

Light must travel from an optically denser medium (higher refractive index

n_1

) toward an optically rarer medium (lower refractive index

n_2

).

This direction allows the ray to bend away from the normal, increasing the angle of refraction relative to the angle of incidence.

2

Determine the required angle of incidence at the boundary.

The angle of incidence

i

must be strictly greater than the critical angle

\theta_c

(where

\sin \theta_c = n_2 / n_1

).

When

i > \theta_c

, no refraction can take place because

\sin r > 1

, forcing all light energy to reflect back into the initial denser medium.

Key Concept

Conditions for Total Internal Reflection

Question 11Question →

A swimming pool has an apparent depth of $1.8\text{ m}$ when viewed vertically from directly above. If the refractive index of water relative to air is $\frac{4}{3}$ , what is the real depth of the pool in meters?

Show answer & explanation

Answer: 2.4

Answer

The real depth of the pool is

2.4\text{ m}

.

The refractive index

n

of a medium is defined as the ratio of the real depth to the apparent depth (

n = \frac{\text{Real Depth}}{\text{Apparent Depth}}

). Multiplying the observed apparent depth of

1.8\text{ m}

by the refractive index

\frac{4}{3}

gives the true real depth of

2.4\text{ m}

.

Step-by-Step Solution

1

Identify the relationship between refractive index, real depth, and apparent depth.

Refractive index

n = \frac{\text{Real Depth}}{\text{Apparent Depth}}

.

Light bending at the boundary causes an object submerged in a denser medium to appear closer to the surface.

2

Rearrange the equation to make Real Depth the subject.

\text{Real Depth} = n \times \text{Apparent Depth}.

To calculate the true depth from the observed apparent depth and the optical density of water.

3

Substitute the given values into the formula.

\text{Real Depth} = \frac{4}{3} \times 1.8\text{ m} = 2.4\text{ m}.

Multiplying the apparent depth by the refractive index yields the actual physical depth.

Key Concept

Refraction of Light and Real/Apparent Depth

Question 12Question →

A small object lies at the bottom of a transparent vessel containing two immiscible liquid layers, $A$ and $B$ . Layer $A$ (top) has a real thickness of $7.0\text{ cm}$ and a refractive index of $1.40$ . Layer $B$ (bottom) has a real thickness of $8.0\text{ cm}$ and a refractive index of $1.60$ . Calculate the apparent displacement of the object, in centimeters, when viewed vertically from directly above.

Show answer & explanation

Answer: 5

Answer

The apparent displacement of the object is

5.0\text{ cm}

.

For multiple refractive layers viewed normally, the total apparent depth is the sum of individual layer apparent depths:

\frac{7.0}{1.40} + \frac{8.0}{1.60} = 5.0 + 5.0 = 10.0\text{ cm}

. Subtracting this total apparent depth from the total real depth of

15.0\text{ cm}

gives an apparent vertical shift (displacement) of

5.0\text{ cm}

.

Step-by-Step Solution

1

Calculate the apparent depth of the top liquid layer (Layer A)

Apparent depth of Layer A =

5.0\text{ cm}

Apparent depth for a single medium is obtained by dividing real depth by its refractive index:

\frac{7.0}{1.40} = 5.0\text{ cm}

.

2

Calculate the apparent depth of the bottom liquid layer (Layer B)

Apparent depth of Layer B =

5.0\text{ cm}

Apparent depth for Layer B is

\frac{8.0}{1.60} = 5.0\text{ cm}

.

3

Calculate total real depth and total apparent depth

Total real depth =

15.0\text{ cm}

; Total apparent depth =

10.0\text{ cm}

Depths in composite media are additive.

4

Calculate vertical apparent displacement

Apparent displacement =

5.0\text{ cm}

Displacement is the difference between total real depth and total apparent depth:

15.0\text{ cm} - 10.0\text{ cm} = 5.0\text{ cm}

.

Key Concept

Refraction through composite media and vertical apparent displacement

Question 13Question →

A light ray traveling inside a transparent glass prism of refractive index $1.50$ strikes the boundary with air. What is the sine of the critical angle ( $\sin C$ ) for total internal reflection to occur at this boundary?

0.67

1.50

0.50

0.87

Show answer & explanation

Answer: $0.67$

Answer

The sine of the critical angle (

\sin C

) for total internal reflection at the glass-air boundary is

0.67

.

For light moving from a medium with refractive index

n

into air, total internal reflection occurs when the angle of incidence exceeds the critical angle

C

. The critical angle satisfies

\sin C = \frac{1}{n}

. Substituting

n = 1.50

gives

\sin C = \frac{1}{1.50} = 0.67

.

Step-by-Step Solution

1

Identify the relationship between the refractive index (

n

) and the critical angle (

C

) when light travels from a medium to air.

\sin C = \frac{1}{n}

By Snell's law, at the critical angle of incidence, the angle of refraction in air is

90^\circ

(so

\sin 90^\circ = 1

).

2

Substitute the given refractive index (

n = 1.50

) into the formula.

\sin C = \frac{1}{1.50} = \frac{2}{3} \approx 0.67

Dividing

1

by

1.50

yields

0.666...

, which rounds to

0.67

.

Key Concept

Critical Angle and Total Internal Reflection

Estimated Time:45s

Question 14Question →

A light ray enters the first face of a glass prism surrounded by air. The prism has a refracting angle of $75^\circ$ and a refractive index of $\sqrt{2}$ . Inside the prism, the ray strikes the second refracting face at the critical angle for total internal reflection. Calculate the angle of incidence, in degrees, at the first face.

Show answer & explanation

Answer: 45

Answer

The angle of incidence at the first face is

45^\circ

.

To find the angle of incidence at the first face, first determine the critical angle

C

at the second glass-air surface using

\sin C = \frac{1}{n} = \frac{1}{\sqrt{2}}

, which yields

C = 45^\circ

. Since the ray strikes the second face at this critical angle,

r_2 = 45^\circ

. Next, using the geometric relationship for a prism

A = r_1 + r_2

, the angle of refraction at the first surface is

r_1 = A - r_2 = 75^\circ - 45^\circ = 30^\circ

. Finally, applying Snell's law at the first face gives

\sin i = n \sin r_1 = \sqrt{2} \sin 30^\circ = \frac{\sqrt{2}}{2}

. Taking the inverse sine yields

i = 45^\circ

.

Step-by-Step Solution

1

Find the critical angle at the second face

Critical angle

C = 45^\circ

, so

r_2 = 45^\circ

Light travels from glass to air at the critical angle, so

\sin C = \frac{1}{n} = \frac{1}{\sqrt{2}}

.

2

Determine the angle of refraction at the first face

r_1 = 30^\circ

The apex angle of a prism satisfies

A = r_1 + r_2

, hence

r_1 = A - r_2 = 75^\circ - 45^\circ = 30^\circ

.

3

Apply Snell's Law at the entry boundary

\sin i = \frac{\sqrt{2}}{2}

Refraction at the first surface gives

\sin i = n \sin r_1 = \sqrt{2} \sin 30^\circ = \sqrt{2} \times 0.5 = \frac{\sqrt{2}}{2}

.

4

Solve for the incident angle

i

i = 45^\circ

\arcsin\left(\frac{\sqrt{2}}{2}\right) = 45^\circ

.

Key Concept

Refraction through a prism combined with total internal reflection critical angle condition

Question 15Question →

A light wave travels through medium A at a speed of $2.25 \times 10^8\text{ m s}^{-1}$ and enters medium B, where its speed decreases to $1.50 \times 10^8\text{ m s}^{-1}$ . What is the value of the sine of the critical angle for total internal reflection between these two media, and in which medium must the light ray originate?

\frac{2}{3}

, originating in medium B

\frac{2}{3}

, originating in medium A

\frac{3}{2}

, originating in medium B

\frac{3}{2}

, originating in medium A

Show answer & explanation

Answer: $\frac{2}{3}$ , originating in medium B

Answer

The sine of the critical angle is

\frac{2}{3}

, and the light ray must originate in medium B.

Total internal reflection occurs when light travels from an optically denser medium to an optically rarer medium. Since the speed of light is lower in medium B (

1.50 \times 10^8\text{ m s}^{-1}

) than in medium A (

2.25 \times 10^8\text{ m s}^{-1}

), medium B is the denser medium. The critical angle

C

satisfies

\sin C = \frac{v_{\text{dense}}}{v_{\text{rare}}} = \frac{1.50 \times 10^8}{2.25 \times 10^8} = \frac{2}{3}

. Therefore, the light must originate in medium B and the sine of the critical angle is

\frac{2}{3}

.

Step-by-Step Solution

1

Determine the relative optical densities of medium A and medium B from wave speed

Medium B has a lower light speed (

1.50 \times 10^8\text{ m s}^{-1}

) than medium A (

2.25 \times 10^8\text{ m s}^{-1}

), so medium B is optically denser than medium A.

Refractive index is inversely proportional to wave speed (

n \propto \frac{1}{v}

).

2

Identify the required direction of light propagation for total internal reflection

The light ray must originate in medium B and travel toward medium A.

Total internal reflection occurs only when light attempts to pass from a medium of higher refractive index (denser) to a medium of lower refractive index (rarer).

3

Calculate the sine of the critical angle

\sin C = \frac{v_B}{v_A} = \frac{1.50 \times 10^8}{2.25 \times 10^8} = \frac{2}{3}

.

By Snell's law at the critical angle,

\sin C = \frac{n_A}{n_B} = \frac{v_B}{v_A}

.

Key Concept

Conditions for Total Internal Reflection and Critical Angle calculation from wave speeds

Estimated Time:1m 30s

Question 16Question →

A glass prism with a refracting angle of $60^\circ$ has a refractive index of $\sqrt{2}$ . What is the angle of minimum deviation, in degrees, experienced by a light ray passing through this prism?

Show answer & explanation

Answer: 30

Answer

The angle of minimum deviation is

30^\circ

.

Using the prism minimum deviation relation

n = \frac{\sin\left((A + D_m)/2\right)}{\sin(A/2)}

, substituting

n = \sqrt{2}

and

A = 60^\circ

yields

\sin\left((60^\circ + D_m)/2\right) = \frac{1}{\sqrt{2}}

. This gives

(60^\circ + D_m)/2 = 45^\circ

, so

D_m = 30^\circ

.

Step-by-Step Solution

1

Apply the prism minimum deviation equation.

n = \frac{\sin\left(\frac{A + D_m}{2}\right)}{\sin\left(\frac{A}{2}\right)}

This relates refractive index, prism refracting angle, and minimum deviation angle.

2

Substitute the given values

A = 60^\circ

and

n = \sqrt{2}

.

\sqrt{2} = \frac{\sin\left(\frac{60^\circ + D_m}{2}\right)}{\sin(30^\circ)}

Dividing the refracting angle

60^\circ

by 2 gives

30^\circ

for the denominator angle.

3

Calculate the numerator sine term.

\sin\left(\frac{60^\circ + D_m}{2}\right) = \frac{1}{\sqrt{2}}

Multiplying

\sqrt{2}

by

\sin(30^\circ) = 0.5

gives

\frac{\sqrt{2}}{2} = \frac{1}{\sqrt{2}}

.

4

Solve for the angle

D_m

.

D_m = 30^\circ

Since

\arcsin\left(\frac{1}{\sqrt{2}}\right) = 45^\circ

, we have

\frac{60^\circ + D_m}{2} = 45^\circ

, leading to

60^\circ + D_m = 90^\circ

.

Key Concept

Minimum Deviation in Triangular Prisms

Question 17Question →

A block of transparent polymer with a refractive index of $1.25$ is placed over a small mark on a table. When viewed vertically from directly above, the mark appears to be shifted upward by $3.0\text{ cm}$ . What is the real thickness of the polymer block in centimeters?

Show answer & explanation

Answer: 15

Answer

The real thickness of the polymer block is

15.0\text{ cm}

.

Refraction at the interface produces an upward shift

s = d\left(1 - \frac{1}{n}\right)

. Substituting

s = 3.0\text{ cm}

and

n = 1.25

gives

3.0 = d(1 - 0.80) = 0.20d

, which solves to a real thickness

d = 15.0\text{ cm}

.

Step-by-Step Solution

1

Relate apparent depth

d'

to real depth

d

using the refractive index.

d' = \frac{d}{n}

Light refracting at the surface of a denser medium causes the object to appear at a shallower depth.

2

Express the apparent displacement (upward shift)

s

in terms of

d

and

n

.

s = d - d' = d\left(1 - \frac{1}{n}\right)

The upward shift is the vertical distance between the true position and the apparent position.

3

Substitute

s = 3.0\text{ cm}

and

n = 1.25

into the equation and solve for

d

.

3.0 = d\left(1 - \frac{1}{1.25}\right) = d(1 - 0.80) = 0.20 d \implies d = 15.0\text{ cm}

Algebraic evaluation yields the exact real thickness of the polymer block.

Key Concept

Apparent depth and upward displacement due to refraction

Question 18Question →

An optical fiber consists of a core with a refractive index of $1.50$ surrounded by a cladding with a refractive index of $1.20$ . Which of the following correctly describes the direction a light ray must travel for total internal reflection to take place at the boundary, as well as the sine of the critical angle?

From core to cladding with

\sin C = 0.80

From cladding to core with

\sin C = 0.80

From core to cladding with

\sin C = 1.25

From cladding to core with

\sin C = 1.25

Show answer & explanation

Answer: From core to cladding with $\sin C = 0.80$

Answer

Light must travel from the core to the cladding with

\sin C = 0.80

Total internal reflection takes place only when light travels from an optically denser medium (

n_1 = 1.50

) to an optically less dense medium (

n_2 = 1.20

). Applying Snell's law at the critical boundary gives

\sin C = \frac{n_2}{n_1} = \frac{1.20}{1.50} = 0.80

.

Step-by-Step Solution

1

Determine the direction requirement for Total Internal Reflection (TIR)

Light must travel from the denser medium (core,

n_1 = 1.50

) toward the less dense medium (cladding,

n_2 = 1.20

).

TIR only occurs when light attempts to pass into a medium of lower optical density so that the refracted ray bends away from the normal.

2

Calculate the sine of the critical angle

C

\sin C = \frac{n_2}{n_1} = \frac{1.20}{1.50} = 0.80

By Snell's law at the critical angle,

n_1 \sin C = n_2 \sin 90^\circ \implies \sin C = \frac{n_2}{n_1}

.

Key Concept

Total Internal Reflection and Critical Angle

Estimated Time:1m 30s

Question 19Question →

A beam of monochromatic light travels through a transparent liquid toward a boundary with air. The speed of light in the liquid is $1.80 \times 10^8\text{ m s}^{-1}$ and the speed of light in air is $3.00 \times 10^8\text{ m s}^{-1}$ . What is the value of the sine of the critical angle ( $\sin C$ ) for total internal reflection at this boundary, and under what condition of propagation can total internal reflection occur?

\sin C = 0.60

, and the light ray must travel from the liquid into air.

\sin C = 0.60

, and the light ray must travel from air into the liquid.

\sin C = 1.67

, and the light ray must travel from the liquid into air.

\sin C = 1.67

, and the light ray must travel from air into the liquid.

Show answer & explanation

Answer: $\sin C = 0.60$ , and the light ray must travel from the liquid into air.

Answer

The sine of the critical angle is

\sin C = 0.60

, and total internal reflection can only occur when light travels from the optically denser liquid into air.

Total internal reflection requires light to originate in the optically denser medium (the liquid) and travel toward the rarer medium (air), at an angle of incidence greater than the critical angle. The sine of the critical angle is calculated directly as the ratio of the speed of light in the medium to the speed of light in air:

\sin C = \frac{v_{\text{liquid}}}{v_{\text{air}}} = \frac{1.80 \times 10^8}{3.00 \times 10^8} = 0.60

.

Step-by-Step Solution

1

Determine the refractive index

n

of the liquid relative to air.

n = \frac{c}{v} = \frac{3.00 \times 10^8\text{ m s}^{-1}}{1.80 \times 10^8\text{ m s}^{-1}} = 1.67

(or

\frac{5}{3}

).

Refractive index is the ratio of the speed of light in vacuum/air to the speed of light in the medium.

2

Calculate the sine of the critical angle

\sin C

.

\sin C = \frac{1}{n} = \frac{v}{c} = \frac{1.80 \times 10^8}{3.00 \times 10^8} = 0.60

.

The critical angle relationship between a medium and air is given by

\sin C = \frac{1}{n}

.

3

Identify the necessary physical condition for total internal reflection to take place.

Light must travel from an optically denser medium (liquid) toward an optically less dense medium (air).

For total internal reflection to happen, the ray must bend away from the normal until the angle of refraction reaches

90^\circ

. This only occurs when moving from a denser to a rarer medium.

Key Concept

Critical Angle and Conditions for Total Internal Reflection

Estimated Time:1m 15s

Question 20Question →

A container holds a layer of water of depth $16.0\text{ cm}$ . An immiscible layer of oil of refractive index $1.20$ and thickness $6.0\text{ cm}$ floats on top of the water. If the refractive index of water is $1.33$ (or $\frac{4}{3}$ ), what is the total apparent depth, in centimeters, of a small object resting at the bottom of the container when viewed normally from directly above?

Show answer & explanation

Answer: 17

Answer

The total apparent depth of the object when viewed normally from directly above is

17.0\text{ cm}

.

When an object at the bottom of a container is viewed normally through multiple transparent media, the overall apparent depth is the sum of the apparent depths produced by each medium individually (

d_{\text{app}} = \sum \frac{d_i}{n_i}

). Substituting the given values gives

\frac{16.0}{4/3} + \frac{6.0}{1.20} = 12.0\text{ cm} + 5.0\text{ cm} = 17.0\text{ cm}

.

Step-by-Step Solution

1

Calculate the apparent depth of the water layer.

Apparent depth of water =

12.0\text{ cm}

Apparent depth in a single medium is given by the real depth divided by its refractive index (

d_{\text{app}} = d / n

). For water,

\frac{16.0}{4/3} = 12.0\text{ cm}

.

2

Calculate the apparent depth of the oil layer.

Apparent depth of oil =

5.0\text{ cm}

Using

d_{\text{app}} = d / n

for the oil layer,

\frac{6.0}{1.20} = 5.0\text{ cm}

.

3

Sum the apparent depths of both media.

Total apparent depth =

17.0\text{ cm}

For multiple parallel transparent layers, the total apparent depth is the sum of the apparent depths of each individual layer.

Key Concept

Apparent depth in composite media layers

Refraction of Light, Total Internal Reflection, and Prisms

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