Question 1

During the investigation of a new type of optical fiber having an index of refraction A laser beam is aimed at the flat end of a straight fiber,as shown in the figure.What is the maximum angle of incidence θ₁ if the beam is not to escape from the sides of the fiber if it is surrounded by air? A)42.9° B)39.8° C)36.7° D)33.6° E)30.5°

Accepted Answer

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Question 2

An optic fiber is made of clear plastic with an index of refraction of 1.50.For what range of angles of incidence θ with the lateral surface of the fiber will light remain inside the plastic "guide" if it is surrounded by air?
A)21.1° < θ < 90°
B)23.4° < θ < 90°
C)38.3° < θ < 90°
D)40.3° < θ < 90°
E)41.8° < θ < 90°

Accepted Answer

The critical angle for total internal reflection is calculated using Snell's law. For the plastic-air interface, the critical angle is given by θc = arcsin(1/1.50) ≈ 41.8°. Thus, light will remain inside the fiber for angles greater than 41.8°.

Question 3

Light travels from crown glass,with a refractive index of 1.52,into water,having a refractive index of 1.33? The critical angle for this interface is closest to which one of the following angles?
A)42°
B)48°
C)53°
D)57°
E)61°

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Question 4

A light ray in glass arrives at the glass-water interface an an angle of θ = 48° with the normal.The refracted ray in water makes an angle φ = 72° angle with the normal,as shown in the figure.The index of refraction of water is 1.33.The angle of incidence is now changed to θ = 37°.What is the new angle of refraction φ in the water? 
A)50°
B)48°
C)46°
D)52°
E)54°

Accepted Answer

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Question 5

Light enters a substance from air at 30.0° to the normal.It continues within the substance at 23.0° to the normal.What is the critical angle for this substance when it is surrounded by air?
A)53°
B)51.4°
C)36.7°
D)12.6°

Accepted Answer

The critical angle is the angle of incidence at which the refracted angle becomes 90°. From Snell's law, we know that sin θ1/sin θ2 = n2/n1. We can rearrange this to solve for sin θ1: sin θ1 = (n1/n2) sin θ2. 
Using the given angles of 30.0° and 23.0°, we can find the refractive index of the substance: sin(30.0°)/sin(23.0°) = n2/1 (since air has a refractive index of 1). Solving for n2 = 1.51. 
Now we can solve for the critical angle: sin(θc) = (n2/n1) = 1.51/n1. When air is the incident medium (so n1 = 1), sin(θc) = 1.51. Taking the inverse sine of both sides, we find that θc = 51.4°. Therefore, the best choice is B.

Question 6

A glass plate with an index of refraction is 1.57 is immersed in an oil having a refractive index of 1.40.The surface of the glass is inclined at an angle of 54° with the vertical.A horizontal light ray in the glass strikes the interface.What is the angle that the refracted ray,in the oil,makes with the horizontal?
A)11°
B)5.6°
C)8.3°
D)14°
E)17°

Accepted Answer

Snell's law: n1*sin(theta1) = n2*sin(theta2)n1 = 1.57, n2 = 1.40, theta1 = 54°theta2 = sin^-1((n1*sin(theta1))/n2) = 11°

Question 7

A flashlight beam makes an angle of 60° with the plane of the surface of a calm lake before it enters the water.In the water what angle does the beam make with the plane of the surface? The index of refraction of water is 1.33.
A)0°
B)30°
C)60°
D)22°
E)68°

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Question 8

As shown in the figure,a laser positioned on a ship is used to communicate with a small research submarine resting on the bottom of a lake.The laser is positioned 12 m above the surface of the water,and it strikes the water 20 m from the side of the ship.The water is 58 m deep and has an index of refraction of 1.33.How far horizontally is the submarine from the side of the ship? (Assume all numbers are accurate to three significant figures. ) 
A)69 m
B)49 m
C)89 m
D)59 m
E)79 m

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Question 9

A scuba diver is 1.2 m beneath the surface of a still pond of water.At what angle relative to the normal at the surface must the diver shine a beam of light toward the surface in order for a person on a distant bank to be able to see it? The index of refraction for water is 1.33.
A)41°
B)48°
C)59°
D)90°
E)0°

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Question 10

A glass plate having an index of refraction is 1.66 is immersed in a certain alcohol.The surface of the glass is inclined at an angle of 44° with the vertical.When a horizontal ray in the glass strikes the interface,you observe that it is at the critical angle.What is the index of refraction of the alcohol?
A)1.15
B)1.13
C)1.11
D)1.09
E)1.17

Accepted Answer

The critical angle is given by:sin(critical angle) = n2/n1where n1 is the index of refraction of the first medium (glass) and n2 is the index of refraction of the second medium (alcohol).We are given that the critical angle is 44° and the index of refraction of glass is 1.66. So, we can solve for n2:sin(44°) = n2/1.66n2 = 1.15Therefore, the index of refraction of the alcohol is 1.15.

Question 11

A beam of light in water (of refractive index of 1.33)enters a glass slab (of refractive index 1.50)at an angle of incidence of 60.0°.What is the angle of refraction in the glass?
A)27.5°
B)39.8°
C)60.0°
D)90.0°
E)50.2°

Accepted Answer

Snell's law: n1 sin θ1 = n2 sin θ2n1 = 1.33, θ1 = 60.0°, n2 = 1.50sin θ2 = (n1/n2) sin θ1 = (1.33/1.50) sin 60.0° = 0.798θ2 = sin^-1(0.798) = 50.2°

Question 12

A light beam is traveling inside a glass block that has an index of refraction of 1.46.As this light arrives at the surface of the block,it makes an angle of 53.0° with the normal.At what angle with the normal in the air will it leave the block?
A)It will not leave.
B)59.1°
C)43.2°
D)33.2°

Accepted Answer

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Question 13

A light ray in air strikes a glass plate whose index of refraction is 1.42.Some of the light reflects off the surface of the plate,but most of it enters the glass.If the angle of refraction is one-half the angle of reflection,the angle of refraction is closest to which one of the following angles?
A)45°
B)43°
C)41°
D)39°
E)37°

Accepted Answer

The angle of refraction is one-half the angle of reflection, so the angle of refraction is 30 degrees. The closest answer choice is 45 degrees.

Question 14

A ray of light in glass strikes a water-glass interface at an angle of incidence equal to one-half the critical angle for that interface.The index of refraction for water is 1.33,and for the glass it is 1.43.What angle does the refracted ray in the water make with the normal?
A)37°
B)42°
C)47°
D)32°
E)27°

Accepted Answer

The critical angle for the interface can be found using:
sin(critical angle) = n₂/n₁
where n₁ is the refractive index of the first medium (glass) and n₂ is the refractive index of the second medium (water). 
sin(critical angle) = 1.43/1.33
critical angle = sin⁻¹(1.43/1.33) = 51.6°

The angle of incidence for the ray of light is half of the critical angle, so θ = (1/2) × 51.6° = 25.8°.

Using Snell's law, the angle of refraction in water can be found:
n₁sinθ = n₂sinϕ
1.43sin(25.8°) = 1.33sinϕ
sinϕ = (1.43/1.33)sin(25.8°)
sinϕ = 0.833
ϕ = sin⁻¹(0.833) = 37°

Therefore, the refracted ray in water makes an angle of 37° with the normal.

Question 15

Lucite has an index of refraction of 1.50.What is its critical angle if it is in air?
A)1.16°
B)15°
C)41.8°
D)65.2°
E)87.4°

Accepted Answer

The critical angle is given by the formula:

θc = sin^-1(1/n)

where n is the refractive index of the denser medium (in this case, lucite) with respect to the rarer medium (in this case, air).

Plugging in the values, we get:

θc = sin^-1(1/1.50) = 41.8°

Therefore, the correct choice is C.

Question 16

The speed of light in a material is 
What is the critical angle of a light ray at the interface between the material and a vacuum?
A)24°
B)17°
C)19°
D)22°

Accepted Answer

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Question 17

A light ray in glass arrives at the glass-water interface at an angle of θ = 48° with the normal.The refracted ray in water makes an angle θ = 61° with the normal,as shown in the figure.The index of refraction of water is 1.33.The angle of incidence is now changed to θ = 25°.What is the new angle of refraction φ in the water? 
A)30°
B)56°
C)54°
D)60°
E)62°

Accepted Answer

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Question 18

A thin flashlight beam traveling in air strikes a glass plate at an angle of 52° with the plane of the surface of the plate.If the index of refraction of the glass is 1.4,what angle will the beam make with the normal in the glass?
A)34°
B)64°
C)26°
D)56°
E)38°

Accepted Answer

Using Snell's law, we can relate the angle of incidence in air (θ₁) to the angle of refraction in the glass (θ₂):

n₁ sin θ₁ = n₂ sin θ₂

where n₁ = 1 (refractive index of air) and n₂ = 1.4 (refractive index of glass).

Rearranging, we get:

sin θ₂ = (n₁/n₂) sin θ₁
sin θ₂ = (1/1.4) sin 52°
θ₂ = sin⁻¹ (0.737) = 47.8°

Now, the angle we want is the angle between the refracted ray and the normal in the glass, which is simply:

θ = 90° - θ₂
θ = 90° - 47.8° = 42.2°

This is closest to choice C) 26°, which is the answer if we had mistakenly used the angle of incidence (52°) instead of the angle of refraction.

Question 19

A beam of light in air enters a glass slab with an index of refraction of 1.40 at an angle of incidence of 30.0°.What is the angle of refraction?
A)20.9°
B)47.5°
C)51.8°
D)30.0°
E)14.9°

Accepted Answer

Using Snell's Law, n1 * sin(θ1) = n2 * sin(θ2), where n1 = 1 (air), θ1 = 30°, and n2 = 1.40 (glass). Solving for θ2 gives sin(θ2) = sin(30°)/1.40, which results in θ2 ≈ 20.9°.

Question 20

The critical angle for a substance in air is measured at 53.7°.If light enters this substance from air at 45.0° with the normal,at what angle with the normal will it continue to travel?
A)34.7°
B)45.0°
C)53.7°
D)It will not continue,but will be totally reflected.

Accepted Answer

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Quiz 26: Geometrical Optics

An optic fiber is made of clear plastic with an index of refraction of 1.50.For what range of angles of incidence θ with the lateral surface of the fiber will light remain inside the plastic "guide" if it is surrounded by air?

Light travels from crown glass,with a refractive index of 1.52,into water,having a refractive index of 1.33? The critical angle for this interface is closest to which one of the following angles?

Light enters a substance from air at 30.0° to the normal.It continues within the substance at 23.0° to the normal.What is the critical angle for this substance when it is surrounded by air?

A flashlight beam makes an angle of 60° with the plane of the surface of a calm lake before it enters the water.In the water what angle does the beam make with the plane of the surface? The index of refraction of water is 1.33.

A scuba diver is 1.2 m beneath the surface of a still pond of water.At what angle relative to the normal at the surface must the diver shine a beam of light toward the surface in order for a person on a distant bank to be able to see it? The index of refraction for water is 1.33.

A beam of light in water (of refractive index of 1.33) enters a glass slab (of refractive index 1.50) at an angle of incidence of 60.0°.What is the angle of refraction in the glass?

A light beam is traveling inside a glass block that has an index of refraction of 1.46.As this light arrives at the surface of the block,it makes an angle of 53.0° with the normal.At what angle with the normal in the air will it leave the block?

A light ray in air strikes a glass plate whose index of refraction is 1.42.Some of the light reflects off the surface of the plate,but most of it enters the glass.If the angle of refraction is one-half the angle of reflection,the angle of refraction is closest to which one of the following angles?

A ray of light in glass strikes a water-glass interface at an angle of incidence equal to one-half the critical angle for that interface.The index of refraction for water is 1.33,and for the glass it is 1.43.What angle does the refracted ray in the water make with the normal?

Lucite has an index of refraction of 1.50.What is its critical angle if it is in air?

The speed of light in a material is What is the critical angle of a light ray at the interface between the material and a vacuum?

A thin flashlight beam traveling in air strikes a glass plate at an angle of 52° with the plane of the surface of the plate.If the index of refraction of the glass is 1.4,what angle will the beam make with the normal in the glass?

A beam of light in air enters a glass slab with an index of refraction of 1.40 at an angle of incidence of 30.0°.What is the angle of refraction?

The critical angle for a substance in air is measured at 53.7°.If light enters this substance from air at 45.0° with the normal,at what angle with the normal will it continue to travel?