Total Internal Reflection
Q. A beam of light in diamond (n=2.42) strikes the surface of water (n=1.33). What is the critical angle for total internal reflection?
A.
25.5°
B.
30.0°
C.
36.9°
D.
42.0°
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Solution
Critical angle θc = sin⁻¹(n2/n1) = sin⁻¹(1.33/2.42) ≈ 36.9°.
Correct Answer: C — 36.9°
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Q. A beam of light in glass (n=1.5) strikes the glass-air interface at an angle of 60°. What happens to the light?
A.
It is refracted into the air.
B.
It undergoes total internal reflection.
C.
It is absorbed by the glass.
D.
It is scattered.
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Solution
Since the angle of incidence (60°) is greater than the critical angle (approximately 41.8° for glass to air), total internal reflection occurs.
Correct Answer: B — It undergoes total internal reflection.
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Q. A beam of light in glass (n=1.5) strikes the glass-air interface at an angle of 60°. Will total internal reflection occur?
A.
Yes
B.
No
C.
Only if the angle is increased
D.
Only if the angle is decreased
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Solution
To determine if total internal reflection occurs, we first find the critical angle using sin(θc) = 1/n = 1/1.5, which gives θc ≈ 41.8°. Since 60° > 41.8°, total internal reflection will not occur.
Correct Answer: B — No
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Q. A beam of light passes from diamond (n=2.42) to air. What is the critical angle?
A.
24.4°
B.
30.0°
C.
36.9°
D.
42.0°
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Solution
Critical angle θc = sin⁻¹(1/2.42) ≈ 24.4°.
Correct Answer: A — 24.4°
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Q. A diamond has a refractive index of 2.42. What is the critical angle for total internal reflection at the diamond-air interface?
A.
24.4°
B.
41.1°
C.
23.6°
D.
17.5°
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Solution
Critical angle θc = sin^(-1)(1/n) = sin^(-1)(1/2.42) ≈ 24.4°.
Correct Answer: A — 24.4°
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Q. A fiber optic cable uses total internal reflection to transmit light. What is the primary requirement for this to work effectively?
A.
The core must have a higher refractive index than the cladding
B.
The cladding must have a higher refractive index than the core
C.
The light must be monochromatic
D.
The cable must be straight
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Solution
For total internal reflection to occur in a fiber optic cable, the core must have a higher refractive index than the cladding.
Correct Answer: A — The core must have a higher refractive index than the cladding
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Q. A fiber optic cable uses total internal reflection. If the refractive index of the core is 1.6 and the cladding is 1.5, what is the critical angle?
A.
38.7°
B.
41.8°
C.
48.6°
D.
60.0°
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Solution
Critical angle θc = sin⁻¹(n2/n1) = sin⁻¹(1.5/1.6) ≈ 38.7°.
Correct Answer: A — 38.7°
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Q. A fiber optic cable uses total internal reflection. If the refractive index of the core is 1.5 and that of the cladding is 1.4, what is the critical angle?
A.
42.0°
B.
48.6°
C.
60.0°
D.
30.0°
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Solution
Critical angle θc = sin⁻¹(n2/n1) = sin⁻¹(1.4/1.5) ≈ 42.0°.
Correct Answer: A — 42.0°
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Q. A fiber optic cable uses total internal reflection. What is the role of the cladding?
A.
To increase the refractive index.
B.
To decrease the refractive index.
C.
To prevent light loss.
D.
To enhance light absorption.
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Solution
The cladding has a lower refractive index than the core, ensuring that light is kept within the core through total internal reflection.
Correct Answer: C — To prevent light loss.
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Q. A light ray in glass (n=1.5) strikes the boundary with air at an angle of 30°. Will it undergo total internal reflection?
A.
Yes
B.
No
C.
Depends on the angle
D.
Not enough information
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Solution
Since the angle of incidence is less than the critical angle (θc = sin⁻¹(1/1.5) ≈ 41.8°), it will not undergo total internal reflection.
Correct Answer: B — No
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Q. A light ray traveling in a medium with a refractive index of 1.6 strikes a boundary with air at an angle of 50°. What will be the outcome?
A.
Total internal reflection occurs.
B.
Light is refracted into the air.
C.
Light is absorbed.
D.
Light is scattered.
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Solution
The critical angle for this scenario is approximately 38.7°. Since 50° is greater than the critical angle, total internal reflection occurs.
Correct Answer: A — Total internal reflection occurs.
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Q. A light ray traveling in a medium with n=1.5 strikes the boundary with n=1.0 at 50°. What occurs?
A.
Total internal reflection
B.
Partial reflection and refraction
C.
Complete refraction
D.
None of the above
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Solution
Since 50° is greater than the critical angle (θc ≈ 41.8°), total internal reflection occurs.
Correct Answer: A — Total internal reflection
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Q. A light ray traveling in a medium with n=1.6 strikes the boundary with air at an angle of 50°. What will happen?
A.
Total internal reflection
B.
Partial reflection and refraction
C.
Complete absorption
D.
No reflection
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Solution
Calculate critical angle: θc = sin^(-1)(1/n) = sin^(-1)(1/1.6) ≈ 38.7°. Since 50° > 38.7°, total internal reflection occurs.
Correct Answer: A — Total internal reflection
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Q. A light ray traveling in a medium with n=2.0 strikes a boundary with air at an angle of incidence of 45°. What will be the angle of refraction in air?
A.
22.5°
B.
45°
C.
60°
D.
90°
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Solution
Using Snell's law, n1 * sin(θ1) = n2 * sin(θ2). Here, n1 = 2.0, θ1 = 45°, and n2 = 1.0 (air). Thus, 2.0 * sin(45°) = 1.0 * sin(θ2) leads to sin(θ2) = √2, which gives θ2 = 90°.
Correct Answer: D — 90°
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Q. A light ray traveling in diamond (n=2.42) strikes the diamond-air interface. What is the critical angle?
A.
24.4°
B.
36.9°
C.
42.5°
D.
49.5°
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Solution
Using sin(θc) = n2/n1, where n1 = 2.42 (diamond) and n2 = 1.0 (air), we find sin(θc) = 1.0/2.42, leading to θc ≈ 24.4°.
Correct Answer: A — 24.4°
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Q. A light ray traveling in diamond (n=2.42) strikes the diamond-air interface. What is the critical angle for total internal reflection?
A.
24.4°
B.
30.0°
C.
36.0°
D.
42.0°
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Solution
Using Snell's law, sin(θc) = n2/n1 = 1.00/2.42. Thus, θc ≈ 24.4°.
Correct Answer: A — 24.4°
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Q. A light ray traveling in glass (n=1.5) strikes the glass-air interface at an angle of 30°. Will it undergo total internal reflection?
A.
Yes
B.
No
C.
Depends on the angle
D.
Not enough information
Show solution
Solution
Since sin(30°) < sin(θc) where θc = sin⁻¹(1/1.5) ≈ 41.8°, it will not undergo total internal reflection.
Correct Answer: B — No
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Q. A ray of light in a medium with n=1.33 strikes the boundary with air at an angle of 45°. What is the behavior of the ray?
A.
Total internal reflection
B.
Partial reflection and refraction
C.
Complete absorption
D.
Total refraction
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Solution
Since 45° < θc ≈ 48.6°, the ray will partially reflect and refract.
Correct Answer: B — Partial reflection and refraction
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Q. A ray of light in glass (n=1.5) strikes the glass-air interface at an angle of 30°. What will be the angle of refraction in air?
A.
60°
B.
30°
C.
45°
D.
90°
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Solution
Using Snell's law, n1 * sin(θ1) = n2 * sin(θ2). Here, n1 = 1.5, θ1 = 30°, n2 = 1.0. Thus, sin(θ2) = (1.5 * sin(30°))/1.0 = 0.75, giving θ2 ≈ 60°.
Correct Answer: A — 60°
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Q. If a light ray enters a medium with a refractive index of 1.33 at an angle of 60°, what is the angle of refraction in the medium?
A.
30°
B.
45°
C.
60°
D.
75°
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Solution
Using Snell's law: n1 * sin(θ1) = n2 * sin(θ2), 1 * sin(60°) = 1.33 * sin(θ2) => sin(θ2) = (sin(60°)/1.33) ≈ 0.577, θ2 ≈ 35.0°.
Correct Answer: B — 45°
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Q. If a light ray in a medium with n=1.5 strikes a boundary with n=1.0 at an angle of 70°, what will happen?
A.
Total internal reflection occurs
B.
Light refracts into the air
C.
Light is absorbed
D.
Light reflects at 70°
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Solution
Since the angle of incidence (70°) is greater than the critical angle (approximately 41.8°), total internal reflection occurs.
Correct Answer: A — Total internal reflection occurs
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Q. If a light ray in diamond (n=2.42) strikes the diamond-air interface at an angle of 70°, will it undergo total internal reflection?
A.
Yes
B.
No
C.
Only if the angle is decreased
D.
Only if the angle is increased
Show solution
Solution
The critical angle for diamond to air is sin⁻¹(1/2.42) ≈ 24.4°. Since 70° > 24.4°, total internal reflection will occur.
Correct Answer: A — Yes
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Q. If the angle of incidence in a medium is 45° and the refractive index of the medium is 1.5, what is the angle of refraction in air?
A.
30°
B.
45°
C.
60°
D.
90°
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Solution
Using Snell's law: n1 * sin(θ1) = n2 * sin(θ2), 1.5 * sin(45°) = 1 * sin(θ2) => sin(θ2) = 1.5 * 0.7071 = 1.0607, which is not possible, hence total internal reflection occurs.
Correct Answer: C — 60°
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Q. If the angle of incidence in a medium is 70° and the refractive index of the medium is 1.5, will total internal reflection occur?
A.
Yes
B.
No
C.
Only at certain angles
D.
Not applicable
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Solution
Since 70° > θc = sin⁻¹(1/1.5) ≈ 41.8°, total internal reflection will occur.
Correct Answer: A — Yes
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Q. If the angle of incidence is 30° in a medium with a refractive index of 1.5, what is the angle of refraction in air?
A.
19.5°
B.
20.0°
C.
22.5°
D.
25.0°
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Solution
Using Snell's law: n1 * sin(θ1) = n2 * sin(θ2). Here, n1 = 1.5, n2 = 1.0, and θ1 = 30°. Thus, sin(θ2) = (1.5 * sin(30°))/1.0 = 0.75, leading to θ2 ≈ 48.6°.
Correct Answer: A — 19.5°
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Q. If the angle of incidence is 70° in a medium with a refractive index of 1.5, what is the angle of refraction in air?
A.
30°
B.
20°
C.
10°
D.
Total internal reflection occurs
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Solution
Since the angle of incidence (70°) is greater than the critical angle (approximately 41.8° for glass to air), total internal reflection occurs.
Correct Answer: D — Total internal reflection occurs
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Q. If the angle of incidence is 70° in a medium with a refractive index of 1.6, will total internal reflection occur when it hits air?
A.
Yes
B.
No
C.
Only if the angle is less than 90°
D.
Not enough information
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Solution
Critical angle θc = sin⁻¹(1/1.6) ≈ 38.7°. Since 70° > 38.7°, total internal reflection occurs.
Correct Answer: A — Yes
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Q. If the angle of incidence is equal to the critical angle, what is the angle of refraction?
A.
0°
B.
45°
C.
90°
D.
It cannot be determined.
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Solution
At the critical angle, the angle of refraction is 90°, meaning the refracted ray travels along the boundary.
Correct Answer: C — 90°
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Q. If the angle of incidence is equal to the critical angle, what is the behavior of the light ray?
A.
It is refracted at 90°.
B.
It is totally internally reflected.
C.
It is absorbed.
D.
It is transmitted.
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Solution
When the angle of incidence equals the critical angle, the light ray is refracted along the boundary at 90°, marking the threshold for total internal reflection.
Correct Answer: A — It is refracted at 90°.
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Q. If the critical angle for a certain material is 30°, what is the refractive index of that material?
A.
1.00
B.
1.15
C.
1.73
D.
2.00
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Solution
The refractive index n can be calculated using n = 1/sin(θc). For θc = 30°, n = 1/sin(30°) = 1/0.5 = 2.00.
Correct Answer: C — 1.73
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