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* These questions are only for students studying for the Optometry Admission Test (OAT). If you are taking the DAT, this section is not applicable to you. There is no physics on the DAT.
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Question 1 of 40Physics  Optics
1. Question
1. A thin lens produces a real image when:Correct / You marked this questionLet’s look at our thin lens equation and associated sign conventions:
I highly recommend memorizing the above information for the OAT. Whether you memorize this information in bullet points or visually depends on your preference; the point is that you shouldn’t be drawing out ray diagrams to derive it during your test. This will save you time and make it more likely that you obtain the correct answer.
You can see that the ONLY case where a real image is produced is if i (the distance of the image from the lens) is positive (+). This means that the image is projected on the OPPOSITE side of the lens (relative to the object).
The only case where i is positive (+) is if the lens is converging and if the object is farther away from the lens than the focal point
(o > f ). Here is the diagram so you can visualize:Incorrect / You marked this questionLet’s look at our thin lens equation and associated sign conventions:
I highly recommend memorizing the above information for the OAT. Whether you memorize this information in bullet points or visually depends on your preference; the point is that you shouldn’t be drawing out ray diagrams to derive it during your test. This will save you time and make it more likely that you obtain the correct answer.
You can see that the ONLY case where a real image is produced is if i (the distance of the image from the lens) is positive (+). This means that the image is projected on the OPPOSITE side of the lens (relative to the object).
The only case where i is positive (+) is if the lens is converging and if the object is farther away from the lens than the focal point
(o > f ). Here is the diagram so you can visualize: 
Question 2 of 40Physics  Optics
2. Question
2. A thin lens produces an inverted image when:Correct / You marked this questionLet’s look at our thin lens equation and associated sign conventions:
I highly recommend memorizing the above information for the OAT. Whether you memorize this information in bullet points or visually depends on your preference; the point is that you shouldn’t be drawing out ray diagrams to derive it during your test. This will save you time and make it more likely that you obtain the correct answer.
You can see that the ONLY case where an inverted image is produced is if the lens is converging. To address a couple of the more compelling incorrect answer options:
If the object distance equals the focal distance, no image will be produced.
The image will only be inverted if the lens is converging. Converging lenses only produce magnified imagesHere is the diagram so you can visualize:
Incorrect / You marked this questionLet’s look at our thin lens equation and associated sign conventions:
I highly recommend memorizing the above information for the OAT. Whether you memorize this information in bullet points or visually depends on your preference; the point is that you shouldn’t be drawing out ray diagrams to derive it during your test. This will save you time and make it more likely that you obtain the correct answer.
You can see that the ONLY case where an inverted image is produced is if the lens is converging. To address a couple of the more compelling incorrect answer options:
If the object distance equals the focal distance, no image will be produced.
The image will only be inverted if the lens is converging. Converging lenses only produce magnified imagesHere is the diagram so you can visualize:

Question 3 of 40Physics  Optics
3. Question
3. Light moves through the air into a piece of glass whose index of refraction is 2. The light hits the glass straight on (i.e the angle of incidence is 90^{o}).
What is the angle of refraction in the piece of glass?Correct / You marked this questionStep 1: Let’s draw a quick sketch and what we are solving for
Solving for: angle of refraction in the piece of glass (θ_{2})
Step 2: Let’s apply Snell’s Law to solve for θ_{2}
Recall the equation for Snell’s Law:
Plug in our given values to solve for θ_{2}:
n_{1}sinθ_{1} = n_{2}sinθ_{2}
(1)(1) = (2)(sinθ_{2})
sinθ_{2} = ^{1}/_{2}
θ_{2} = 30^{o}Incorrect / You marked this questionStep 1: Let’s draw a quick sketch and what we are solving for
Solving for: angle of refraction in the piece of glass (θ_{2})
Step 2: Let’s apply Snell’s Law to solve for θ_{2}
Recall the equation for Snell’s Law:
Plug in our given values to solve for θ_{2}:
n_{1}sinθ_{1} = n_{2}sinθ_{2}
(1)(1) = (2)(sinθ_{2})
sinθ_{2} = ^{1}/_{2}
θ_{2} = 30^{o} 
Question 4 of 40Physics  Optics
4. Question
4. Light comes down from the sun and hits your face at a 30^{o} angle relative to the normal. You, however, came prepared, and have a thick layer of zinc oxide sunscreen protecting you. The index of refraction for the sunscreen on your face provides a index of refraction of 2.5.
What is the sin value of the corresponding angle of refraction (θ_{2}) ?Correct / You marked this questionStep 1: Let’s draw a quick sketch and what we are solving for
Solving for: the sin value of angle of refraction (θ_{2}) in the zinc oxide sunscreen
Step 2: Let’s apply Snell’s Law to solve for sinθ_{2}
Recall the equation for Snell’s Law:
n_{1}sinθ_{1} = n_{2}sinθ_{2}
(1)(sin30) = (2.5)(sinθ_{2})
^{1}/_{2} = (2.5)(sinθ_{2})
sinθ_{2} = 0.2Real World Application: As you can see, the greater the index of refraction, the smaller the value of sinθ_{2}. This equates to the light slowing down, and thus becoming less powerful/ damaging to your skin.
Incorrect / You marked this questionStep 1: Let’s draw a quick sketch and what we are solving for
Solving for: the sin value of angle of refraction (θ_{2}) in the zinc oxide sunscreen
Step 2: Let’s apply Snell’s Law to solve for sinθ_{2}
Recall the equation for Snell’s Law:
n_{1}sinθ_{1} = n_{2}sinθ_{2}
(1)(sin30) = (2.5)(sinθ_{2})
^{1}/_{2} = (2.5)(sinθ_{2})
sinθ_{2} = 0.2Real World Application: As you can see, the greater the index of refraction, the smaller the value of sinθ_{2}. This equates to the light slowing down, and thus becoming less powerful/ damaging to your skin.

Question 5 of 40Physics  Optics
5. Question
5. Zinc oxide is an ingredient commonly used in sunscreens. The speed of light in zinc oxide is 1.25 x 10^{8} m/s What is the index of refraction in zinc oxide?Correct / You marked this questionStep 1: Let’s apply our index of refraction equation
n = c / v
n = (3 x 10^{8} m/s) / (1.25 x 10^{8})
n = 2.4Incorrect / You marked this questionStep 1: Let’s apply our index of refraction equation
n = c / v
n = (3 x 10^{8} m/s) / (1.25 x 10^{8})
n = 2.4 
Question 6 of 40Physics  Optics
6. Question
6. Scenario 1: You are looking at an object through a thin converging lens with a focal length of 1 cm. The image is projected at 4 cm.
Scenario 2: You are now using a lens with a focal length half of the original lens, but the image is still projected at 4 cm.
What does this imply about the object distance in Scenario 2 relative to Scenario 1?Correct / You marked this questionStep 1: Let’s break out our thin lens equation and associated sign conventions, and write out what we are solving for:
Solving for: distance of the object from the lens (o_{2}) in Scenario 2 relative to Scenario 1 (o_{1})
Step 2: Analyze the object distance (o) in each scenario using the thin lens equation
Scenario 1:
^{1}/_{o} + ^{1}/_{i} = ^{1}/_{f}
^{1}/_{o} + ^{1}/_{4 cm} = ^{1}/_{1 cm}
^{1}/_{o} = ^{3}/_{4 cm}
o = ^{4}/_{3} cmRemember, by convention, when an o value is positive, the object is to the left side of our lens in our sketch. In Scenario 1, the object is ^{4}/_{3} cm away from the lens.
Scenario 2:
^{1}/_{o} + ^{1}/_{i} = ^{1}/_{f}
^{1}/_{o} + ^{1}/_{4 cm} = ^{1}/_{0.5 cm}
^{1}/_{o} + ^{1}/_{4 cm} = 2 cm
^{1}/_{o} = ^{7}/_{4 cm}
o = ^{4}/_{7} cm^{4}/_{7} cm < ^{4}/_{3} cm. Thus, the object is closer in Scenario 2 than Scenario 1
Incorrect / You marked this questionStep 1: Let’s break out our thin lens equation and associated sign conventions, and write out what we are solving for:
Solving for: distance of the object from the lens (o_{2}) in Scenario 2 relative to Scenario 1 (o_{1})
Step 2: Analyze the object distance (o) in each scenario using the thin lens equation
Scenario 1:
^{1}/_{o} + ^{1}/_{i} = ^{1}/_{f}
^{1}/_{o} + ^{1}/_{4 cm} = ^{1}/_{1 cm}
^{1}/_{o} = ^{3}/_{4 cm}
o = ^{4}/_{3} cmRemember, by convention, when an o value is positive, the object is to the left side of our lens in our sketch. In Scenario 1, the object is ^{4}/_{3} cm away from the lens.
Scenario 2:
^{1}/_{o} + ^{1}/_{i} = ^{1}/_{f}
^{1}/_{o} + ^{1}/_{4 cm} = ^{1}/_{0.5 cm}
^{1}/_{o} + ^{1}/_{4 cm} = 2 cm
^{1}/_{o} = ^{7}/_{4 cm}
o = ^{4}/_{7} cm^{4}/_{7} cm < ^{4}/_{3} cm. Thus, the object is closer in Scenario 2 than Scenario 1

Question 7 of 40Physics  Optics
7. Question
7. Scenario 1: A lens yields an image that is virtual and upright.
Scenario 2: A lens yields an image that is real and inverted.
What can be inferred to have changed from Scenario 1 to Scenario 2?Correct / You marked this questionStep 1: Let’s recall our thin lens equation and associated sign conventions:
Step 2: Let’s analyze Scenario 1 vs. Scenario 2 using our thin lens equation
Scenario 1:
We are told that the image is virtual and upright. This implies that i is negative (). We can see from the additional information underneath our thin lens equation that this could be the case for a Converging OR Diverging lens.Scenario 2:
We are told that image is real and inverted. This can only be the case for a converging lens system in which the object is farther from the lens than the focal point. We can rule out option B.Though it may be less clear, we can also rule out option A. As described above, we cannot deduce with certainty that the original lens was diverging. We could be moving from one converging lens to another, and only changing the object distance.
We can definitely rule out option C. Nothing tells us that the object distance from the original focal point is smaller in magnitude.
You may now be leaning towards answer option D. However …
Consider if Scenario 1 was a diverging lens. For any object distance, the diverging lens will produce a virtual, upright image. Thus, we could certainly have: Scenario 2: Converging Lens with object distance greater than focal distance
 Scenario 1: Diverging Lens with object distance greater than object distance in Scenario 2
 Thus, we can rule out option D
We cannot infer ANY of the answer options AD with certainty.
Incorrect / You marked this questionStep 1: Let’s recall our thin lens equation and associated sign conventions:
Step 2: Let’s analyze Scenario 1 vs. Scenario 2 using our thin lens equation
Scenario 1:
We are told that the image is virtual and upright. This implies that i is negative (). We can see from the additional information underneath our thin lens equation that this could be the case for a Converging OR Diverging lens.Scenario 2:
We are told that image is real and inverted. This can only be the case for a converging lens system in which the object is farther from the lens than the focal point. We can rule out option B.Though it may be less clear, we can also rule out option A. As described above, we cannot deduce with certainty that the original lens was diverging. We could be moving from one converging lens to another, and only changing the object distance.
We can definitely rule out option C. Nothing tells us that the object distance from the original focal point is smaller in magnitude.
You may now be leaning towards answer option D. However …
Consider if Scenario 1 was a diverging lens. For any object distance, the diverging lens will produce a virtual, upright image. Thus, we could certainly have: Scenario 2: Converging Lens with object distance greater than focal distance
 Scenario 1: Diverging Lens with object distance greater than object distance in Scenario 2
 Thus, we can rule out option D
We cannot infer ANY of the answer options AD with certainty.

Question 8 of 40Physics  Optics
8. Question
8. Scenario 1: A lens yields an image that is real and inverted.
Scenario 2: A lens yields an image that is virtual and upright.
What can be inferred from the given image?Correct / You marked this questionStep 1: Let’s recall our thin lens equation and associated sign conventions:
Step 2: Let’s analyze Scenario 1 vs. Scenario 2 using our thin lens equation
Scenario 1:
We are told that image is real and inverted. This can only be the case for a converging lens system in which the object is farther from the lens than the focal point. Thus, we can quickly see that the answer is D.Step 3: For thoroughness, let’s analyze the rest of the answer options:
Based on our analysis of Scenario 1, we can rule out answer option A, C, & E.
Scenario 2:
We are told that the image is virtual and upright. This implies that i is negative (). We can see from the additional information underneath our thin lens equation that this could be the case for a Converging OR Diverging lens. Thus, we can rule out answer option BIncorrect / You marked this questionStep 1: Let’s recall our thin lens equation and associated sign conventions:
Step 2: Let’s analyze Scenario 1 vs. Scenario 2 using our thin lens equation
Scenario 1:
We are told that image is real and inverted. This can only be the case for a converging lens system in which the object is farther from the lens than the focal point. Thus, we can quickly see that the answer is D.Step 3: For thoroughness, let’s analyze the rest of the answer options:
Based on our analysis of Scenario 1, we can rule out answer option A, C, & E.
Scenario 2:
We are told that the image is virtual and upright. This implies that i is negative (). We can see from the additional information underneath our thin lens equation that this could be the case for a Converging OR Diverging lens. Thus, we can rule out answer option B 
Question 9 of 40Physics  Optics
9. Question
9. An image is projected 5 cm from a diverging lens. The focal length of the lens is 10 cm. What is the magnification of the lens?Correct / You marked this questionStep 1: First, let’s break out our magnification equation and write out what we are solving for:
Solving for: the magnification of the lens
Let’s recall our magnification equation:
We are given i. We must solve for o.
Step 2: Solve for o using the thin lens equation for a diverging lens:
^{1}/_{o} + ^{1}/_{i} = ^{1}/_{f}
^{1}/_{o} + ^{1}/_{5 cm} = ^{1}/_{10 cm}
^{1}/_{o} = ^{1}/_{10 cm}
o = 10 cmNote: you will get this problem WRONG if you forget the sign convention for i. With a diverging lens, the image (i) is ALWAYS negative.
Step 3: Solve for magnification
M = – i / o
M = – (5 cm) / 10 cm
M = ^{1}/_{2} = 0.5Incorrect / You marked this questionStep 1: First, let’s break out our magnification equation and write out what we are solving for:
Solving for: the magnification of the lens
Let’s recall our magnification equation:
We are given i. We must solve for o.
Step 2: Solve for o using the thin lens equation for a diverging lens:
^{1}/_{o} + ^{1}/_{i} = ^{1}/_{f}
^{1}/_{o} + ^{1}/_{5 cm} = ^{1}/_{10 cm}
^{1}/_{o} = ^{1}/_{10 cm}
o = 10 cmNote: you will get this problem WRONG if you forget the sign convention for i. With a diverging lens, the image (i) is ALWAYS negative.
Step 3: Solve for magnification
M = – i / o
M = – (5 cm) / 10 cm
M = ^{1}/_{2} = 0.5 
Question 10 of 40Physics  Optics
10. Question
10. A real image is projected 12 cm from a converging lens. The focal length of the lens is 3 cm. What is the magnification of the lens?Correct / You marked this questionStep 1: First, let’s break out our magnification equation and write out what we are solving for:
Solving for: the magnification of the lens
Let’s recall our magnification equation:
We are given i. We must solve for o.
Step 2: Solve for o using the thin lens equation for a converging lens
Note: remember the sign convention for i. A real image implies that i is positive.
^{1}/_{o} + ^{1}/_{i} = ^{1}/_{f}
^{1}/_{o} + ^{1}/_{12 cm} = ^{1}/_{3 cm}
^{1}/_{o} = ^{1}/_{4 cm}
o = 4 cmStep 3: Solve for magnification
M = – i / o
M = – (12 cm) / 4 cm
M = – 3Incorrect / You marked this questionStep 1: First, let’s break out our magnification equation and write out what we are solving for:
Solving for: the magnification of the lens
Let’s recall our magnification equation:
We are given i. We must solve for o.
Step 2: Solve for o using the thin lens equation for a converging lens
Note: remember the sign convention for i. A real image implies that i is positive.
^{1}/_{o} + ^{1}/_{i} = ^{1}/_{f}
^{1}/_{o} + ^{1}/_{12 cm} = ^{1}/_{3 cm}
^{1}/_{o} = ^{1}/_{4 cm}
o = 4 cmStep 3: Solve for magnification
M = – i / o
M = – (12 cm) / 4 cm
M = – 3 
Question 11 of 40Physics  Optics
11. Question
11. An object is viewed using a thin diverging lens. The image produced will be:Correct / You marked this questionStep 1: First, let’s recall our thin lens equation and associated sign conventions:
I highly recommend memorizing the above information for the OAT. Whether you memorize this information in bullet points or visually depends on your preference; the point is that you shouldn’t be drawing out ray diagrams to derive it during your test. This will save you time and make it more likely that you obtain the correct answer.
Step 2: Let’s go through our answer options individually:
[A] Real, magnified, & inverted. This only is the case for a converging lens
[B] Virtual, upright, & reduced. This is the correct answer. All images produced by a diverging lens will be virtual, same side, upright, and reduced
[C] Magnified. This only is the case for a converging lens
[D] On the opposite side of the object. This only is the case for a converging lens
[E] Real if object distance > focal distance. This only is the case for a converging lens
Incorrect / You marked this questionStep 1: First, let’s recall our thin lens equation and associated sign conventions:
I highly recommend memorizing the above information for the OAT. Whether you memorize this information in bullet points or visually depends on your preference; the point is that you shouldn’t be drawing out ray diagrams to derive it during your test. This will save you time and make it more likely that you obtain the correct answer.
Step 2: Let’s go through our answer options individually:
[A] Real, magnified, & inverted. This only is the case for a converging lens
[B] Virtual, upright, & reduced. This is the correct answer. All images produced by a diverging lens will be virtual, same side, upright, and reduced
[C] Magnified. This only is the case for a converging lens
[D] On the opposite side of the object. This only is the case for a converging lens
[E] Real if object distance > focal distance. This only is the case for a converging lens

Question 12 of 40Physics  Optics
12. Question
12. An object is viewed using a thin diverging lens. The object is twice as far from the lens as the focal point. The object is in front of the diverging lens.
What will the corresponding image distance be?Correct / You marked this questionStep 1: First, let’s recall our thin lens equation and associated sign conventions:
I highly recommend memorizing the above information for the OAT. Whether you memorize this information in bullet points or visually depends on your preference; the point is that you shouldn’t be drawing out ray diagrams to derive it during your test. This will save you time and make it more likely that you obtain the correct answer.
Step 2: Let’s analyze the scenario with the thin lens equation:
Remember, by convention, for a diverging lens, both f and i are ALWAYS negative. I recommend using some example values here, instead of working only with variables.
We are told that o = 2f. Let’s set object distance (o) at 2 cm, and focal distance (f) at 1 cm. Because this is a diverging lens, and we are told the object is “in front of” the lens:
o = + 2 cm
f = – 1 cm
i = negative^{1}/_{o} + ^{1}/_{i} = ^{1}/_{f}
^{1}/_{i} = ^{1}/_{f} – ^{1}/_{o}
^{1}/_{i} = ^{1}/_{1 cm} – ^{1}/_{2 cm}
^{1}/_{i} = ^{3}/_{2 cm}
i = –^{2}/_{3} cmYou can see that i is ^{2}/_{3} of the focal distance
Incorrect / You marked this questionStep 1: First, let’s recall our thin lens equation and associated sign conventions:
I highly recommend memorizing the above information for the OAT. Whether you memorize this information in bullet points or visually depends on your preference; the point is that you shouldn’t be drawing out ray diagrams to derive it during your test. This will save you time and make it more likely that you obtain the correct answer.
Step 2: Let’s analyze the scenario with the thin lens equation:
Remember, by convention, for a diverging lens, both f and i are ALWAYS negative. I recommend using some example values here, instead of working only with variables.
We are told that o = 2f. Let’s set object distance (o) at 2 cm, and focal distance (f) at 1 cm. Because this is a diverging lens, and we are told the object is “in front of” the lens:
o = + 2 cm
f = – 1 cm
i = negative^{1}/_{o} + ^{1}/_{i} = ^{1}/_{f}
^{1}/_{i} = ^{1}/_{f} – ^{1}/_{o}
^{1}/_{i} = ^{1}/_{1 cm} – ^{1}/_{2 cm}
^{1}/_{i} = ^{3}/_{2 cm}
i = –^{2}/_{3} cmYou can see that i is ^{2}/_{3} of the focal distance

Question 13 of 40Physics  Optics
13. Question
13. Total internal reflection could occur in which the of following scenarios:Correct / You marked this questionStep 1: Let’s recall our Critical Angle Formula:
Step 2: Let’s analyze our answer options using the information above
By definition, if the angle of incidence is > than critical angle, light will undergo total internal reflection. Thus, we can rule out options D & E.
When light passes from a less dense medium (lower n) to a more dense medium (higher n) light is refracted TOWARDS the normal.
Thus, we can rule out option option B.When light passes from a one medium to another, which both have the same index of refraction (n_{2} = n_{1}) NO REFRACTION will occur. Thus, we can rule out option C.
When light passes from a more dense medium (higher n) to a less dense medium (lower n) light is refracted AWAY FROM the normal. This does not GUARANTEE Total Internal Reflection; however, for total internal reflection to occur, light must pass from medium 1 to a medium with a LOWER index of refraction. That is, n_{1} must be > n_{2}
Graphically:
Angle Incidence < Critical Angle
Angle Incidence = Critical Angle
Angle Incidence > Critical Angle
Incorrect / You marked this questionStep 1: Let’s recall our Critical Angle Formula:
Step 2: Let’s analyze our answer options using the information above
By definition, if the angle of incidence is > than critical angle, light will undergo total internal reflection. Thus, we can rule out options D & E.
When light passes from a less dense medium (lower n) to a more dense medium (higher n) light is refracted TOWARDS the normal.
Thus, we can rule out option option B.When light passes from a one medium to another, which both have the same index of refraction (n_{2} = n_{1}) NO REFRACTION will occur. Thus, we can rule out option C.
When light passes from a more dense medium (higher n) to a less dense medium (lower n) light is refracted AWAY FROM the normal. This does not GUARANTEE Total Internal Reflection; however, for total internal reflection to occur, light must pass from medium 1 to a medium with a LOWER index of refraction. That is, n_{1} must be > n_{2}
Graphically:
Angle Incidence < Critical Angle
Angle Incidence = Critical Angle
Angle Incidence > Critical Angle

Question 14 of 40Physics  Optics
14. Question
14. Total internal reflection will always occur in which the of following scenarios:Correct / You marked this questionStep 1: Let’s recall our Critical Angle Formula:
The Critical Angle is the incident angle beyond which light is completely reflected. The corresponding formula is:
Step 2: Let’s analyze our answer options using the information above
When light passes from a less dense medium (lower n) to a more dense medium (higher n) light is refracted TOWARDS the normal.
Thus, we can rule out option option B.When light passes from a one medium to another, which both have the same index of refraction (n_{2} = n_{1}) NO REFRACTION will occur. Thus, we can rule out option C.
When light passes from a more dense medium (higher n) to a less dense medium (lower n) light is refracted AWAY FROM the normal. For total internal reflection to occur, light must pass from medium 1 to a medium with a LOWER index of refraction. HOWEVER, this does not GUARANTEE Total Internal Reflection. Thus, we can rule out option A.
When the angle of incidence is = the critical angle, the light will be refracted at 90^{o}, NOT reflected. Thus, we can rule out option E. By definition, if the angle of incidence is > than critical angle, light will undergo total internal reflection.
Graphically:
Angle Incidence < Critical Angle
Angle Incidence = Critical Angle
Angle Incidence > Critical Angle
Incorrect / You marked this questionStep 1: Let’s recall our Critical Angle Formula:
The Critical Angle is the incident angle beyond which light is completely reflected. The corresponding formula is:
Step 2: Let’s analyze our answer options using the information above
When light passes from a less dense medium (lower n) to a more dense medium (higher n) light is refracted TOWARDS the normal.
Thus, we can rule out option option B.When light passes from a one medium to another, which both have the same index of refraction (n_{2} = n_{1}) NO REFRACTION will occur. Thus, we can rule out option C.
When light passes from a more dense medium (higher n) to a less dense medium (lower n) light is refracted AWAY FROM the normal. For total internal reflection to occur, light must pass from medium 1 to a medium with a LOWER index of refraction. HOWEVER, this does not GUARANTEE Total Internal Reflection. Thus, we can rule out option A.
When the angle of incidence is = the critical angle, the light will be refracted at 90^{o}, NOT reflected. Thus, we can rule out option E. By definition, if the angle of incidence is > than critical angle, light will undergo total internal reflection.
Graphically:
Angle Incidence < Critical Angle
Angle Incidence = Critical Angle
Angle Incidence > Critical Angle

Question 15 of 40Physics  Optics
15. Question
15. An object is 10 cm in front of thin lens and yields an image 5 cm n front of the lens. What can we deduce about the lens?Correct / You marked this questionStep 1: First, let’s recall our thin lens equation and associated sign conventions:
I highly recommend memorizing the above information for the OAT. Whether you memorize this information in bullet points or visually depends on your preference; the point is that you shouldn’t be drawing out ray diagrams to derive it during your test. This will save you time and make it more likely that you obtain the correct answer.
Step 2: Let’s analyze the scenario with the thin lens equation:
You can assume the “in front of” and “to the left of” mean the same thing. Thus, we can deduce:
o = + 10 cm
i = – 5 cm^{1}/_{o} + ^{1}/_{i} = ^{1}/_{f}
^{1}/_{10 cm} + ^{1}/_{5 cm} = ^{1}/_{f}
– ^{1}/_{10 cm} = ^{1}/_{f}
f = – 10 cmBy convention, only a diverging lens has a negative focal length. This rules out answer options A, C, D, & E.
Incorrect / You marked this questionStep 1: First, let’s recall our thin lens equation and associated sign conventions:
I highly recommend memorizing the above information for the OAT. Whether you memorize this information in bullet points or visually depends on your preference; the point is that you shouldn’t be drawing out ray diagrams to derive it during your test. This will save you time and make it more likely that you obtain the correct answer.
Step 2: Let’s analyze the scenario with the thin lens equation:
You can assume the “in front of” and “to the left of” mean the same thing. Thus, we can deduce:
o = + 10 cm
i = – 5 cm^{1}/_{o} + ^{1}/_{i} = ^{1}/_{f}
^{1}/_{10 cm} + ^{1}/_{5 cm} = ^{1}/_{f}
– ^{1}/_{10 cm} = ^{1}/_{f}
f = – 10 cmBy convention, only a diverging lens has a negative focal length. This rules out answer options A, C, D, & E.

Question 16 of 40Physics  Optics
16. Question
16. An object is 4 cm in front of thin lens and yields an image 8 cm behind the lens. What can we deduce about the lens?Correct / You marked this questionStep 1: First, let’s recall our thin lens equation and associated sign conventions:
I highly recommend memorizing the above information for the OAT. Whether you memorize this information in bullet points or visually depends on your preference; the point is that you shouldn’t be drawing out ray diagrams to derive it during your test. This will save you time and make it more likely that you obtain the correct answer.
Step 2: Let’s analyze the scenario with the thin lens equation:
You can assume the “in front of” and “to the left of” mean the same thing. Thus, we can deduce:
o = + 4 cm
i = + 8 cmWe know from our sign conventions that a positive image distance (i) can only be the case for a converging lens. Thus, options A is the correct answer.
Step 3: For thoroughness, we’ll disprove our other answer options:
[B] We have already proven the lens must be converting
[C] A converging lens always magnifies the image
[D] By convention, a converging lens always has a positive focal length. If we solve out for focal length:^{1}/_{o} + ^{1}/_{i} = ^{1}/_{f}
^{1}/_{4 cm} + ^{1}/_{8 cm} = ^{1}/_{f}
^{3}/_{8 cm} = ^{1}/_{f}
f = ^{8}/_{3} cm[E] We were able to deduce information about the lens.
Incorrect / You marked this questionStep 1: First, let’s recall our thin lens equation and associated sign conventions:
I highly recommend memorizing the above information for the OAT. Whether you memorize this information in bullet points or visually depends on your preference; the point is that you shouldn’t be drawing out ray diagrams to derive it during your test. This will save you time and make it more likely that you obtain the correct answer.
Step 2: Let’s analyze the scenario with the thin lens equation:
You can assume the “in front of” and “to the left of” mean the same thing. Thus, we can deduce:
o = + 4 cm
i = + 8 cmWe know from our sign conventions that a positive image distance (i) can only be the case for a converging lens. Thus, options A is the correct answer.
Step 3: For thoroughness, we’ll disprove our other answer options:
[B] We have already proven the lens must be converting
[C] A converging lens always magnifies the image
[D] By convention, a converging lens always has a positive focal length. If we solve out for focal length:^{1}/_{o} + ^{1}/_{i} = ^{1}/_{f}
^{1}/_{4 cm} + ^{1}/_{8 cm} = ^{1}/_{f}
^{3}/_{8 cm} = ^{1}/_{f}
f = ^{8}/_{3} cm[E] We were able to deduce information about the lens.

Question 17 of 40Physics  Optics
17. Question
17. An object is 4 cm in front of thin lens and yields an image 4 cm behind the lens. What is the focal length of the lens?Correct / You marked this questionStep 1: First, let’s recall our thin lens equation and associated sign conventions:
I highly recommend memorizing the above information for the OAT. Whether you memorize this information in bullet points or visually depends on your preference; the point is that you shouldn’t be drawing out ray diagrams to derive it during your test. This will save you time and make it more likely that you obtain the correct answer.
Step 2: Let’s analyze the scenario with the thin lens equation:
You can assume the “in front of” and “to the left of” mean the same thing. Thus, we can deduce:
o = + 4 cm
i = + 4 cm^{1}/_{o} + ^{1}/_{i} = ^{1}/_{f}
^{1}/_{4 cm} + ^{1}/_{4 cm} = ^{1}/_{f}
^{1}/_{2 cm} = ^{1}/_{f}
f = 2 cmIncorrect / You marked this questionStep 1: First, let’s recall our thin lens equation and associated sign conventions:
I highly recommend memorizing the above information for the OAT. Whether you memorize this information in bullet points or visually depends on your preference; the point is that you shouldn’t be drawing out ray diagrams to derive it during your test. This will save you time and make it more likely that you obtain the correct answer.
Step 2: Let’s analyze the scenario with the thin lens equation:
You can assume the “in front of” and “to the left of” mean the same thing. Thus, we can deduce:
o = + 4 cm
i = + 4 cm^{1}/_{o} + ^{1}/_{i} = ^{1}/_{f}
^{1}/_{4 cm} + ^{1}/_{4 cm} = ^{1}/_{f}
^{1}/_{2 cm} = ^{1}/_{f}
f = 2 cm 
Question 18 of 40Physics  Optics
18. Question
18. An object is 40 mm in front of thin lens with a focal length of +2 cm. What is the image distance?Correct / You marked this questionStep 1: First, let’s recall our thin lens equation and associated sign conventions:
I highly recommend memorizing the above information for the OAT. Whether you memorize this information in bullet points or visually depends on your preference; the point is that you shouldn’t be drawing out ray diagrams to derive it during your test. This will save you time and make it more likely that you obtain the correct answer.
Step 2: Let’s analyze the scenario with the thin lens equation:
You can assume the “in front of” and “to the left of” mean the same thing. Thus, we can deduce:
o = + 40 mm
f = + 2 cm = 20 mm^{1}/_{o} + ^{1}/_{i} = ^{1}/_{f}
^{1}/_{40 mm} + ^{1}/_{i} = ^{1}/_{20 mm}
^{1}/_{i} = ^{1}/_{40 mm}
i = 40 mmNote: make sure you keep your units consistent throughout your use of the thin lens equation! As long as you do this, the equation holds true if you use cm or mm (or, any other units of length, but those are the most common you’ll see).
Incorrect / You marked this questionStep 1: First, let’s recall our thin lens equation and associated sign conventions:
I highly recommend memorizing the above information for the OAT. Whether you memorize this information in bullet points or visually depends on your preference; the point is that you shouldn’t be drawing out ray diagrams to derive it during your test. This will save you time and make it more likely that you obtain the correct answer.
Step 2: Let’s analyze the scenario with the thin lens equation:
You can assume the “in front of” and “to the left of” mean the same thing. Thus, we can deduce:
o = + 40 mm
f = + 2 cm = 20 mm^{1}/_{o} + ^{1}/_{i} = ^{1}/_{f}
^{1}/_{40 mm} + ^{1}/_{i} = ^{1}/_{20 mm}
^{1}/_{i} = ^{1}/_{40 mm}
i = 40 mmNote: make sure you keep your units consistent throughout your use of the thin lens equation! As long as you do this, the equation holds true if you use cm or mm (or, any other units of length, but those are the most common you’ll see).

Question 19 of 40Physics  Optics
19. Question
19. An object is 500 mm in front of a thin lens. The image is projected 500 mm behind the lens. What is the power of the lens?Correct / You marked this questionStep 1: Let’s work backwards. How do we solve for the power of a lens?
Let’s recall our power equation:
We must solve for the focal length (f) in order to solve for power.
Step 2: Solve for the focal length (f) of the lens
Recall our thin lens equation and associated sign conventions:
We can deduce that the object distance (o) and image distance (i) are both positive. This is a converging lens.
^{1}/_{o} + ^{1}/_{i} = ^{1}/_{f}
^{1}/_{500 mm} + ^{1}/_{500 mm} = ^{1}/_{f}
^{1}/_{250 mm} = ^{1}/_{f}
f = 250 mm = 0.25 mP = ^{1}/_{f}
P = ^{1}/_{0.25 m}
P = 4 dioptersIncorrect / You marked this questionStep 1: Let’s work backwards. How do we solve for the power of a lens?
Let’s recall our power equation:
We must solve for the focal length (f) in order to solve for power.
Step 2: Solve for the focal length (f) of the lens
Recall our thin lens equation and associated sign conventions:
We can deduce that the object distance (o) and image distance (i) are both positive. This is a converging lens.
^{1}/_{o} + ^{1}/_{i} = ^{1}/_{f}
^{1}/_{500 mm} + ^{1}/_{500 mm} = ^{1}/_{f}
^{1}/_{250 mm} = ^{1}/_{f}
f = 250 mm = 0.25 mP = ^{1}/_{f}
P = ^{1}/_{0.25 m}
P = 4 diopters 
Question 20 of 40Physics  Optics
20. Question
20. An object is 100 cm in front a of thin lens. The image is projected 100 cm behind the lens. What is the power of the lens?Correct / You marked this questionStep 1: Let’s work backwards. How do we solve for the power of a lens?
Let’s recall our power equation:
We must solve for the focal length (f) in order to solve for power.
Step 2: Solve for the focal length (f) of the lens
Recall our thin lens equation and associated sign conventions:
We can deduce that the object distance (o) and image distance (i) are both positive. This is a converging lens.
^{1}/_{o} + ^{1}/_{i} = ^{1}/_{f}
^{1}/_{100 cm} + ^{1}/_{100 cm} = ^{1}/_{f}
^{1}/_{50 cm} = ^{1}/_{f}
f = 50 cm = 0.5 mP = ^{1}/_{f}
P = ^{1}/_{0.5 m}
P = 2 dioptersIncorrect / You marked this questionStep 1: Let’s work backwards. How do we solve for the power of a lens?
Let’s recall our power equation:
We must solve for the focal length (f) in order to solve for power.
Step 2: Solve for the focal length (f) of the lens
Recall our thin lens equation and associated sign conventions:
We can deduce that the object distance (o) and image distance (i) are both positive. This is a converging lens.
^{1}/_{o} + ^{1}/_{i} = ^{1}/_{f}
^{1}/_{100 cm} + ^{1}/_{100 cm} = ^{1}/_{f}
^{1}/_{50 cm} = ^{1}/_{f}
f = 50 cm = 0.5 mP = ^{1}/_{f}
P = ^{1}/_{0.5 m}
P = 2 diopters 
Question 21 of 40Physics  Optics
21. Question
21. An image is projected 10 cm behind a converging lens. The object distance is 5 cm. What is the magnification of the lens?Correct / You marked this questionStep 1: First, let’s break out our magnification equation and write out what we are solving for:
Solving for: the magnification of the lens
We are given both i & o.
Step 3: Solve for magnification
M = – i / o
M = – (10 cm / 5 cm)
M = – 2xIncorrect / You marked this questionStep 1: First, let’s break out our magnification equation and write out what we are solving for:
Solving for: the magnification of the lens
We are given both i & o.
Step 3: Solve for magnification
M = – i / o
M = – (10 cm / 5 cm)
M = – 2x 
Question 22 of 40Physics  Optics
22. Question
22. What distinguishes an object that appears blue from an object that appears green?Correct / You marked this questionThis is a conceptual problem.
An object appears as a different color due to the wavelength of light reflected back into our eyes.
An object that appears blue reflects light with a wavelength of ~ 470 nm. Light of all other wavelengths (i.e. all of the other colors) is absorbed by the object.
An object that appears green reflects light with a wavelength of ~ 540 nm. Light of all other wavelengths (i.e. all of the other colors) is absorbed by the object.
The polarization and amplitude of the light reflected or absorbed by an object may be the same for objects that appear two different colors.
Incorrect / You marked this questionThis is a conceptual problem.
An object appears as a different color due to the wavelength of light reflected back into our eyes.
An object that appears blue reflects light with a wavelength of ~ 470 nm. Light of all other wavelengths (i.e. all of the other colors) is absorbed by the object.
An object that appears green reflects light with a wavelength of ~ 540 nm. Light of all other wavelengths (i.e. all of the other colors) is absorbed by the object.
The polarization and amplitude of the light reflected or absorbed by an object may be the same for objects that appear two different colors.

Question 23 of 40Physics  Optics
23. Question
23. What type of lens is a magnifying glass?Correct / You marked this questionThis is a conceptual problem.
A magnifying glass is a type of converging (convex) lens used to produce a magnified image. A diverging (concave) lens will only produced a reduced image. Though multiple lens systems can be used for magnification purposes, a magnifying glass generally uses only a single lens.
Incorrect / You marked this questionThis is a conceptual problem.
A magnifying glass is a type of converging (convex) lens used to produce a magnified image. A diverging (concave) lens will only produced a reduced image. Though multiple lens systems can be used for magnification purposes, a magnifying glass generally uses only a single lens.

Question 24 of 40Physics  Optics
24. Question
24. When you are looking through a magnifying glass, what would cause the image to invert?Correct / You marked this questionThis is a conceptual problem.
Only a converging (aka convex) lens can produce a magnified image. A diverging (concave) lens will always produce a reduced image. When the object distance is greater than the focal distance, the image produced will be: real, opposite side, & inverted.
Incorrect / You marked this questionThis is a conceptual problem.
Only a converging (aka convex) lens can produce a magnified image. A diverging (concave) lens will always produce a reduced image. When the object distance is greater than the focal distance, the image produced will be: real, opposite side, & inverted.

Question 25 of 40Physics  Optics
25. Question
25. An image is projected 5 cm behind a converging lens. The object distance is 10 cm. The object height is 20 cm. What is the magnitude of the image’s height?Correct / You marked this questionStep 1: Let’s work backwards. How do we solve for image height?
Our magnification equation relates image height to object height:
Magnification also relates image distance to object distance. Conveniently, we are given both image distance (i) & object distance (o).
Step 2: Solve for magnification using image and object distance
M = – i / o
M = – (5 cm / 10 cm)
M = – ^{1}/_{2}xThe negative magnification implies that the image is inverted, which doesn’t change its height.
Step 3: Solve for the magnitude of image height using the magnification equation
M = h_{i} / h_{o}
^{1}/_{2} = h_{i} / 20 cm
h_{i} = 10 cmIncorrect / You marked this questionStep 1: Let’s work backwards. How do we solve for image height?
Our magnification equation relates image height to object height:
Magnification also relates image distance to object distance. Conveniently, we are given both image distance (i) & object distance (o).
Step 2: Solve for magnification using image and object distance
M = – i / o
M = – (5 cm / 10 cm)
M = – ^{1}/_{2}xThe negative magnification implies that the image is inverted, which doesn’t change its height.
Step 3: Solve for the magnitude of image height using the magnification equation
M = h_{i} / h_{o}
^{1}/_{2} = h_{i} / 20 cm
h_{i} = 10 cm 
Question 26 of 40Physics  Optics
26. Question
26. What type of lens are glasses?Correct / You marked this questionThis is a conceptual problem.
Glasses are used to alter the refraction of light into our eyes. Different types of lenses are prescribed for people who are either nearsighted or farsighted:
Nearsighted people see close objects well, but have difficulty seeing objects far away. This is because the eye is overrefracting light from distant objects, and thus the image is formed in front of the retina.
In order to correct this overrefraction, a diverging lens is used.
Farsighted people see far objects well, but have difficulty seeing objects up close. They want to magnify the objects. This is because the eye does not have enough refractive power, and thus the image is formed behind the retina.
In order to correct this overrefraction, a converging lens is used.
Incorrect / You marked this questionThis is a conceptual problem.
Glasses are used to alter the refraction of light into our eyes. Different types of lenses are prescribed for people who are either nearsighted or farsighted:
Nearsighted people see close objects well, but have difficulty seeing objects far away. This is because the eye is overrefracting light from distant objects, and thus the image is formed in front of the retina.
In order to correct this overrefraction, a diverging lens is used.
Farsighted people see far objects well, but have difficulty seeing objects up close. They want to magnify the objects. This is because the eye does not have enough refractive power, and thus the image is formed behind the retina.
In order to correct this overrefraction, a converging lens is used.

Question 27 of 40Physics  Optics
27. Question
27. A pair of glasses has a strength of 2 diopters. The image is projected 300 cm behind the lens. What is the object distance from the lens?Correct / You marked this questionStep 1: Let’s work backwards. How do we solve for the object distance from a lens?
Recall our thin lens equation and associated sign conventions:
As the image is projected behind the lens, we can deduce that image distance (i) is positive (aka “real image”). This is a converging lens.
We are given image distance (i) but we must solve for f (focal length) in order to solve for object distance (o)
Step 2: Use our power equation to solve for focal distance (f)
P = ^{1}/_{f}
2 = ^{1}/_{f}
f = 0.5 m = 50 cmStep 3: Use the thin lens equation to solve for object distance (o)
^{1}/_{o} + ^{1}/_{i} = ^{1}/_{f}
^{1}/_{o} + ^{1}/_{300 cm} = ^{1}/_{50 cm}
^{5}/_{300 cm} = ^{1}/_{o}
o = 60 cmIncorrect / You marked this questionStep 1: Let’s work backwards. How do we solve for the object distance from a lens?
Recall our thin lens equation and associated sign conventions:
As the image is projected behind the lens, we can deduce that image distance (i) is positive (aka “real image”). This is a converging lens.
We are given image distance (i) but we must solve for f (focal length) in order to solve for object distance (o)
Step 2: Use our power equation to solve for focal distance (f)
P = ^{1}/_{f}
2 = ^{1}/_{f}
f = 0.5 m = 50 cmStep 3: Use the thin lens equation to solve for object distance (o)
^{1}/_{o} + ^{1}/_{i} = ^{1}/_{f}
^{1}/_{o} + ^{1}/_{300 cm} = ^{1}/_{50 cm}
^{5}/_{300 cm} = ^{1}/_{o}
o = 60 cm 
Question 28 of 40Physics  Optics
28. Question
28. By what mechanism does a glass prism demonstrate dispersion?Correct / You marked this questionThis is a conceptual question.
When you think of a prism, you probably think of a triangular block of glass splitting a single beam of light into a rainbow of colors on the other side. This phenomenon is called dispersion.
What causes dispersion to occur in a prism?
The speed of light is dependent on the medium through which it is traveling. When light travels in between two mediums with different densities, the light if refracted, thus changing its angle of travel.
Dispersion in a glass prism is based on the index of refraction of glass. Light of different wavelengths (i.e. different color light) travel at different speeds through glass. This causes index of refraction for glass to vary based on the wavelength of the light that is entering it.
As such, red light will bend differently from blue light, or green light, etc. and will exit the prism at a different angle. This is especially pronounced due to the fact that the sides of a prism are not parallel – which causes the net refraction to be even greater.
Incorrect / You marked this questionThis is a conceptual question.
When you think of a prism, you probably think of a triangular block of glass splitting a single beam of light into a rainbow of colors on the other side. This phenomenon is called dispersion.
What causes dispersion to occur in a prism?
The speed of light is dependent on the medium through which it is traveling. When light travels in between two mediums with different densities, the light if refracted, thus changing its angle of travel.
Dispersion in a glass prism is based on the index of refraction of glass. Light of different wavelengths (i.e. different color light) travel at different speeds through glass. This causes index of refraction for glass to vary based on the wavelength of the light that is entering it.
As such, red light will bend differently from blue light, or green light, etc. and will exit the prism at a different angle. This is especially pronounced due to the fact that the sides of a prism are not parallel – which causes the net refraction to be even greater.

Question 29 of 40Physics  Optics
29. Question
29. What is distinguishes polarized from unpolarized light?Correct / You marked this questionThis is a conceptual question.
Light is an electromagnetic wave.
Polarized Light is is a wave whose electric field vectors are all oscillating in one direction.
Unpolarized Light is a wave whose electric field vectors are oscillating in multiple directions.
Incorrect / You marked this questionThis is a conceptual question.
Light is an electromagnetic wave.
Polarized Light is is a wave whose electric field vectors are all oscillating in one direction.
Unpolarized Light is a wave whose electric field vectors are oscillating in multiple directions.

Question 30 of 40Physics  Optics
30. Question
30. An object is placed outside of the focal length of a converging lens. The image produced will be:Correct / You marked this questionRecall our Thin Lens Sign Conventions
As you can see: IF o > f → i (+)
A positive image (+) is real, on the opposite side of the object, and inverted.
Incorrect / You marked this questionRecall our Thin Lens Sign Conventions
As you can see: IF o > f → i (+)
A positive image (+) is real, on the opposite side of the object, and inverted.

Question 31 of 40Physics  Optics
31. Question
31. An object is placed closer than the focal length of a converging lens. The image produced will be:Correct / You marked this questionRecall our Thin Lens Sign Conventions
As you can see: IF o < f → i ()
A negative image () is virtual, on the same side of the object, and upright.
Incorrect / You marked this questionRecall our Thin Lens Sign Conventions
As you can see: IF o < f → i ()
A negative image () is virtual, on the same side of the object, and upright.

Question 32 of 40Physics  Optics
32. Question
32. An object is placed at the focal point of a converging lens. The image produced will be:Correct / You marked this questionRecall our Thin Lens Sign Conventions
As you can see: IF o = f → no image is formed
Incorrect / You marked this questionRecall our Thin Lens Sign Conventions
As you can see: IF o = f → no image is formed

Question 33 of 40Physics  Optics
33. Question
33. A beam of light is shone from the air into two different mediums with indexes of refraction n_{1} and n_{2} as shown below.
Which of the following is true?
Correct / You marked this questionIn Medium 1: the light beam is refracted TOWARD the normal. Light is refracted toward the normal when light passes from a less dense medium (lower n) to a more dense medium (higher n).
In Medium 2: the light beam is refracted AWAY FROM the normal. Light is refracted away from the normal when light passes from a more dense medium (higher n) to a less dense medium (lower n).
Both light beams start in the same medium (air: n = 1). Thus, Medium 1 is more dense (higher n) than air and Medium 2 is less dense (lower n) than air.
Thus: Medium 1 is more dense (higher n) than Medium 2 (lower n). n_{1} > n_{2}
Incorrect / You marked this questionIn Medium 1: the light beam is refracted TOWARD the normal. Light is refracted toward the normal when light passes from a less dense medium (lower n) to a more dense medium (higher n).
In Medium 2: the light beam is refracted AWAY FROM the normal. Light is refracted away from the normal when light passes from a more dense medium (higher n) to a less dense medium (lower n).
Both light beams start in the same medium (air: n = 1). Thus, Medium 1 is more dense (higher n) than air and Medium 2 is less dense (lower n) than air.
Thus: Medium 1 is more dense (higher n) than Medium 2 (lower n). n_{1} > n_{2}

Question 34 of 40Physics  Optics
34. Question
34. A beam of light is shone from the air into two different mediums with indexes of refraction n_{1} and n_{2} as shown below.
Which of the following is true?
Correct / You marked this questionIn Medium 1: the light beam is not refracted (i.e. does not change direction). This implies that the light is passing into a medium which has the same index of refraction as the original medium. That is: n_{1} = n_{AIR} ≅ 1
In Medium 2: the light beam is not refracted (i.e. does not change direction). This implies that the light is passing into a medium which has the same index of refraction as the original medium. That is: n_{2} = n_{AIR} ≅ 1
Thus: n_{1} = n_{2}
Incorrect / You marked this questionIn Medium 1: the light beam is not refracted (i.e. does not change direction). This implies that the light is passing into a medium which has the same index of refraction as the original medium. That is: n_{1} = n_{AIR} ≅ 1
In Medium 2: the light beam is not refracted (i.e. does not change direction). This implies that the light is passing into a medium which has the same index of refraction as the original medium. That is: n_{2} = n_{AIR} ≅ 1
Thus: n_{1} = n_{2}

Question 35 of 40Physics  Optics
35. Question
35. The index of refraction of light in water is 1.33. What is the speed of light in water?Correct / You marked this questionStep 1: Let’s apply our index of refraction equation
n = c / v
1.33 = (3 x 10^{8} m/s) / v
^{4}/_{3} = (3 x 10^{8} m/s) / v
v = (^{3}/_{4})(3 x 10^{8} m/s)
v = 2.25 x 10^{8} m/sNote: we can immediately rule out option E, as nothing can move faster than the speed of light.
Incorrect / You marked this questionStep 1: Let’s apply our index of refraction equation
n = c / v
1.33 = (3 x 10^{8} m/s) / v
^{4}/_{3} = (3 x 10^{8} m/s) / v
v = (^{3}/_{4})(3 x 10^{8} m/s)
v = 2.25 x 10^{8} m/sNote: we can immediately rule out option E, as nothing can move faster than the speed of light.

Question 36 of 40Physics  Optics
36. Question
36. Light comes from the sun towards the Earth. The light moves through the air and hits the Arctic Ocean 60^{o} angle relative to vertical. The index of refraction for the arctic ocean is √3.
What is the value of the angle of refraction (θ_{2})?Correct / You marked this questionStep 1: Let’s draw a quick sketch and what we are solving for
Solving for: the value of angle of refraction (θ_{2}) in the Arctic Ocean
Step 2: Let’s apply Snell’s Law to solve for θ_{2}
Recall the equation for Snell’s Law:
Plug in our given values to solve for θ_{2}:
n_{1}sinθ_{1} = n_{2}sinθ_{2}
(1)(sin60) = √3(sinθ_{2})
^{√3}/_{2} = √3sinθ_{2}
sinθ_{2} = ^{1}/_{2}
θ_{2} = 30^{o}Sanity Check: we know that √3 > 1 Thus n_{2} > n_{1}. When light passes from a less dense medium (lower n) to a more dense medium (higher n) light is refracted TOWARDS the normal.
In this case, this means that θ_{2} < θ_{1}. We can see this is the case, as 30^{o} < 60^{o}
Knowing the sin and cos values of 0, 30, 45, 60, and 90 is very important throughout the OAT. Commit them to memory.
Incorrect / You marked this questionStep 1: Let’s draw a quick sketch and what we are solving for
Solving for: the value of angle of refraction (θ_{2}) in the Arctic Ocean
Step 2: Let’s apply Snell’s Law to solve for θ_{2}
Recall the equation for Snell’s Law:
Plug in our given values to solve for θ_{2}:
n_{1}sinθ_{1} = n_{2}sinθ_{2}
(1)(sin60) = √3(sinθ_{2})
^{√3}/_{2} = √3sinθ_{2}
sinθ_{2} = ^{1}/_{2}
θ_{2} = 30^{o}Sanity Check: we know that √3 > 1 Thus n_{2} > n_{1}. When light passes from a less dense medium (lower n) to a more dense medium (higher n) light is refracted TOWARDS the normal.
In this case, this means that θ_{2} < θ_{1}. We can see this is the case, as 30^{o} < 60^{o}
Knowing the sin and cos values of 0, 30, 45, 60, and 90 is very important throughout the OAT. Commit them to memory.

Question 37 of 40Physics  Optics
37. Question
37. Light comes towards Planet X through the gas which surrounds it, which has an index of refraction of √2 (~ 1.4). The light hits a pool of liquid nitrogen on the planet at a 45^{o} angle relative to the normal. The index of refraction for the liquid nitrogen is ^{2}/_{√3} (1.15). What is the value of the angle of refraction (θ_{2})?Correct / You marked this questionStep 1: Let’s draw a quick sketch and what we are solving for
Solving for: the value of angle of refraction (θ_{2}) in the liquid nitrogen
Step 2: Let’s apply Snell’s Law to solve for θ_{2}
Recall the equation for Snell’s Law:
Plug in our given values to solve for θ_{2}:
n_{1}sinθ_{1} = n_{2}sinθ_{2}
(√2)(sin45) = ^{2}/_{√3}(sinθ_{2})
(√2)(^{1}/_{√2}) = (^{2}/_{√3})sinθ_{2}
1 = ^{2}/_{√3}sinθ_{2}
sinθ_{2} = ^{√3}/_{2}
θ_{2} = 60^{o}Sanity Check: we know that √2 (~1.41) > ^{2}/_{√3} (~1.15). Thus n_{1} > n_{2}. When light passes from a more dense medium (higher n) to a less dense medium (lower n) light is refracted AWAY FROM the normal.
In this case, this means that θ_{2} > θ_{1} . We can see this is the case, as 60^{o} > 45^{o}
Knowing the sin and cos values of 0, 30, 45, 60, and 90 is very important throughout the OAT. Commit them to memory.
Incorrect / You marked this questionStep 1: Let’s draw a quick sketch and what we are solving for
Solving for: the value of angle of refraction (θ_{2}) in the liquid nitrogen
Step 2: Let’s apply Snell’s Law to solve for θ_{2}
Recall the equation for Snell’s Law:
Plug in our given values to solve for θ_{2}:
n_{1}sinθ_{1} = n_{2}sinθ_{2}
(√2)(sin45) = ^{2}/_{√3}(sinθ_{2})
(√2)(^{1}/_{√2}) = (^{2}/_{√3})sinθ_{2}
1 = ^{2}/_{√3}sinθ_{2}
sinθ_{2} = ^{√3}/_{2}
θ_{2} = 60^{o}Sanity Check: we know that √2 (~1.41) > ^{2}/_{√3} (~1.15). Thus n_{1} > n_{2}. When light passes from a more dense medium (higher n) to a less dense medium (lower n) light is refracted AWAY FROM the normal.
In this case, this means that θ_{2} > θ_{1} . We can see this is the case, as 60^{o} > 45^{o}
Knowing the sin and cos values of 0, 30, 45, 60, and 90 is very important throughout the OAT. Commit them to memory.

Question 38 of 40Physics  Optics
38. Question
38. What is the formula used to determine the power of a lens?Correct / You marked this questionThis is simply something you must memorize.
Incorrect / You marked this questionThis is simply something you must memorize.

Question 39 of 40Physics  Optics
39. Question
39. The formula for power is: P = 1/f. For power to yield units of diopters, in what unit must focal length (f) be?Correct / You marked this questionOften times, we can use different units in an equation, as long as we keep things consistent throughout. This, however, is an exception! In the power equation, focal length MUST be in units of meters for power to yield units of diopters.
Incorrect / You marked this questionOften times, we can use different units in an equation, as long as we keep things consistent throughout. This, however, is an exception! In the power equation, focal length MUST be in units of meters for power to yield units of diopters.

Question 40 of 40Physics  Optics
40. Question
40. Which of the following is the formula for Magnification? Where i is the distance of the image from the lens and o is the distance of the object from the lens.Correct / You marked this questionThis is simply something you must memorize. Here are the details of the magnification equation:
Incorrect / You marked this questionThis is simply something you must memorize. Here are the details of the magnification equation:
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