Test Your Knowledge On Optic Lights!
Test your knowledge on this optic lights quiz and see how you do and compare your score too.
Questions and Answers
- 1.
A survivor from a ship wreck sees an image of a fish in the water. To catch it with her spear, she must
- A.
Aim above the image of the fish
- B.
Aim below the image of the fish
- C.
Aim at the image of the fish
- D.
Aim behind the fish
- E.
Put away the spear and use a fishing rod
Correct Answer
B. Aim below the image of the fishExplanation
When light travels from one medium to another, it changes its speed and direction. This phenomenon is called refraction. When the survivor sees the image of the fish in the water, the light rays from the fish have already undergone refraction. The image appears higher than the actual position of the fish. To compensate for this, the survivor needs to aim below the image of the fish in order to actually hit it with her spear.Rate this question:
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- 2.
A light ray has an angle of incidence of 34º. The reflected ray will make what angle with the reflecting surface?
- A.
0º
- B.
34º
- C.
56º
- D.
66º
- E.
74º
Correct Answer
C. 56ºExplanation
When a light ray reflects off a surface, the angle of incidence is equal to the angle of reflection. In this case, the angle of incidence is 34º, so the angle of reflection will also be 34º. Since the reflected ray is on the opposite side of the normal line, we subtract the angle of reflection from 90º (the angle between the normal line and the reflecting surface) to find the angle between the reflected ray and the reflecting surface. Therefore, the angle between the reflected ray and the reflecting surface is 90º - 34º = 56º.Rate this question:
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- 3.
The critical angle for diamond (n = 2.42) submerged in water (n = 1.33) is
- A.
33°
- B.
49°
- C.
24°
- D.
17°
- E.
Does not exist
Correct Answer
A. 33°Explanation
The critical angle is the angle of incidence at which light traveling from a denser medium to a rarer medium is refracted at an angle of 90 degrees. In this case, the diamond (with a refractive index of 2.42) is submerged in water (with a refractive index of 1.33). The critical angle can be calculated using the formula sin(critical angle) = n2/n1, where n2 is the refractive index of the initial medium (water) and n1 is the refractive index of the final medium (diamond). Plugging in the values, sin(critical angle) = 1.33/2.42, which gives a critical angle of approximately 33°.Rate this question:
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- 4.
Calculate the index of refraction for an object in which light travels at 1.97 x 108 m/s.
- A.
1.97 m/s
- B.
0.66 m/s
- C.
1.52 m/s
- D.
1.95
- E.
1.52
Correct Answer
E. 1.52Explanation
The index of refraction is a measure of how much light is slowed down when it passes through a medium compared to its speed in a vacuum. In this question, the given speed of light in the object is 1.97 x 108 m/s. To calculate the index of refraction, we need to divide the speed of light in a vacuum (3 x 108 m/s) by the speed of light in the object. When we do this calculation, we get an index of refraction of approximately 1.52.Rate this question:
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- 5.
The critical angle of zircon is 31°. Which of the following incident angles would result in total internal reflection?
- A.
17°
- B.
34°
- C.
42°
- D.
A and C
- E.
B and C
Correct Answer
E. B and CExplanation
The critical angle of a material is the angle at which light is incident on the surface and refracts at an angle of 90 degrees. Any incident angle greater than the critical angle will result in total internal reflection, meaning that all of the light will be reflected back into the material and none will be transmitted through the surface. In this case, the critical angle of zircon is 31 degrees. Therefore, incident angles of 34 degrees and 42 degrees (options B and C) are greater than the critical angle and would result in total internal reflection.Rate this question:
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- 6.
When a light ray travelling in glass is incident on an air surface,
- A.
It will refract away from the normal
- B.
Some of the light may be reflected
- C.
All of the light may be reflected
- D.
All of A, B, and C
- E.
Two of A, B, and C
Correct Answer
D. All of A, B, and CExplanation
When a light ray travels from a denser medium like glass to a less dense medium like air, it undergoes refraction. Refraction causes the light ray to change direction and bend away from the normal (an imaginary line perpendicular to the surface of the medium). This explains why the light ray refracts away from the normal when incident on an air surface. Additionally, some of the light may also be reflected at the air-glass interface due to the difference in refractive indices between the two mediums. This reflection can occur partially or completely, depending on the angle of incidence and the refractive indices involved. Therefore, all of the options A, B, and C are correct.Rate this question:
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- 7.
Light travels from medium X into medium Y. Medium Y has a higher index of refraction. Consider each statement below:(i) The light travels faster in X. (ii) The light will bend towards the normal. (iii) The light will speed up. (iv) The light will bend away from the normal.
- A.
(ii)
- B.
(iii) and (iv)
- C.
(i)
- D.
(i) and (ii)
- E.
(ii), (iii), and (iv)
Correct Answer
D. (i) and (ii)Explanation
When light travels from a medium with a lower index of refraction to a medium with a higher index of refraction, it slows down and bends towards the normal. Therefore, statement (ii) is correct. However, statement (iii) is incorrect because the light actually slows down when it enters the medium with a higher index of refraction. Statement (i) is correct because the light travels slower in medium X compared to medium Y. Therefore, the correct answer is (i) and (ii).Rate this question:
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- 8.
For a converging lens, a light ray that is travelling parallel to the principal axis refracts
- A.
Through the principal focus
- B.
Through the secondary focus
- C.
Through the optical centre
- D.
Parallel to the principal axis
- E.
In line with the principal focus
Correct Answer
A. Through the principal focusExplanation
A converging lens causes parallel rays of light to converge at a point called the principal focus. When a light ray is traveling parallel to the principal axis, it will refract (bend) and pass through the principal focus. This is because the lens is designed to bring parallel rays of light to a focus at this point. Therefore, the correct answer is "through the principal focus."Rate this question:
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- 9.
A object is placed between f and 2f for a diverging lens. The image will be located
- A.
Between f and 2f
- B.
Between the lens and f
- C.
Farther than 2f
- D.
A or B
- E.
There is insufficient information to answer the question.
Correct Answer
B. Between the lens and fExplanation
When an object is placed between the focal point (f) and twice the focal point (2f) for a diverging lens, the image formed is virtual, upright, and located on the same side as the object. Since the image is on the same side as the object, it will be located between the lens and the focal point (f). Therefore, the correct answer is between the lens and f.Rate this question:
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- 10.
Light that travels into the eye passes through several parts to get to the retina. The correct order is
- A.
Cornea, vitreous humour, lens, pupil
- B.
Lens, cornea, pupil, vitreous humour
- C.
Cornea, lens, pupil, vitreous humour
- D.
Pupil, cornea, lens, vitreous humour
- E.
Cornea, pupil, lens, vitreous humour
Correct Answer
E. Cornea, pupil, lens, vitreous humourExplanation
Light first enters the eye through the cornea, which is the clear outer layer of the eye. It then passes through the pupil, which is the opening in the center of the iris. The lens, located behind the iris, helps to focus the light onto the retina at the back of the eye. Finally, the light passes through the vitreous humor, a clear gel-like substance that fills the space between the lens and the retina. Therefore, the correct order is cornea, pupil, lens, vitreous humor.Rate this question:
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- 11.
The proper name for the inability to focus in different planes is
- A.
Hyperopia
- B.
Presbyopia
- C.
Astigmatism
- D.
Myopia
- E.
Glaucoma
Correct Answer
C. AstigmatismExplanation
Astigmatism is the correct answer because it refers to a condition where the eye does not focus light evenly onto the retina, causing blurred or distorted vision at all distances. This inability to focus in different planes is characteristic of astigmatism. Hyperopia refers to farsightedness, presbyopia is the loss of near vision with age, myopia is nearsightedness, and glaucoma is a condition where increased pressure in the eye damages the optic nerve.Rate this question:
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- 12.
The proper name for farsightedness is
- A.
Hyperopia
- B.
Presbyopia
- C.
Astigmatism
- D.
Myopia
- E.
Glaucoma
Correct Answer
A. HyperopiaExplanation
Hyperopia, also known as farsightedness, is a common vision condition where objects that are far away can be seen clearly, but close-up objects appear blurry. This occurs when the eyeball is shorter than normal or when the cornea has too little curvature. As a result, light entering the eye focuses behind the retina instead of directly on it, causing distant objects to be clearer than near objects. Presbyopia, astigmatism, myopia, and glaucoma are all different eye conditions that are unrelated to farsightedness.Rate this question:
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- 13.
A block of glass is pushed into the path of the light, as shown below.The point where the light rays cross will
- A.
Stay in the same place
- B.
Move to the left
- C.
Move to the right
- D.
Shift up
- E.
Shift down
Correct Answer
C. Move to the rightExplanation
When a block of glass is placed in the path of light, it causes the light rays to change direction due to refraction. Refraction occurs because the speed of light is different in glass compared to air. In this case, the light rays will bend towards the normal (a line perpendicular to the surface of the glass) as they enter the glass, and then bend away from the normal as they exit the glass. This change in direction causes the point where the light rays cross to move to the right.Rate this question:
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- 14.
The focal length of a converging lens is 15 cm. An object is placed 45 cm away from the lens. The image will be
- A.
Smaller and real
- B.
Larger and real
- C.
The same size and real
- D.
Smaller and virtual
- E.
Larger and virtual
Correct Answer
A. Smaller and realExplanation
When an object is placed beyond the focal point of a converging lens, the image formed is real, inverted, and smaller than the object. In this case, the object is placed 45 cm away from the lens, which is beyond the focal length of 15 cm. Therefore, the image formed will be smaller and real.Rate this question:
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- 15.
The focal length of a diverging lens is 12 cm. An object is placed 5.0 cm away from the lens. The image will be
- A.
Smaller and real
- B.
Larger and real
- C.
The same size and real
- D.
Smaller and virtual
- E.
Larger and virtual
Correct Answer
D. Smaller and virtualExplanation
When an object is placed in front of a diverging lens, the image formed is always virtual and smaller than the object. This is because a diverging lens causes light rays to spread out, resulting in the formation of a virtual image that cannot be projected onto a screen. The size of the image is determined by the ratio of the object distance to the focal length of the lens. In this case, the object is placed 5.0 cm away from the lens, which is less than the focal length of 12 cm. Thus, the image will be smaller and virtual.Rate this question:
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Current Version
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Mar 19, 2023Quiz Edited by
ProProfs Editorial Team -
May 17, 2012Quiz Created by
Crackiitjee
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