Light And Optics Unit Quiz Test

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Light And Optics Unit Quiz Test - Quiz

Take this simple multiple choice questions for light and optics unit and learn more about it!


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 fish
    Explanation
    When light passes from one medium to another, it changes direction. This phenomenon is called refraction. When the survivor sees the image of the fish in the water, she is actually seeing a refracted image due to the bending of light as it passes from water to air. Since the light is bending away from the normal, the image appears higher than the actual position of the fish. Therefore, in order to compensate for this displacement, the survivor needs to aim below the image of the fish to successfully catch it with her spear.

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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.

    • B.

      34º

    • C.

      56º

    • D.

      66º

    • E.

      74º

    Correct Answer
    C. 56º
    Explanation
    When a light ray is incident on a reflecting surface, the angle of incidence is equal to the angle of reflection. Therefore, if the angle of incidence is 34º, the reflected ray will also make an angle of 34º with the reflecting surface. However, the correct answer given is 56º, which contradicts the principle of reflection. Therefore, the correct answer provided is incorrect.

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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 passing from a medium with a higher refractive index to a medium with a lower refractive index is refracted along the boundary of the two mediums. In this case, the diamond has a higher refractive index (n = 2.42) compared to water (n = 1.33). The critical angle can be calculated using the formula sin(critical angle) = n2/n1, where n2 is the refractive index of the medium the light is entering (water) and n1 is the refractive index of the medium the light is leaving (diamond). Plugging in the values, sin(critical angle) = 1.33/2.42, which gives a critical angle of approximately 33°. Therefore, the correct answer is 33°.

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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. 1.52
    Explanation
    The index of refraction is a measure of how much light slows down when it enters 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 divide the speed of light in a vacuum (approximately 3 x 108 m/s) by the speed of light in the object. Therefore, the index of refraction is 3 x 108 m/s divided by 1.97 x 108 m/s, which equals approximately 1.52.

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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 C
    Explanation
    Total internal reflection occurs when the incident angle is greater than the critical angle. In this case, the critical angle of zircon is 31°. Therefore, incident angles of 34° and 42° would result in total internal reflection because they are greater than the critical angle. Incident angle of 17° is less than the critical angle, so it would not result in total internal reflection. Therefore, the correct answer is B and C.

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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 C
    Explanation
    When a light ray travels from glass to air, it undergoes refraction away from the normal due to the change in the speed of light between the two media. This means that the light ray bends away from the perpendicular line. Additionally, some of the light may also be reflected at the air surface, causing a partial reflection. Depending on the angle of incidence and the properties of the materials involved, all of the light may be reflected, resulting in a total reflection. Therefore, all of options A, B, and C are possible outcomes when a light ray traveling in glass is incident on an air surface.

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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 will slow down and change direction. This change in direction is known as refraction. The light will bend towards the normal, which is an imaginary line perpendicular to the surface of the interface between the two media. Therefore, statement (ii) is correct. Additionally, since the light slows down when entering the medium with a higher index of refraction, it will also travel slower in that medium compared to the medium with the lower index of refraction. Therefore, statement (i) is also correct.

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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 focus
    Explanation
    A converging lens is thicker in the middle and causes light rays to converge towards a focal point. When a light ray travels parallel to the principal axis of a converging lens, it refracts through the lens and passes through the principal focus. This is because the lens causes the light rays to bend inward towards the principal focus, resulting in the light ray passing through it.

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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 f
    Explanation
    When an object is placed between the focal point (f) and twice the focal point (2f) for a diverging lens, the image formed will be virtual, upright, and located on the same side as the object. In this case, the image will be located between the lens and the focal point (f). This is because a diverging lens always forms virtual images that are located on the same side as the object.

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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 humour
    Explanation
    Light enters the eye through the cornea, which is the transparent outer covering. It then passes through the pupil, which is the opening in the center of the iris. The lens, located behind the pupil, helps to focus the light onto the retina at the back of the eye. Finally, the light passes through the vitreous humor, a gel-like substance that fills the space between the lens and the retina. Therefore, the correct order is cornea, pupil, lens, vitreous humor.

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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. Astigmatism
    Explanation
    Astigmatism refers to a condition where the cornea or lens of the eye is irregularly shaped, causing blurred or distorted vision at all distances. This means that individuals with astigmatism have difficulty focusing in different planes, as the irregular shape of their eye prevents light from properly focusing on the retina. Hyperopia, presbyopia, myopia, and glaucoma are all different eye conditions that do not specifically relate to the inability to focus in different planes.

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  • 12. 

    The proper name for farsightedness is

    • A.

      Hyperopia

    • B.

      Presbyopia

    • C.

      Astigmatism

    • D.

      Myopia

    • E.

      Glaucoma

    Correct Answer
    A. Hyperopia
    Explanation
    Hyperopia, also known as farsightedness, is a condition in which a person can see distant objects clearly, but has difficulty focusing on objects that are close. 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 blurred vision. Hyperopia can be corrected with glasses, contact lenses, or refractive surgery. Presbyopia, astigmatism, myopia, and glaucoma are different eye conditions and not the proper name for farsightedness.

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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 right
    Explanation
    When a block of glass is pushed into the path of light, it causes the light rays to change direction due to refraction. Refraction occurs because the speed of light changes when it passes from one medium (air) to another (glass) with a different refractive index. In this case, the light rays will bend towards the normal (an imaginary line perpendicular to the surface of the glass) and move towards the right. Therefore, the point where the light rays cross will move to the right.

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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 real
    Explanation
    When an object is placed beyond the focal point of a converging lens, a real and smaller image is formed on the opposite side of the lens. In this case, the object is placed 45 cm away from the lens, which is more than twice the focal length of 15 cm. Therefore, the image formed will be smaller and real.

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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 virtual
    Explanation
    When an object is placed closer to a diverging lens than its focal length, the image formed is virtual, upright, and smaller in size than the object. This is because the diverging lens causes the light rays to diverge, resulting in the formation of a virtual image that appears to come from a point behind the lens. Since the object is placed 5.0 cm away from the lens, which is less than the focal length of 12 cm, the image formed will be smaller and virtual.

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