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MCAT Exam  >  MCAT Tests  >  Physics for MCAT  >  Test: Reflection and Refraction - MCAT MCQ

Test: Reflection and Refraction - MCAT MCQ


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10 Questions MCQ Test Physics for MCAT - Test: Reflection and Refraction

Test: Reflection and Refraction for MCAT 2024 is part of Physics for MCAT preparation. The Test: Reflection and Refraction questions and answers have been prepared according to the MCAT exam syllabus.The Test: Reflection and Refraction MCQs are made for MCAT 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Test: Reflection and Refraction below.
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Test: Reflection and Refraction - Question 1
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Which of the following is true of reflected light rays from a surface?
I. The angle of incidence is equal to the angle of reflection for diffuse reflections.
II. The angle measured between the surface and the ray is the same magnitude as the angle of reflection
III. The angle of incidence is found by subtracting the angle made with the surface from the normal

Detailed Solution for Test: Reflection and Refraction - Question 1
  • The angle of incidence is in fact found by taking the difference between the normal and the angle of reflection made at the surface
  • The angle between the surface and the ray will not be equal to the angle of reflection
  • The angle of incidence is always equal to the angle of reflection, even for diffuse reflections. Therefore I + III are both correct.
Test: Reflection and Refraction - Question 2
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In the figure below, from P → Q a ray of light passes from air through glass with a refraction index of 1.5. If the angle of reflection is 30° , what is sin of the angle of refraction?

Detailed Solution for Test: Reflection and Refraction - Question 2
  • To find the sin of the angle of refraction you must use Snell’s law:
  • The angle of reflection will be equal to the angle of incidence, so we can simply use 30° as θ1.
  • We can basically treat the index of refraction of air (n1) as 1. 
  • Plugging in our values into Snell’s law we get
    sin(30) = 1.5 sin θ2 therefore, 
    0.5 = 1.5 sin (θ2)
    sin (θ2) = 1/3
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Test: Reflection and Refraction - Question 3
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A diamond has an index of refraction of 2.4, point, 4. What is the speed of light within the diamond?

Detailed Solution for Test: Reflection and Refraction - Question 3
  • The formula for the index of refraction is n = c/v, where n is the index, c is the speed of light in a vacuum (3 × 108 m/s), and v is the velocity of light in the other medium.
  • We can immediately disregard 7.2 × 108 m/s because that would mean light is moving faster than the speed of light in a vacuum, which isn’t possible.
  • To solve, we simply divide 3.2 × 108 m/s by 2.4 A quick math trick to help solve this is to block out the extra zeroes and realize that 2.4 can’t go into 3.0 more than once, and therefore only 1.25 × 108 m/s could work as an answer.
Test: Reflection and Refraction - Question 4
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A ray of light refracts as it passes from glass into a vacuum. It’s angle of incidence is 30°, and its angle of refraction is 60° what is the index of refraction of the glass?
Detailed Solution for Test: Reflection and Refraction - Question 4
  • Use Snell’s law to calculate the index of reflection of the glass. Snell’s law: n1 sin (θ1) = n2 sin (θ2)
  • The index of refraction for light in a vacuum is 1, therefore n2 = 1 and we simply must solve for n1.
  • The index of refraction can never be less than 1, therefore  cannot be an answer.
  • Plugging in our values:
    n1 sin (30) = 1 sin (60)

    n1 = √3
Test: Reflection and Refraction - Question 5
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A prism is used to separate white light into a rainbow. Which color of the visible spectrum would have the smallest angle of refraction (the angle between the ray and a line normal to the surface of the prism)?
Detailed Solution for Test: Reflection and Refraction - Question 5
  • Wavelength of light and index of refraction are inversely proportional; i.e. the index of refraction increases as the wavelength of light decreases.
  • Violet light would have the smallest wavelength of visible light, therefore the highest index of refraction.
  • The larger the index of refraction, the smaller the angle of refraction (use Snell's Law to convince yourself of this!), therefore violet should have the smallest angle of refraction for visible light.
Test: Reflection and Refraction - Question 6
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What would the critical angle be for a flare lit underneath the surface ocean, if the index of refraction for ocean water is 1.33?
Detailed Solution for Test: Reflection and Refraction - Question 6
  • To find the critical angle, we must use the formula 
  • The index of refraction for air can be considered 1, and since the example is light passing into air from water, our n2 should be 1.
  • Plugging our values into the formula, we should get sin 
Test: Reflection and Refraction - Question 7
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Above is a schematic of a fiber optic cable, which maintains total internal reflection throughout the length of the cable. The index of refraction of the cladding of a cable is generally 1.52, while the core has an index of refraction of 1.62 Choosing from the list below, what is the smallest angle of incidence light could have to enter from the acceptance cone that would be able to ‘escape’ the fiber optic cable?
Detailed Solution for Test: Reflection and Refraction - Question 7
  • There are multiple ways to come to an answer. You can use the critical angle formula sin θc = n2/n1 but this requires some math gymnastics.
  • One important realization to make is that the larger an angle of incidence is, the more likely it will be internally reflected. The angle of incidence is measured between the ray of light and the normal of the inside of the cable, therefore in the figure, the rays coming in from outside the acceptance cone would have the lowest angle of incidence.
  • Since we’re looking for the smallest angle of incidence possible to escape the cable, we should simply pick 30°
  • The harder way, if we use the formula, would give the result of sin θc = 1.52/1.62. We probably would need a calculator to solve this directly, but we should know the sine of 30°, 45°, and 60° , which are the available answer choices. Additionally, we know we want to find the smallest angle, so we can focus on 30°
  • sin (30°) = 1/2, therefore sin (30°) is much smaller than the sin (θc), which is all we need to know. Remember, if an angle is smaller than the critical angle, it will not be internally reflected! Since 30° is our smallest incident angle choice, even if the other angles can ‘escape’ the cable, 30° must be the correct answer. 
Test: Reflection and Refraction - Question 8
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A laser beam is refracted in a prism with an index of refraction (n2) of √3/√2. The angle of refraction is measured to be 45°. If the prism is in a vacuum, what is the angle between the incoming ray of light and the surface of the prism?
Detailed Solution for Test: Reflection and Refraction - Question 8
  • Note that the angle of incidence is found by subtracting the angle made with the surface from the normal, and note how this is different from what the question is asking for.
  • Use Snell’s law to find the angle of incidence, since it is in a vacuum, n1 = 1. 
  • We are trying to find θ1 Therefore:


    θ = 60°
  • Don’t let that answer fool you! That is simply our angle of incidence! Since our angle of incidence is measured from the normal of the surface, this means that the angle between the ray and the surface of the prism must be 30° (90 - 60 = 30)
Test: Reflection and Refraction - Question 9
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Which of the following is true concerning total internal reflection?
I. Light traveling from a lower index of refraction to a higher one can experience total internal refraction.
II. If the angle of incidence equals the critical angle then light will not be refracted
III. If an angle of incidence is larger than the critical angle, it will not experience total internal reflection.
Detailed Solution for Test: Reflection and Refraction - Question 9
  • Total internal reflection can only occur if the angle of incidence is larger than the critical angle.
  • Total internal reflection occurs only when light travels from a higher index of refraction to a lower one.
  • If the angle of incidence exactly equals the critical angle, the angle of refraction will be 90°, meaning it will skirt along the boundary between the two media.
Test: Reflection and Refraction - Question 10
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A ray of light passing through a gel with an index of refraction of 1.5 has an incidence angle of 30° as it transitions through a pane of glass. The angle of refraction is 45°. What is the speed of light within the glass?
Detailed Solution for Test: Reflection and Refraction - Question 10
  • To find the speed of light in a medium, we have to know it’s index of refraction.
  • First we must find the index of refraction for the glass (n2) using Snell’s Law:




  • Next we take the index of refraction for glass and put it into our index of refraction formula:
    n = c/v





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