Worked Example: Using Snell’s Law, Apparent Depth, physics 10

Physics 10.VI Geometrical Optics T.Trường 9/6/17 1,244 0
  1. Worked Example: Using Snell’s Law, Apparent Depth, physics 10
    1/ Worked Example 1: Using Snell’s Law
    Question:
    A light ray with an angle of incidence of 35◦ passes from water to air. Find the angle of refraction using Snell’s Law and Table 7.4.1. Discuss the meaning of your answer.
    Answer
    Step 1 : Determine the refractive indices of water and air

    From Table 7.4.1, the refractive index is 1,333 for water and about 1 for air. We know the angle of incidence, so we are ready to use Snell’s Law.
    Step 2 : Substitute values
    According to Snell’s Law:
    [​IMG]
    Step 3 : Discuss the answer
    The light ray passes from a medium of high refractive index to one of low refractive index. Therefore, the light ray is bent away from the normal.

    2/ Worked Example 2: Using Snell’s Law
    Question:
    A light ray passes from water to diamond with an angle of incidence of 75◦. Calculate the angle of refraction. Discuss the meaning of your answer.
    Answer
    Step 1 : Determine the refractive indices of water and air

    From Table 7.4.1, the refractive index is 1,333 for water and 2,42 for diamond. We know the angle of incidence, so we are ready to use Snell’s Law.
    Step 2 : Substitute values and solve
    According to Snell’s Law:
    [​IMG]
    Step 3 : Discuss the answer
    The light ray passes from a medium of low refractive index to one of high refractive index. Therefore, the light ray is bent towards the normal.

    If n2 > n1 then from Snell’s Law, sin θ1 > sin θ2.
    For angles smaller than 90◦, sin θ increases as θ increases. Therefore, θ1 > θ2.
    This means that the angle of incidence is greater than the angle of refraction and the light ray is bent toward the normal. Similarly, if n$_{2 }$< n1 then from Snell’s Law, sin θ1 < sin θ2.
    For angles smaller than 90◦, sin θ increases as θ increases. Therefore, θ1 < θ2.
    This means that the angle of incidence is less than the angle of refraction and the light ray is away toward the normal.
    Both these situations can be seen in Figure 7.12.
    [​IMG]
    Figure 7.12: Refraction of two light rays. (a) A ray travels from a medium of low refractive index to one of high refractive index. The ray is bent towards the normal. (b) A ray travels from a medium with a high refractive index to one with a low refractive index. The ray is bent away from the normal.

    What happens to a ray that lies along the normal line? In this case, the angle of incidence is 0 ◦ and sin θ2 = n1/n2 sin θ1 = 0 => θ2 = 0.
    This shows that if the light ray is incident at 0◦, then the angle of refraction is also 0◦. The ray passes through the surface unchanged, i.e. no refraction occurs.

    Activity :: Investigation : Snell’s Law 1
    The angles of incidence and refraction were measured in five unknown media and recorded in the table below. Use your knowledge about Snell’s Law to identify each of the unknown media A - E. Use Table 7.4.1 to help you.
    [​IMG]
    Activity :: Investigation : Snell’s Law 2
    Zingi and Tumi performed an investigation to identify an unknown liquid. They shone a beam of light into the unknown liquid, varying the angle of incidence and recording the angle of refraction. Their results are recorded in the following table:
    [​IMG]
    1. Write down an aim for the investigation.
    2. Make a list of all the apparatus they used.
    3. Identify the unknown liquid.
    4. Predict what the angle of refraction will be for 70◦, 75◦, 80◦and 85◦.

    3/ Apparent Depth
    Imagine a coin on the bottom of a shallow pool of water. If you reach for the coin, you will miss it because the light rays from the coin are refracted at the water’s surface.

    Consider a light ray that travels from an underwater object to your eye. The ray is refracted at the water surface and then reaches your eye. Your eye does not know Snell’s Law; it assumes light rays travel in straight lines. Your eye therefore sees the image of the at coin shallower location. This shallower location is known as the apparent depth.

    The refractive index of a medium can also be expressed as
    [​IMG]
    Worked Example 3: Apparent Depth 1
    Question:
    A coin is placed at the bottom of a 40 cm deep pond. The refractive index for water is 1,33. How deep does the coin appear to be?
    Answer
    Step 1 : Identify what is given and what is asked

    n = 1,33
    real depth = 40 cm
    apparent depth = ?
    Step 2 : Substitute values and find answer
    [​IMG]
    The coin appears to be 30,08 cm deep.

    Worked Example 4: Apparent Depth 2
    Question:
    A R1 coin appears to be 7 cm deep in a colourless liquid. The depth of the liquid is 10,43 cm.
    1. Determine the refractive index of the liquid.
    2. Identify the liquid.
    Answer
    Step 1 : Identify what is given and what is asked

    real depth = 7 cm
    apparent depth = 10,43 cm
    n = ?
    Identify the liquid.
    Step 2 : Calculate refractive index
    [​IMG]
    Step 3 : Identify the liquid
    Use Table 7.4.1. The liquid is an 80% sugar solution.

    High School Students Studying the Sciences Physics
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