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Geometrical Optics Practice Questions

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Updated on Sep 27, 2011

Review these concepts at: Geometrical Optics Study Guide

Practice Questions

  1. In the previous example, find the angle the outgoing ray (Ray 3) makes with the vertical (N1).
  2. Two planes are set such that they form an angle of 90° with each other. A light is incident on the first surface at an angle of 70°. Find the angle of the reflected light from the second plane with respect to the normal to that plane.
  3. In Figure 18.5, the propagation of the incident ray is such that the angle of the reflected ray with the horizontal is 25°. Find the incident angle.

    Reflection of Light

  1. Determine the angle of refraction for a light ray that travels from air to water at an angle of incidence of 42°.
  2. In Figure 18.7, draw the light ray propagation of the two parallel rays through air, in glass, and back in air.
  3. Geometrical Optics

  4. In Figure 18.8, draw the light ray propagation of the two parallel rays through air in glass and back in air. Consider that inside the glass, the surface is such that, instead of refraction, reflection will occur.
  5. Geometrical Optics

  6. A light ray propagates from a medium with a small index of refraction to a medium with a larger index of refraction. Is the light going to be bent toward or away from the normal? Explain.
  7. For Figure 18.9, fill in the missing numbers and find the index of refraction for the second medium.
  8. Geometrical Optics

  9. Imagine a setup where the light has to travel from air to glass and to go out in ice. Draw the diagram representing light propagation through these media and show the angle of incidence and refraction on each surface.
  10. For Figure 18.10, find the angle of refraction for the first surface of separation and the angle of refraction for the ray that propagates outside the glass (flint) and into the air again.
  11. Geometrical Optics

  12. For Figure 18.11, find the angle of refraction for the ray that propagates outside the glass (flint) and into the air again.
  1. A light source is placed at the bottom of a swimming pool, and a light ray directed toward the surface of the water makes an incident angle of 68° with the normal. Find the angle of refraction in air.
  2. A light ray has an optical path, as shown in Figure 18.13. Determine if the light will suffer a total internal reflection in the glass or not. The angle between surfaces 1 and 2 is 20°.

Internal Reflection

Answers

  1. 10° west of the normal N1
  2. 20° with respect to the normal on the second mirror
  3. θi = 65°
  4. r = 30°
  5. The light rays will encounter two surfaces of separation and refraction will happen at the outside of glass and inside. The refracted light rays are parallel to the incident ones.
  6. Geometrical Optics

  7. The light rays will suffer refraction, reflection, and refraction back into air. The light returns to air at the same angle as the incident angle but shifted toward right due to the path covered in glass.
  8. Geometrical Optics

  9. Because the product of the index of refraction to the sine of the angle is constant and sine is a function increasing with angle between 0° and 90°, then a decrease in the index of refraction means an increase in the refraction angle, so the refracted light will be bent away from the normal.
  10. We apply Snell's law and find the index is: n = sin 55 ° I sin 37 ° = 1.4.
  11. Geometrical Optics

  12. There are two refraction processes in between three different media, so the angles made by the light with the normals are going to be different according to the index of refraction:
  13. Geometrical Optics

    So, from air to glass, the light bends toward the normal, whereas from glass to ice, it bends away from normal but less than it would bend in air. According to the incidence angle, there also can be total internal reflection at the surface between glass and ice.

    Geometrical Optics

  14. The angle of refraction on the first surface is 20°, and the angle of refraction at the second surface is the same as the angle of initial incidence 35°.
  15. The angle of the outgoing ray is 20°.
  16. The light ray will totally reflect water because the first term of Snell's law can not be reproduced in air regardless of the angle of refraction: nwater · sin θ = constant and 1.33 · sin 68° = 1.23. Because the light ray is supposed to travel in air, n = 1, then the sin θr would have to be greater than 1!
  17. The light will be refracted into the water at an angle of refraction of 60° with respect to the second surface of separation.
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