n = c/v

λ_of = c       (Equation 1)

λ_nf = v      (Equation 2)

λ_o/ λ_n = c/v = n  or  λ_n= λ_o/n.

1. Angle of incidence Θ_i is the angle between the incident ray and the normal to the surface.
   
   
2. Angle of reflection Θ_r is the angle between the incident ray and the normal to the surface.
   
   
3. Angle of refraction or transmission Θ_t is the angle between the refracted (transmitted) ray and the normal to the surface.
   
   

n_i sin Θ_i = n_t sin Θ_t
sin Θ_t = n_i sin Θ_i / n_t

1. If the transmitted medium is more dense than the incident medium

   (n_t > n_i),
   sin Θ_t < sin Θ_i,
   Θ_t < Θ_i,

   and the light is bent toward the normal. See Fig. 1a below.
   
   
   
   
2. If the transmitted medium is less dense than the incident medium

   (n_t < n_i),
   sin Θ_t > sin Θ_i,
   Θ_t > Θ_i,

   and the light is bent away from the normal. See Fig. 1b below.
   
   
   
   
3. When an incident ray hits a surface normally, the angle between the incident ray and the normal is 0^o and, by definition, the angle of incidence is 0^o. For an angle of incidence of 0^o, the angle of reflection is 0^o, and the angle of refraction is 0^o regardless of the relative values of n_i and n_t, as shown in Fig. 2 below.
   
   
   
   
   

1. For a real image, the two rays actually meet at a point.
   
   
2. For a virtual image, you trace the two rays back until you find the point at which they appear to meet.
   
   

1. In Fig. 4a, a ray from the top of the object O hits the mirror normally. The angle of incidence is 0^o and the angle of reflection is 0^o. The incident and reflected ray are shown with one arrow.
   
   
2. In Fig. 4b, an incident ray at angle Θ_i and a reflected wave ray at angle Θ_r are shown with two arrows.
   
   
3. In Fig. 4c, only the two reflected rays are shown. The eye "thinks" light travels in straight lines. It traces the two rays back and finds they appear to meet at the location of the top of the image I. The foot of the object, which rests on the normal to the surface, is automatically positioned below the top of the arrow.
   
   
4. For a plane mirror, the image is as far behind the mirror as the image is in front of the mirror. The object distance s_o = the image distance s_i. See solution to Problem 4 in 106 Solution Set for Geometric Optics.
   
   
   

1. In Fig. 5a, two rays are shown leaving the object O. The ray with one arrow hits the surface normally and continues into the other medium with no refraction. The ray at incident angle Θ_i in the medium of index of refraction n_i> n_t is refracted at angle Θ_t in the medium of index of refraction n_t.
   
   
2. In Fig. 5b, the eye is shown tracing the rays back to the image I.
   
   
3. In Fig. 5c , the angle of incidence for the second ray is drawn small so the approximation of sin Θ ≈ tan Θ can be used.
   From Fig. 5c, we see tan Θ_t = x/s_o and tan Θ₁ = x/s_i.
   From Snell's law, n_i sin Θ_i = n_t sin Θ_t.
   For our approximation, n_i x/s_o = n_t x/s_i  or  s_i = n_t s_o /n_i.
   For n_t = 1,  s_i = s_o /n_i.
   
   

1/ s_o + 1/ s_i = 1/f.

1. Focal length is positive. The center of curvature is in front of the mirror.
   
   
2. For object distance s_o > f,  image is real and inverted.
   
   
3. For s_o > 2f,  image is diminished in size.
   2f < s_o < f,  image is enlarged.
   
   In Fig. 6a below, ray 1 (one arrow) goes through the focal point F and is reflected back parallel to the principal axis. Ray 2 goes through the center of Curvature C, hits the mirror normally and is reflected back along it. Ray 3 is parallel to the principal axis and reflected back through the focal point. These three rays meet at the top of the image at I’. The foot of the object O rests on the principal axis and its image I also rests there in a line with the top of the image. Image is real, inverted, diminished in size.
   
   
   
   
4. For s_o < f, image is virtual, erect, and magnified. The same incident rays used in Fig. 6a above are used in Fig. 6b below. The eye traces back the reflected rays, which appear to meet at the top of the image at I’.
   
   
   
   
   
   

1. Focal length for convex mirror is negative. The center of curvature is at the back of the mirror.
   
   
2. Images of a convex mirror are always virtual, erect, and diminished in size.
   
   
3. In Fig. 7 below, Incident Ray 1 (one arrow) is drawn to the focal point F behind the mirror. It is reflected back parallel to the principal axis. Incident Ray 2 is drawn to the center of curvature C behind the mirror and reflected back on itself. Incident Ray 3 is drawn parallel to the principal axis and is reflected back as though it had come from the focal point F behind the mirror. The eye traces the rays back to find the head of the image I’ back of the mirror.
   
   
   
   
4. For Fig. 7,  let f = -3.4 cm  and  s_o = 5.1 cm.
   1/ s_i = 1/f – 1/s_o.
   1/s_i = -1/3.4 cm – 1/5.1 cm = 0.490 cm^-1.
   s_i = -2.0 cm.
   The negative value of s_i indicates that it is a virtual image.
   The magnification is m = - s_i/s_o = - (-2.0 cm)/5.1 cm = +0.39.
   The image is erect and diminished in size.
   
   

1/s_o + 1/s_i = 1/f.

1. Focal length is positive.
   
   
2. For s_o > f,  the image is real and inverted.
   
   
3. For s_o > 2f,  the image is diminished in size.
   For s_o = 2f,  m = 1.
   
   
4. For 2f > s_o > f,  the image is enlarged.
   
   
5. In Fig. 8a below, Ray 1 (one arrow) is drawn through the focal point F and is refracted parallel to the principle axis on the left. Ray 2 is drawn through the center of the lens and passes through it without being refracted. Ray 3 is drawn parallel to the principal axis and is refracted through the focal point F on the right. The three rays from the top of the object meet at the top of the image I’. The foot of the image I is on the principal axis. The image is real, inverted, and diminished in size.
   
   
   
   
6. For s_o < f,  the image is virtual, erect, and magnified.
   
   In Fig. 8b above, the same three rays are drawn, but these rays diverge on the left side. The eye traces the rays and “sees” them as though the top of the image was at I’. For a converging lens and virtual image, the image distance is negative but greater than the object distance. The image forms behind the object.
   
   
   

1. The focal length is negative. For all object distances, the image is virtual, erect and diminished in size. The image distance is negative and less than the object distance. The image forms in front of the object.
   
   
   
   
2. In Fig. 9 above ray 1 is drawn so that it would go through the focal point on the right side of the lens. It is diverged so that it comes out parallel to the principal axis. Ray 2 continues through the center of the lens. Ray 3 is parallel to the principal axis and diverges as though it had come from the focal point to the left of the lens. For all values of s_o, the rays diverge as they pass through the lens and do not form an image on the right. To the eye these refracted rays appear to come from the top of the image at I’.
   
   

1. Nearsightedness
   
   
   1. Light rays originating from a long distance away (s_o = infinity), focus behind the retina. The largest distance the person can see is at her far point. For the contact lens or glasses that correct nearsightedness the image distance s_i = - far point and the object distance = s_o = ∞. With this correction, she will see the object at infinity as though it had come from her far point.
      
      Since 1/f = 1/s_o + 1/s_i = 1/∞ - 1/far point = 0 - 1/far point
      = -1/far point  and  f = - far point.
      
      To correct nearsightedness, you must use a diverging lens.
      
      
   2. Notice the image distance at the far point is smaller than the object distance at infinity so we could have guessed she needed a diverging lens.
      
      
2. Farsightedness
   
   
   1. The ideal nearness of an object that can be seen by the eye is taken to be 25.4 cm. A person who is farsighted needs to have the object further away than 25.4 cm. The distance for which a farsighted person can see comfortably is called her near point. In correcting for farsightedness, s_i = - near point. The image distance is negative because it is a virtual image and s_o = 25.4 cm.  Now 1/f = 1/s_o + 1/s_i = 1/25.4 cm - 1/near point.
      
      Since the near point distance is greater than 25.4 cm,
      1/near point < 1/25.4 cm  and  f > 0,  so that the corrective lens will be a converging lens.
      
      
   2. Notice the image distance is greater than the object distance so you would expect to use a converging lens.
      
      Caution: Remember to express the dioptric power in m^-1.
      
      

1. A magnifying glass uses a converging lens.
   
   
2. Magnification is defined as Θ_a/Θ_u, where Θ_u is the angle the unaided eye makes with the object of height h (Fig. 10a above), and Θ_a is the angle the eye makes with the object of height h when assisted by a lens (Fig. 10b above). The lens allows you to move the object closer to the eye than 25.4 cm.
   
   From Fig. 10a, you see s_o is approximately equal to the focal length f of the lens. Also from Fig. 10a, we find tan Θ_u = h/25.4 cm and tan Θ_a = h/f. For small angles, the angle in radians is approximately equal to the tangent of the angle.
   
   m = Θ_a/Θ_u ≈ tan Θ_a/Θ_u = (h/f)/(h/25.4 cm) = 25.4 cm/f = magnification of magnifying glass.
   
   
   

1. A microscope consists of two lenses. The one closest to your eye (lens 2 in Fig. 11) is called the eyepiece and the other lens is called the objective (lens 1 in Fig. 11). In Fig. 11, you locate the image of the objective with ray 1 through the center of the lens and ray 2 through the focal point of lens 1 which refracts parallel to the principal axis. When ray 2 goes through the eyepiece, it is refracted through the focal point of the eyepiece. You can then draw ray 3 because every ray that comes from the top of the object must arrive at the top of the image. You draw ray 3 so it will pass through the center of the eyepiece.
   
   
2. Light enters through the objective. You want the image of the objective
   
   
   1. to be a real image, and
      
      
   2. to be formed at the focal point of the eyepiece.
      
      
   3. The image of the objective becomes the object for the eyepiece. Since the object of the eyepiece is at its focal point, the image of the eyepiece will be at infinity. These rays are the most restful on your eyes because your eye is set for this distance and does not need accommodation.
      
      
   4. The image of the objective is found from 1/f_o = 1/s_o1 + 1/s_i1, where f_o is the focal length for the objective lens.
      
      
   5. Magnification
      
      
      1. The magnification of the objective m_o = - s_i1/s_o1
         
         
      2. Another formula for m_o can be found in the following way taking the "length" L of the tube as s_i1 – f_o.
         First solve for s_i1 from 1/f_o = 1/s_o1 + 1/s_i1.
         1/s_o1 = 1/f_o -1/s_i1 = (s_i1 - f_o)/ s_i1f_o   or
            s_o1 = s_i1f_o /(s_i1 - f_o) .
         Then m_o = - s_i1/s_o1 = - s_i1(s_i1 - f_o)/s_i1f_o = - L/f_o.
         
         
      3. The magnification of the eyepiece is the magnification of the magnifying glass.

         m_E = 24.5 cm/f_E
         

      4. Total M = m_o m_E = - s_i1/s_o1(24.5 cm/f_E) = - L/f_o(24.5 cm/f_E).
         
         
         
         

1. Parallel rays from a distant object, but not parallel to the principal axis, are brought to focus in the focal plane. Note top of image of height h in Fig. 12 above is in the focal plane but not at the focal point.
   
   
2. This image is formed at the focal distance f_E of the eyepiece. I have not shown the origin of the two rays that go from the top of the image through the center of the eyepiece and another parallel to the principal axis of the eyepiece emerging through the focal point of the eyepiece. The final image is at infinity.
   
   
3. The magnification of a telescope is defined as the

   ratio of angle Θ_E to Θ_o,

   as shown in Fig. 12.  For small angles the ratio of these angles equals the ratio of their tangents.  From Fig. 12, we see that

   m = tan Θ_E/tan Θ_o = (h/f_E)/ (h/f_o) = f_o/f_E.