Rules of ray tracing for convex lenses

  • Any incident ray traveling parallel to the principal axis of a converging lens will refract through the lens and travel through the focal point on the opposite side of the lens.
  • Any incident ray traveling through the focal point on the way to the lens will refract through the lens and travel parallel to the principal axis.
  • An incident ray that passes through the center of the lens will in effect continue in the same direction that it had when it entered the lens.
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    1. Pick a point on the top of the object and draw three incident rays traveling towards the lens.

    Using a straight edge, accurately draw one ray so that it passes exactly through the focal point on the way to the lens. Draw the second ray such that it travels exactly parallel to the principal axis. Draw the third incident ray such that it travels directly to the exact center of the lens. Place arrowheads upon the rays to indicate their direction of travel.


    2. Once these incident rays strike the lens, refract them according to the Snell's laws.

    The ray that passes through the focal point on the way to the lens will refract and travel parallel to the principal axis. Use a straight edge to accurately draw its path. The ray that traveled parallel to the principal axis on the way to the lens will refract and travel through the focal point. And the ray that traveled to the exact center of the lens will continue in the same direction. Place arrowheads upon the rays to indicate their direction of travel. Extend the rays past their point of intersection.

     

    3. Mark the image of the top of the object.

    The image point of the top of the object is the point where the three refracted rays intersect. All three rays should intersect at exactly the same point. This point is merely the point where all light from the top of the object would intersect upon refracting through the lens. Of course, the rest of the object has an image as well and it can be found by applying the same three steps to another chosen point.


    4. Repeat the process for the bottom of the object.

    One goal of a ray diagram is to determine the location, size, orientation, and type of image that is formed by the double convex lens. Typically, this requires determining where the image of the upper and lower extreme of the object is located and then tracing the entire image. After completing the first three steps, only the image location of the top extreme of the object has been found. Thus, the process must be repeated for the point on the bottom of the object. If the bottom of the object lies upon the principal axis (as it does in this example), then the image of this point will also lie upon the principal axis and be the same distance from the mirror as the image of the top of the object. At this point the entire image can be filled in.

    Rules for Ray Tracing Provided by Physics Classroom 

    Example 

     

    A 4.00-cm tall light bulb is placed a distance of 45.7 cm from a double convex lens having a focal length of 15.2 cm. Determine the image distance and the image size.

    Like all problems in physics, begin by the identification of the unknown information.

    ho = 4.00 cm
    do = 45.7 cm
    f = 15.2 cm

    Next identify the unknown quantities that you wish to solve for.

    di = ???
    hi = ???

    To determine the image distance, the lens equation must be used. The following lines represent the solution to the image distance; substitutions and algebraic steps are shown.

    1/f = 1/do + 1/di

    1/(15.2 cm) = 1/(45.7 cm) + 1/di

    0.0658 cm-1 = 0.0219 cm-1 + 1/di

    0.0439 cm-1 = 1/di

    d= 22.8 cm

    The numerical values in the solution above were rounded when written down, yet unrounded numbers were used in all calculations. The final answer is rounded to the third significant digit.

    To determine the image height, the magnification equation is needed. Since three of the four quantities in the equation (disregarding the M) are known, the fourth quantity can be calculated. The solution is shown below.

    hi/ho = - di/do

    h/(4.00 cm) = - (22.8 cm)/(45.7 cm)

    h= - (4.00 cm) • (22.8 cm)/(45.7 cm)

    h= -1.99 cm

    Example Problem provided by Physics Classroom