Fundamental of Optical Engineering Lecture 2

 In order to locate the image, the 2 rays are needed as ◦ The parallel ray: parallel to the axis and then, after refraction, passes through another focal point. ◦ The focal ray: passes through a first focal point then, after refraction, is parallel to the axis. ◦ The chief ray goes through the center of the lens without deviation since the lens is thin.

 L.H.S.  R.H.S.

 Take ratio (1) = (2), we have

 Multiply both sides by 1/ss’f, this yields  Finally, we have

 Magnification,  From (2)  From (3) x s’

 So that,

 Light travels from left to right.  All distances are measured from the plane of a lens.  Object to the left of a lens is called object distance s, and s is negative.  The image distance s’ could be positive (real image) and negative (virtual image).

 The focal length, f ◦ f is ‘+’ for converging lens. ◦ f is ‘-’ for diverging lens.  y’ is the height of an image. ◦ y’ is ‘+’ for upright image. ◦ y’ is ‘-’ for inverted image.  Magnification M ◦ M is ‘+’ for upright image. ◦ M is ‘-’ for inverted image.

 Find s’ and M for these cases of object distances. ss’M -∞-∞ f0 -2f -3f/2 -f -f/2 -f/4

 In case of off-axis rays incident on a converging lens, the displacement of spot in focal plane can be found as

 d = ftanθ  If θ is small, then tan θ  θ.  d  fθ …… Displacement of spot in focal plane.

 In case of diverging lens, focal length is negative in GTLE.  That leads an image distance to be ‘-’.  This means we always have a virtual image to the left of the lens.

 For diverging lens, find s’ and M for various s. ss’M -∞-∞ -f0 -10f -2f -f -f/2

 We may use the combinations of lenses to form desired images.

 When an object is placed 75 mm in front of a converging lens, its image is 3 times as far away from the lens as when the object is at infinity. What is the focal length of the lens?