Lenses

Lenses

1. LENSES

2. THIN LENS
•A lens is an optical system with two
refracting surfaces
•The simplest lens has two spherical
surfaces close enough together that we
can neglect the distance between them
(the thickness of the lens)

4. CONVERGING LENS (CONVEX)
•Has the property that when a beam of
rays parallel to the axis passes through
the lens, the rays converge to a point F2
and form a real image at that point.
•A positive lens (F1 and F2 are both
positive)

7. DIVERGING LENS (CONCAVE)
•The beam of parallel rays incident on
this lens diverges after refraction.
•Is called a negative lens
•The focal length of a diverging lens is
a negative quantity

12. CURVED MIRRORS
•Concave
 positive focal length
 real or virtual image
•Convex
 negative focal length
 virtual image
LENSES
•Concave
negative focal length
 virtual image
•Convex
 positive focal length
 real or virtual image

14. GRAPHICAL METHOD FOR LENSES

15. GRAPHICAL METHOD FOR LENSES

16. GRAPHICAL METHOD FOR LENSES

22. CONVEX LENS (CONVERGING)

23. CONCAVE LENS (DIVERGING)

24. Sample Problem
•An object 5.0 cm high is placed 24
cm away from a converging lens of
focal length 8.0 cm.
a.Calculate the location of the image
b.Calculate the height of the image

25. APPLICATIONS

28. •Increasing the focal length
increases the image size

30. THE EYE AS A CAMERA

31. •Far point is described as the
farthest point from the eye at which
images are clear.
•Near point is described as the
nearest point from the eye at which
images are clear.

32. Far point is infinity

33. JUST LIKE AN OPTOMETRIST
MEASURE THE NEAR POINT

38. SW (1/2 CW)
Do not copy the problem. Show complete solution.

39. MATHEMATICAL METHOD
1.An object 5 cm high is placed 24 cm
away from a concave lens of focal
length -8 cm.

a.) Calculate the location of the image
b.) Calculate the height of the image

40. GRAPHICAL METHOD
2. An object 2 cm high is placed 4 cm
away from a convex lens of focal
length 3 cm. Determine the image
distance and the image height.

41. SHORT QUIZ
OPEN NOTES (1/2 CW) DO NOT COPY THE PROBLEM…
SHOW GIVEN AND COMPLETE SOLUTION

42. 1. An object 5.0 cm high is placed 24.0 cm away