Problem 5: An object 2 cm in height is located 10 cm to the left of a positive l
ID: 1769992 • Letter: P
Question
Problem 5: An object 2 cm in height is located 10 cm to the left of a positive lens with a focal length of 5 cm and a diameter of 5 cm. A 2 cm diameter aperture is located 2 cm to the left of the lens. The drawing below shows the situation but is not to scale. Li K 10cm - 2 cm 2 cm 5 cm 2 cm Complete the following. (a) Identify the aperture stop for the system. (b) Find the size and location of the Entrance Pupil. (c) Find the size and location of the Exit Pupil. (d) Find the image of the object its location and its transverse magnification. (e) Sketch the ray tracing for your system and then draw the chief ray from the tip of the object to the tip of the image Problem 6: Repeat Problem 5 with the following changes: (1) the object is now 4 cm in height and is located 14 cm to the left of lens LI; (2) the lens Li is still 5 cm in diameter, but now has a focal length of +6 cm; (3) the aperture is still 2 cm in diameter centered on the optical axis but is now 2.5 cm to the right of LiExplanation / Answer
given object height h = 2cm
object distance u = -10 cm
focal length f = 5 cm
diameter D = 5 cm
aperture diameter d = 2 cm
distance of aperture from the lens a = 2 cm
a. aperture stop for the system is
at distance x to the left of the lens
then
(x - 2)/2 = x/5
5x - 10 = 2x
3x = 10
x = 3.33 cm
b. let location of entrance pupil be x cm from the lens to its right
then
1/x - 1/u = 1/f
where
u = -2 cm
1/x = 1/5 - 1/2
x = -3.33 cm
hence the entrance pupil is -3.33 cm to the left of the lens
its size = 3.33*2/2 = 3.33 cm
c. let location of exit pupil be x cm from the lens to its right
then
1/x - 1/u = 1/f
where
u = 2 cm
1/x = 1/5 + 1/2
x = 1.428 cm
hence the entrance pupil is 1.428 cm to the right of the lens
its size = 1.428*2/2 = 1.428 cm
d.image of object is at distance v to the right of the lens
1/v - 1/10 = 1/5
1/v = 1/5 + 1/10
v = 3.333 m
magnification = 3.333*2/10 = 0.6667