Tough Question in Electo-Optics

A converging lens held near a flashlight is used to put an image of the bulb
on a wall that is 1.60m from the lens. With the flashlight fixed, the lens
is moved closer to the wall and a sharp image is again formed when the lens
is 15cm from the wall.

a. What is the focal length of the lens?
b. How far is the flashlight bulb from the wall?

I cannot figure it out and am getting very frustrated. Any help would be
appreciated.

Colin

"Andrea Zammit" <[Only registered users see links. ]> wrote in message news:<EB16d.2998$[Only registered users see links. ]>...

The equation you need is here (equation 4):
[Only registered users see links. ]

in ASCII:

1/f = 1/So1 + 1/Si2

Draw a diagram:

source lens image
*------------()-----------|
So1 Si2
<------------B------------>

You actually have 2 cases here:
case I-

1/f = 1/(B-1.6) + 1/(1.6)

where Si2 = 1.6 meters, the distance from the lens to the image, and B
is the answer to part b, the distance from the light source to the
image.

case II-

1/f = 1/(B-0.15) + 1/(0.15)

where Si2 is now 0.15 meters

Are you up to the algebra of solving 2 equations in two unknowns?

HINT: 1/f = 1/(B-1.6) + 1/(1.6) = 1/(B-0.15) + 1/(0.15)


Tom Davidson
Richmond, VA