Question

Given:

Source Position: 4 cm

Convex Lens Position: 25 cm

Convex Focal Length: 13 cm

Concave Lens Position: 43 cm

Concave Focal Length: -20 cm

Object Heght: 3 cm

Find:

a) screen position for focused object?

b) image height on screen?

Please Help me

Answer 1

Locate the image and determine the magnification of lens 1.

$p = (25-4)cm = 21cm$

Apply the thin-lens equation to lens 1:

$\frac{1}{p} + \frac{1}{q}= \frac{1}{f} \Rightarrow \frac{1}{21cm}+ \frac{1}{q} = \frac{1}{13cm}$

$q = +34.125cm$

Compute the magnification of lens 1:

$M_1 = -\frac{q}{p} = -\frac{34.125cm}{21cm} = -1.625$

Locate the final image, and total magnification.

The image formed by lens 1 becomes the object for lens 2. Compute the object distance for lens 2:

$p = 18cm - 34.125cm = -16.125cm$

Once again apply the thin-lens equation to lens 2 to locate the final image:

$\frac{1}{-16.125cm}+ \frac{1}{q} = \frac{1}{-20cm}$

screen position for focused object

$\boxed{q=+83.225cm}$

Calculate the magnification of lens 2:

$M_2 = -\frac{q}{p} = +5.161$

Multiply the two magnifications to get the overall magnification of the system:

$M = M_1\cdot M_2 = -8.387$

image height on screen

$h = 3cm\cdot (-8.387)$

$\boxed{h = -25.161[cm]}$