# A Two Lens System

A cell phone of length 10.0cm sits 50.0cm in front of a lens of focal length 30.0cm.

a. Without equations, use a diagram to determine whether the image of the cell phone is real or virtual, upright or inverted, and magnified or reduced. Estimate the magnification from your diagram.

b. Now calculate the magnification and check it against your part (a) answer.

c. A student holds his eye close to the lens and looks through it at the cell phone. Instead of a sharp image, he sees a blur. Why?

d. Now the student backs up until his eye is 80cm from the lens. He sees a clear image, but it looks much bigger than he expected, give the answer to part (b). Why?

https://brainmass.com/physics/lenses/two-lens-system-572550

## SOLUTION This solution is **FREE** courtesy of BrainMass!

Please see the attached solution for explanation and diagrams.

a) I have made a rough sketch of the Lens system as below

Construction draw 2 focal planes 30cm either side of the lens, Shown by black dotted lines

Draw object scaled 10cm in length on left hand side of lens

Draw horizontal reference plane going through centre O of the lenses, bold black line

Draw horizontal ray line originating from top of object and parallel with reference line. This is the blue ray trace

Draw ray line from top of object through left hand side focal plane and centre of lens O, this is the red dotted ray trace

You will see if you place objects, focal planes all scaled to one another then the image is

Real (as it's formed the opposite side of the lens to the object)

Inverted as the top of the object ray is below the reference line

Magnified as the image arrow line is larger than the object line

Measuring the lengths (sizes) of the image and object one gets the magnification from

Magnification = image height/object height ~ 1.5 (if you measure these)

b) To mathematically calculate the magnification we first need to determine the image distance (distance v)

We use the lens equation where

1/f = 1/u = 1/v

f = 30cm the focal length, u = object distance = 50cm

When we do this we can find

1/v = 1/f - 1/u

1/v = (u - f) / uf

So

v = uf / (u - f) = (50 x 30) / (50 - 30) = 1500/20 = 75cm

So the image distance is 75cm from the lens

Now we can use the magnification equation that

Magnification M is given by

M = image height /object height = image distance / object distance = v / u

Further and to be precise we need to reflect that the image is inverted so we strictly say

Magnification = - v / u = -75 / 50 = - 1.5

c) Image looks blurred as the student is viewing the object inside the focal length plane of the lens.

d) If you look at the diagram this is telling one that the student is looking at the object to the right of point v, or 80cm from the lens. If you extend the rays out to this distance you will see that the image becomes even larger. I try to show this below

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