# Lens Help Needed

1. Feb 22, 2009

### dsptl

Lens Help Needed!!

A converging lens with a focal length of 50cm and a diverging lens with a focal length of -52 cm are 227cm apart. A 3.9-cm-tall object is 70cm in front of the converging lens.

Calculate the distance between image and diverging lens.
Express your answer using two significant figures.
d = 26cm
correct

Part B
Calculate the image height.
Express your answer using two significant figures.
h = 1.3 wrong
one attempt remaing.

Work:
for final image distance:

1/f1 = 1/p1 + 1/q1
1/50 - 1/70 = 1/q1
q1 = 175 cm

227-175=52cm

1/f2 = 1/p2 + 1/q2

-1/52 - 1/52 = 1/q2

q2 = -26 -----> 26 cm final image distance

for image height:

M1 = -q1/p1 = 2.5

l-2.5l = h1(image)/h1(ogject)

9.75cm = h1(image)

M2 = -q2/p2 = .5

.5 = h2(image)/h1(image)

4.9 = h2(image)

is this right? I have only one attempt left

2. Feb 22, 2009

### dsptl

Re: Lens Help Needed!!

3. Feb 23, 2009

### Redbelly98

Staff Emeritus
Re: Lens Help Needed!!

4.9 cm is the correct magnitude.

However, note that M1 should be negative (the positive lens forms an inverted image). I don't know if you are expected you to bother with +/- signs to indicate upright/inverted images.

4. Feb 23, 2009

### dsptl

Re: Lens Help Needed!!

is it suppose to be positve b/c M =absolute[h(image)/h(object)]

or is it neg.?

5. Feb 24, 2009

### Redbelly98

Staff Emeritus
Re: Lens Help Needed!!

M = -q/p

If q and p are both positive or both negative, M is negative.

If q and p are of opposite sign, M is positive.

In this case, I get that the final image height is negative.

6. Feb 25, 2009

### dsptl

Re: Lens Help Needed!!

the answer was positive

Thanks for all Help guys!!!

7. Feb 25, 2009

### Redbelly98

Staff Emeritus
Re: Lens Help Needed!!

You're welcome!

Sounds like they were interested in the absolute value.