Question

1. A candle is placed 36 cm from the screen. Where between the candle and the screen should a converging lens with a focal length of 8.0 cm be placed to produce a sharp image on the screen?

2. A lens has a focal length of +20cm and a magnification of 4. How far apart are the object and the image?

Answer 1

You can figure out the first question using the lens/mirror equation:

\[{1 \over f}={1 \over d_o}+{1 \over d_i}\]
Where f is the focal length, d(o) is the distance from the object to the lens, and d(i) is the distance from the image to the lens.

This is a converging (convex) lens, so by convention, f will be positive, d(o) is positive, and d(i) is positive in this case (since the image is real and on the opposite side of the lens). To have a sharp image, you want the screen to be placed at d(i).

The other thing you need to realize is that the lens is dividing up the space between the object and the screen into two parts. If you place the lens a distance x away from the object, then the distance from the lens to the screen (the image) is 36-x.

What do you get as your answer to the first question?

Answer 2

So it's
di=
1
_________
(1/8)- (1/36)

would that get me d(i)?

Answer 3

Let's take a step back for a minute.

What's f?
What's d(o)?
What's d(i)?

Answer 4

f= 8.o cm
d(o)= 36 cm
but d(i) is unknown

Answer 5

I agree with your value for f.

But for d(o) and d(i), read this part of my answer carefully:

"The other thing you need to realize is that the lens is dividing up the space between the object and the screen into two parts. If you place the lens a distance x away from the object, then the distance from the lens to the screen (the image) is 36-x."

Basically, the object is 36 cm away from the SCREEN, but not 36 cm away from the LENS. So what are the values of d(o) and d(i)?

Answer 6

Well umm we were given this formula for it but I have no idea how to work this then it was
2/36-d(o)+1/d(o)=(1/8)

Answer 7

*correction
1/36-d(o)+1/d(o)=(1/8)

Answer 8

Yup that's the same as what I said! You're just using d(o) instead of x in your equation, but I defined x as d(o) in what I explained above. d(i) is whatever distance is left from the lens to the screen, so 36-d(o).

So if you solve for d(o), what do you get?

Answer 9

That's the thing, I don't know how to make this formula work, like what to move and where to move where.

Answer 10

Ok let's do it step by step because it can be a bit tricky. This is what you're starting with:

\[{1 \over 36-d_o}+{1 \over d_o}={1 \over 8}\]
We want to isolate d(o), so we need to first get it out of the denominator. That means you need to multiply both sides by 36-d(o), then d(o). I'm going to do it in one step instead of two to make things a bit quicker:

\[{(36-d_o)(d_o) \over 36-d_o}+{(36-d_o)(d_o) \over d_o}={(36-d_o)(d_o) \over 8}\]\[d_o+(36-d_o)={(36-d_o)(d_o) \over 8}\]
Do you think you can take it from there?

Answer 11

Are we still trying to isolate the d(o)?

Answer 12

Because then would it be
36=(36-do/8)+d(o)-d(o)?

Answer 13

Yup always trying to isolate the d(o). So what's your final value for d(o)?

Answer 14

Um would the
36=(36-do/8) * do + do - do
would the + do and the - do cancel out?

Answer 15

So then would it be
36= (36-do/8) * do

Answer 16

From the last step I wrote out above, you can simplify the left side:

\[36={(36-d_o)(d_o) \over 8}\]
Then multiply both sides by 8 to get rid of the denominator:
\[288=(36-d_o)(d_o)\]
Then distribute the d(o) on the other side:
\[288=36d_o-d_o^2\]
Then move everything over to the left side:
\[d_o^2-36d_o+288=0\]
Now do you think you can solve it?

Answer 17

So would this be correct so far?
d2(o)-36d(o)=-288?

Answer 18

Well yes, but that doesn't get you closer to the answer. Just use the quadratic formula on my last step to find the value of d(o)!

Answer 19

So would then d(o)= -8?

Answer 20

If you apply the quadratic formula, you should find that d(o)=12 or d(o)=24. Both seem like reasonable numbers, but there's one thing you need to realize. The focal length of the lens is 8 cm, meaning its centre of curvature is 16 cm away.

Now you need to remember some facts about lens.

If the object is located 12 cm from the lens, it's between the centre of curvature and the focal point. What do we know about the image that is formed?

If the object is located 24 cm from the lens, it's past the centre of curvature. What do we know about the image that is formed?

If you are able to answer these questions, you'll know which value is the right value of d(o)!

Answer 21

May I ask how did you do the quadratic formula? I'm pretty sure it's
x=-b+or- √ b^2-4ac over 2a

Answer 22

That's right! In this case, a=1, b=-36, c=288. Have a look at my last equation I wrote above.

Answer 23

Ok I got the answer 12 and 24 now! So if the object is in between the focal or in front the image is real and inverted and smaller. If it's beyond the focal point then the image becomes virtual and upright, also bigger.

Answer 24

Thank you so much!!!

Answer 25

You're right! So we know the image MUST be real because it's on a screen on the opposite side of the lens. So which value do we use for d(o) - 12 cm or 24 cm?

Answer 26

The 12 correct?

Answer 27

By the way sorry this is late and thank you so much!

Answer 28

12 is absolutely right! And glad I could help you understand!

Answer 29

And if you need help with the second question let me know

Answer 30

I'll repost the second question