# Find the Focal Length Of a lens

1. Apr 30, 2013

### AyooNisto

1. The problem statement, all variables and given/known data
When an object is placed 69.0cm from a certain converging lens, it forms a real image. When the object is moved to 39.0cm from the lens, the image moves 19.0cm farther from the lens.

Find the focal length of this lens.

2. Relevant equations
1/do + 1/di = 1/f

3. The attempt at a solution
1/69.0cm + 1/di = 1/f

1/39.0cm + 1/di + 19cm = 1/f

1/69.0cm + 1/di = 1/39.0cm + 1/di + 19cm

which gives di = 20cm

1/69.0 cm + 1/20.0 cm = 1/f

f = 15.5??? this is not the right answer though, where am i going wrong?

Last edited: Apr 30, 2013
2. Apr 30, 2013

### TSny

Parentheses need to be used here: 1/(di + 19cm)

Likewise here.

The algebra is a little tedious, so you will need to be careful soving for di.

3. Apr 30, 2013

### AyooNisto

Even with the parenthesis in the equations wouldnt di still be 20, 39-19 is 20 either way

4. Apr 30, 2013

### TSny

The equation is

$\frac{1}{69} + \frac{1}{d_i} = \frac{1}{39} + \frac{1}{d_i+19}$

Do you see why you can't simply subtract the 39 and 19?

5. Apr 30, 2013

### AyooNisto

not really im having a little trouble understanding this problem

6. Apr 30, 2013

### TSny

$\frac{1}{69} + \frac{1}{d_i} = \frac{1}{39} + \frac{1}{d_i+19}$

There are sevaral approaches to solving this equation for $d_i$.

You might start by rearrainging the equation so that the two terms containing $d_i$ are on the left side and the two fractions without the unknown are on the right.

7. Apr 30, 2013

### AyooNisto

okay so i have

1/di - 1/di + 19 = 1/39 - 1/69

di + 19/di = 69/39

39di + 741 = 69di

741 = 30di

24.7 = di....is this correct?

8. Apr 30, 2013

### SammyS

Staff Emeritus
You continue to ignore the required parentheses !

1/di - 1/di + 19 = 1/39 - 1/69​

should be:
1/di - 1/(di + 19) = 1/39 - 1/69​

The next step is wrong.
So what follows that is pointless.

What is the common denominator for $\displaystyle\ \frac{1}{d_i}-\frac{1}{d_i+19}\ ?$

What is the smallest common denominator for $\displaystyle\ \frac{1}{39}-\frac{1}{69}\ ?$

9. May 1, 2013

### AyooNisto

my apologies for ignoring the parenthesis again, i believe the common denominators are

di2+19di and 897

so then it would be 19/di2+19di = 36/897

which would give you di equal .0401337793

1/.0401337793 = 24.91666667

10. May 1, 2013

### TSny

Can you combine $\displaystyle\ \frac{1}{d_i}-\frac{1}{d_i+19}$ to make one fraction?

11. May 1, 2013

### AyooNisto

i believe i did that in the post above it gives you 19/di2 + 19di. is this wrong?

12. May 1, 2013

### TSny

Oh, I didn't see it! Yes, that's right if you include the parentheses!:

19/(di2 + 19di)

[EDIT:The 36/897 is incorrect.]

13. May 1, 2013

### AyooNisto

okay how is the 36/897 wrong? The first number that 39 and 69 go into evenly is 897. 13*23 and 69*13 is 897. which would igve you 23/897 and 13/897

14. May 1, 2013

### TSny

Yes, but remember that you are subtracting the fractions.

15. May 1, 2013

### AyooNisto

Ohh okay so it would be 19/(di2 + 19di) = 10/897

16. May 1, 2013

### TSny

Yes, that's correct.