Find the Focal Length Of a lens

Join the discussion
Registration is free. Start your own thread to ask a follow-up.
15 replies · 4K views
AyooNisto
Messages
16
Reaction score
0

Homework Statement


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.

Homework Equations


1/do + 1/di = 1/f

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:
Physics news on Phys.org
AyooNisto said:
1/39.0cm + 1/di + 19cm = 1/f

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

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

Likewise here.

The algebra is a little tedious, so you will need to be careful soving for di.
 
Even with the parenthesis in the equations wouldn't di still be 20, 39-19 is 20 either way

TSny said:
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.
 
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?
 
not really I am having a little trouble understanding this problem

TSny said:
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?
 
##\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.
 
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?

TSny said:
##\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.
 
AyooNisto said:
okay so i have

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

You continue to ignore the required parentheses !

Your equation:
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.
di + 19/di = 69/39
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}\ ? ##
 
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

SammyS said:
You continue to ignore the required parentheses !

Your equation:
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}\ ? ##
 
Can you combine ##\displaystyle\ \frac{1}{d_i}-\frac{1}{d_i+19}## to make one fraction?
 
i believe i did that in the post above it gives you 19/di2 + 19di. is this wrong?

TSny said:
Can you combine ##\displaystyle\ \frac{1}{d_i}-\frac{1}{d_i+19}## to make one fraction?
 
AyooNisto said:
i believe i did that in the post above it gives you 19/di2 + 19di. is this wrong?

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

19/(di2 + 19di)

[EDIT:The 36/897 is incorrect.]
 
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

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

19/(di2 + 19di)

[EDIT:The 36/897 is incorrect.]
 
AyooNisto said:
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

Yes, but remember that you are subtracting the fractions.
 
Ohh okay so it would be 19/(di2 + 19di) = 10/897



TSny said:
Yes, but remember that you are subtracting the fractions.
 
AyooNisto said:
Ohh okay so it would be 19/(di2 + 19di) = 10/897

Yes, that's correct.