Find the Focal Length Of a lens

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Homework Help Overview

The problem involves a converging lens and the relationship between object distance, image distance, and focal length. The original poster describes a scenario where an object is placed at two different distances from the lens, resulting in a change in the image position.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the application of the lens formula and the importance of parentheses in the equations. There are attempts to rearrange the equations to isolate the image distance, di, and questions about the validity of certain algebraic manipulations.

Discussion Status

The discussion is ongoing with participants providing feedback on each other's algebraic steps. There is a focus on ensuring correct application of the lens formula and clarifying misunderstandings regarding the use of parentheses in the equations.

Contextual Notes

Some participants express confusion about the algebra involved and the implications of the distances given in the problem. There is a recognition of the need for careful handling of fractions and common denominators in the calculations.

AyooNisto
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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?
 
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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}\ ? ##
 
  • #10
Can you combine ##\displaystyle\ \frac{1}{d_i}-\frac{1}{d_i+19}## to make one fraction?
 
  • #11
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?
 
  • #12
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.]
 
  • #13
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.]
 
  • #14
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.
 
  • #15
Ohh okay so it would be 19/(di2 + 19di) = 10/897



TSny said:
Yes, but remember that you are subtracting the fractions.
 
  • #16
AyooNisto said:
Ohh okay so it would be 19/(di2 + 19di) = 10/897

Yes, that's correct.
 

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