Find the Focal Length Of a lens

In summary: Now, what does that tell you about ##d_i##?well, i know that 897/19 = 47.21 and 897/10 = 89.7 so i would think that di2 + 19di would be 47.21 and 10/897 would be diNow, what does that tell you about ##d_i##?You are close, but you have the numerator and denominator of the left side reversed. Also, you don't need the exact values; you just need to know the relationship between ##d_i## and ##d_i^2 + 19d_i##.okay so it would be
  • #1
AyooNisto
16
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:
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  • #2
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.
 
  • #3
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.
 
  • #4
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
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?
 
  • #6
##\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
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.
 
  • #8
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}\ ? ##
 
  • #9
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.
 

FAQ: Find the Focal Length Of a lens

What is the focal length of a lens?

The focal length of a lens is the distance between the lens and the point where the light converges to a single point. It is usually measured in millimeters (mm) and determines the magnification and field of view of the lens.

How do you find the focal length of a lens?

The focal length of a lens can be found by using the lens formula, which is 1/f = 1/u + 1/v, where f is the focal length, u is the object distance, and v is the image distance. You can also use a lensometer or a ruler to measure the distance from the lens to the point where the image is formed.

What is the difference between the focal length of a lens and its aperture?

The focal length and aperture of a lens are two different properties. The focal length determines the magnification and field of view, while the aperture determines the amount of light that can pass through the lens. A longer focal length results in a narrower field of view, while a larger aperture allows more light to enter the lens.

How does the focal length affect the image quality?

The focal length of a lens can affect the image quality in several ways. A shorter focal length can result in distortions, such as barrel distortion, while a longer focal length can minimize these distortions. Additionally, a longer focal length can also result in sharper images and a shallower depth of field.

Can the focal length of a lens be changed?

The focal length of a lens is a fixed property determined by the lens design. However, some lenses, such as zoom lenses, have variable focal lengths that can be adjusted to change the magnification and field of view. Prime lenses, on the other hand, have a fixed focal length and cannot be changed.

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