Optics Problem -- possible values for the object distance?

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SUMMARY

The discussion revolves around calculating the possible values for the object distance in a lens system where an object and its real image are 2.4 m apart, with a lens focal length of 55 cm. The lens formula used is 1/s + 1/s' = 1/f, where s is the object distance and s' is the image distance. The user attempted to solve the problem by considering two scenarios for the image's position relative to the object, leading to equations that resulted in s^2 = 295 and s^2 = -185. The latter indicates an error in the calculations, as negative distances are not physically meaningful.

PREREQUISITES
  • Understanding of the lens formula: 1/s + 1/s' = 1/f
  • Basic knowledge of optics, particularly real images and object distances
  • Familiarity with algebraic manipulation of equations
  • Concept of focal length in lens systems
NEXT STEPS
  • Review the derivation of the lens formula and its applications in optics
  • Learn about the significance of focal length in determining image formation
  • Practice solving lens problems using different object and image configurations
  • Explore the concept of virtual images and how they differ from real images
USEFUL FOR

Students studying optics, physics educators, and anyone interested in understanding lens systems and image formation principles.

TomInPhysics
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Homework Statement


An object and its lens-produced real image are 2.4 m apart. If the lens has 55-cm focal length, what are the possible values for the object distance?

Homework Equations


1/s+1/s'=1/f

The Attempt at a Solution


The object and the image are 2.4m apart so I thought "Hey, that is the same as 240cm apart, cool."

So now I have an image and an object that are 240cm apart and I thought, "well... that is basically the same as saying the image is either 240cm to the left of the object or 240cm to the right of the object. Since we don't know for sure we'll have to solve it both ways."

At this point I'm thinking: s'=s-240 OR s'=s+240

So now I'm like "Sweet, let's plug that baby into the lens equation!"

1/s+(1/s-240)=1/55 OR 1/s+(1/s+240)=1/55

Then I flipped everything since it was all over one to make it easier on my eyeballs to see...

s+2-240=55 OR s+s+240=55

Then I was all like "wham blam math mojo stuff" and finished with

295=s^2 OR -185=s^2

And then from their I thought I was wrong.

So any help would be great, thanks!
 
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TomInPhysics said:

Homework Statement


An object and its lens-produced real image are 2.4 m apart. If the lens has 55-cm focal length, what are the possible values for the object distance?

Homework Equations


1/s+1/s'=1/f

The Attempt at a Solution


The object and the image are 2.4m apart so I thought "Hey, that is the same as 240cm apart, cool."

So now I have an image and an object that are 240cm apart and I thought, "well... that is basically the same as saying the image is either 240cm to the left of the object or 240cm to the right of the object. Since we don't know for sure we'll have to solve it both ways."

At this point I'm thinking: s'=s-240 OR s'=s+240

So now I'm like "Sweet, let's plug that baby into the lens equation!"

1/s+(1/s-240)=1/55 OR 1/s+(1/s+240)=1/55

Then I flipped everything since it was all over one to make it easier on my eyeballs to see...

s+2-240=55 OR s+s+240=55

Then I was all like "wham blam math mojo stuff" and finished with

295=s^2 OR -185=s^2

And then from their I thought I was wrong.

So any help would be great, thanks!
It might surprise you to know that ##\displaystyle \ \frac{1}{\displaystyle\ \frac{1}{a}+\frac{1}{b}\ }\ne a+b \ . ##

Try it with some numbers.

You need a common denominator to simplify the expressions on the left hand side of your equations.
 

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