Optics Problem -- possible values for the object distance?

In summary: Do you see how to get one?In summary, the object and its lens-produced real image are 2.4 m apart. To find the possible values for the object distance, the lens equation 1/s+1/s'=1/f is used, where f is the focal length of the lens. To solve for s, the object distance, the equation is simplified by finding a common denominator.
  • #1
TomInPhysics
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0

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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  • #2
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.
 

Related to Optics Problem -- possible values for the object distance?

1. What is the object distance in optics?

The object distance in optics refers to the distance between the object and the lens or mirror in an optical system. It is typically measured in meters or centimeters.

2. How is the object distance related to the image distance?

The object distance and image distance are inversely related in optics. This means that as the object distance increases, the image distance decreases, and vice versa.

3. What are the possible values for the object distance?

The possible values for the object distance depend on the specific optical system being used. In general, the object distance can range from very small values (close to the lens or mirror) to very large values (far from the lens or mirror).

4. How does the object distance affect the size of the image?

The object distance has a direct impact on the size of the image in optics. As the object distance increases, the size of the image also increases, and vice versa.

5. Can the object distance be negative?

Yes, in some optical systems, the object distance can be negative. This occurs when the object is located behind the lens or mirror, and in these cases, the image will be virtual and upright.

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