How to Solve a Converging Lens Problem?

[kikko]

Homework Statement

There is an object 20 cm from a converging lens with a 10 cm focal length.

Homework Equations

Is the image upright or inverted?
Calculate the image distance.
Calculate the image height.

The Attempt at a Solution

I just had this on a college test in a testing center, I could find the first two with the thin lens equation and diagrams, not sure how I was supposed to get the last. I feel some info was missing, like object height. Any way my feeling was wrong?

 

[Dadface]

Your feeling wasn't wrong since there is not enough information to find the image height.For this question the image height equals the object height (magnification equals one) but the object height was not given.

 

[SammyS]

Hi kikko.

I agree, there's not enough information for an absolute answer, but you could give a relative answer, like, 2 × the object height, or 1/3 the object height or the same as the object height.

 

[Liquidxlax]

Your feeling wasn't wrong since there is not enough information to find the image height.For this question the image height equals the object height (magnification equals one) but the object height was not given.

its possible that they meant magnification, because i had a similar problem on midterms when i was in optics

 

[rwisz]

I would approach this problem by first understanding the basics of converging lenses and their properties. I would then use the thin lens equation, which states that 1/f = 1/do + 1/di, where f is the focal length, do is the object distance, and di is the image distance.

Based on the given information, we know that f = 10 cm and do = 20 cm. By substituting these values into the equation, we can solve for di. This will give us the image distance, which is the distance from the lens to the image formed.

To determine if the image is upright or inverted, we can use the sign convention for lenses. A converging lens has a positive focal length, so the image distance will also be positive. If the image distance is positive, then the image is real and inverted. If the image distance is negative, then the image is virtual and upright.

Lastly, to calculate the image height, we need to use the magnification equation, which is given by M = -di/do. By substituting the values for di and do, we can calculate the magnification. The image height is then given by the product of the magnification and the object height.

In conclusion, with the given information, we can determine that the image formed by the converging lens is real and inverted, with an image distance of 6.67 cm and a magnification of -0.33. Without knowing the object height, we cannot calculate the exact image height, but we can determine its relative size compared to the object.