Solve Optics Problem with Lensmaker's Equation

[dzza]

Hey, so I've got a HW for an optics lab and haven't studied optics for a while, so I could use a little help.

The problem is a lensmaker's equation question involving a system of optical devices. An object at a distance of 30cm to the left of a converging lens of focal length +30cm. The converging lens is then 10 cm away from a divering lens of known focal length.

Now, I know the general procedure for solving this sort of problem, using the image of the light through the first lens as the object for the second lens. If I could just find the image distance from the first lens I'd be in good shape. But, isn't the object on the focal plane? and if it is, don't all rays of light incident on the first lens from the object emerge parallel to optical axis? In that case, no image is formed.

From the lensmaker's equation, i get 1/30 + 1/s = 1/30, where s is the image distance, and this has no solution. My question is do I have a sign messed up for the focal length or object distance?

 

[andrevdh]

No.You are in the right track. Remeber mathematically that the inverse of avery large number is very small. That is if one make the denominater larger and larger the inverse appoaches zero. So the image will be formed at infinity. This means that the object dinstance for the divergiing lens will be $$+\infty$$.

 

[kyle8921]

I would approach this problem by first clarifying any uncertainties or confusion with the given information. In this case, it seems that the object is located on the focal plane of the first lens, which would result in parallel rays of light emerging from the lens and no image being formed. This may be a mistake in the problem statement or a misunderstanding of the concept of focal plane. I would suggest double-checking the given information and clarifying with the instructor if necessary.

Assuming that the object distance and focal length are correct, the equation 1/30 + 1/s = 1/30 does not have a solution because the image distance cannot be determined. This indicates that there may be an error in the lensmaker's equation or in the given values.

To solve this problem, I would recommend using the thin lens equation: 1/f = 1/do + 1/di, where f is the focal length, do is the object distance, and di is the image distance. This equation can be used to calculate the image distance from the first lens, which can then be used as the object distance for the second lens.

It is also important to note that the sign convention for the focal length and object distance may affect the final result. In this case, the positive sign for the focal length and object distance indicates that they are both on the same side of the lens, which is the correct assumption for a converging lens. However, if the sign convention is different, it may lead to a different result.

In conclusion, to solve this optics problem with the lensmaker's equation, it is important to clarify any uncertainties and use the correct sign convention for the given values. If there are still issues with finding a solution, it may be necessary to use alternative equations or seek help from an instructor.