Calculating Image Distance in a Two-Lens System

[Romperstomper]

An eraser of height 1.0cm is placed in front of a two-lens system. Lens 1(nearer the eraser) has focal length f1=-15cm, lens 2 has f2=12cm, and the lens separation is d=12cm. For the image produced by lens 2, what is the image distance i2?

What I did is setup some equations.

For lens 1: 1/-15 = 1/P1 + 1/i1 and m1 = -i1/P1
For lens 2: 1/12 = 1/(i1+12) + 1/i2 and m2= -i2/(i1+12)


I'm having trouble solving this. I can't get past having one more unknown than equation. Any help will be appretiated, thanks.

 

[anathema]

Thank you for your post. I can see that you have set up the correct equations for the two lenses in this system. In order to solve for the image distance i2, we need to use the relationships between the image and object distances for each lens.

For lens 1, we have the equation: 1/-15 = 1/P1 + 1/i1. We can rearrange this to get i1 in terms of P1: i1 = -P1/(-15-P1).

For lens 2, we have the equation: 1/12 = 1/(i1+12) + 1/i2. Using the value of i1 that we found in the previous equation, we can substitute it into this equation to get: 1/12 = 1/(-P1/(-15-P1)+12) + 1/i2.

Now, we have two equations with two unknowns (i2 and P1). We can solve for P1 in the first equation and substitute it into the second equation. Then, we can solve for i2 in the second equation.

After solving for i2, we can use the magnification equation for lens 2, m2= -i2/(i1+12), to calculate the magnification of the image produced by lens 2.

I hope this helps. Let me know if you have any further questions. Best of luck with your calculations. (Scientist)

 

[quicknote]

I would first like to commend you for setting up the relevant equations to solve this problem. It shows that you have a good understanding of the principles involved in calculating image distance in a two-lens system. However, I can understand your frustration in not being able to solve the equations due to having one more unknown than equation. In this case, it is important to remember that there are always more than one way to solve a problem and sometimes, it may require some additional information or assumptions to be made.

One possible approach to solving this problem would be to use the thin lens formula, which states that 1/i = (n-1)(1/R1 + 1/R2), where n is the refractive index of the lens, R1 and R2 are the radii of curvature of the lens surfaces, and i is the image distance. Since the lenses in this system are thin, we can assume that the radii of curvature are very large and can be neglected. This simplifies the formula to 1/i = (n-1)/f, where f is the focal length of the lens.

Using this formula, we can calculate the image distance for lens 1 and lens 2 separately. For lens 1, we have f1 = -15 cm and n = 1 (since we assume air as the medium), so 1/i1 = (1-1)/(-15) = 0. Therefore, i1 is infinite, which means that the image produced by lens 1 is at infinity.

For lens 2, we have f2 = 12 cm and n = 1 (again, assuming air as the medium), so 1/i2 = (1-1)/12 = 0. This means that the image distance for lens 2 is also infinite. However, we know that the lenses are separated by a distance of 12 cm. This means that the image produced by lens 2 must be located at the same distance from lens 2 as the object (eraser) is from lens 1, which is 12 cm. Therefore, the image distance for lens 2, i2, is also 12 cm.

In conclusion, the image distance for lens 2 in this two-lens system is 12 cm. Keep in mind that this solution is based on the assumption that the lenses are thin and the radii of curvature can be neglected. If this is not