Question

Can someone explain in detail the solution to this problem as well as the equations involved in how to plug in the figures in these equations please.

An object is located 30 cm left of a convergent lens of focal length 10 cm. A divergent lens of focal length 20 cm is located 5.0 cm to the right. Fully describe all images and draw a ray diagram for the first image only 130 + 1/1 = 1/10, so i = +15 cm, m =-15/30 =-05. This is a real, half-sized, inverted image located 15 cm to the right of the convergent lens, 10 cm to the right of the divergent lens. Since the rays are intercepted by the divergent lens before the image is actually formed, this represents a virtual object for the divergent lens. For the second image .. I/-10 + 1/is 1/-20, so i = +20 cm,

Answer 1

According to the problem,

The key to this kind of problem is that the 1st (left) lens's (intermediate) image is the 2nd (right) lens's object. Then using notation from (+uN is object distance to left of lens N and +vN is image distance to right of lens N),

we have initially: u1 = 0.30 m; f1 = 0.10 m; sep (separation) = 0.05 m; f2 = 0.20 m

Solving for v2,

v1 = 1/(1/f1-1/u1) = 0.15 m

u2 = sep - v1 = - 0.1 m

v2 = 1/(1/f2-1/u2) = 0.067 m,

that is, 6.7 cm to the left of the second lens. It's a virtual image,

and magnification is v1v2/(u1u2) = -0.335.

The ray diagram is below,

I think you understand my explanation, and hope you got your solution and glad I can help .Rate me good if you like the solution.

Thank you .