How Do You Solve a Combination Lens Problem in Optics?

[Darth Frodo]

Homework Statement

An object is placed 400 mm in front of a convex lens of focal length 80 mm. Find the position of the image formed. State the nature of this image.

A second concave lens of focal length 25 mm is placed 80 mm behind the first convex lens. Find the position of the final image formed and state its nature.

The Attempt at a Solution

$\frac{1}{U} + \frac{1}{V} = \frac{1}{F}$

$\frac{1}{V} = \frac{1}{80} - \frac{1}{400}$

$V = 100 mm$



So the Image from lens one forms 100 mm to the right of lens one. BUT, lens 2 is only 80 mm to the right of lens 1. I don't know what to do here since no image has been formed to create a new object for lens 2.

Any help?

 

[SalsaOnMyTaco]

Which means the Object´s image will appear to the right (or behind) of the concave lense.
80-100= -20
U=-20 solve for V

 

[Darth Frodo]

$\frac{1}{-20} + \frac{1}{V_{2}} = \frac{1}{F}$

$\frac{1}{V_{2}} = \frac{1}{20} - \frac{1}{25} = \frac{1}{100}$

$V = 100$

 

[Darth Frodo]

Is this correct?

 

[Nickie]

I can provide a response to the above content by explaining the concept of combination lenses and how to approach this problem.

Combination lenses are made up of two or more individual lenses placed in close proximity to each other. These lenses work together to form an image of an object. In this case, we have a convex lens and a concave lens placed in close proximity to each other.

To solve this problem, we can use the thin lens equation: 1/U + 1/V = 1/F, where U is the distance of the object from the lens, V is the distance of the image from the lens, and F is the focal length of the lens.

First, we can find the image formed by the convex lens. Placing the values given in the equation, we get 1/V = 1/80 - 1/400, which gives us V = 100 mm. This means that the image formed by the convex lens is 100 mm to the right of the lens.

Next, we need to consider the concave lens. Since the image from the convex lens is already formed 100 mm to the right of the lens, we can consider this as the object for the concave lens. Placing the values in the thin lens equation again, we get 1/V + 1/V' = 1/F, where V is the distance of the object from the concave lens and V' is the distance of the final image from the concave lens. We know that the focal length of the concave lens is -25 mm, so we can solve for V' by substituting the values and rearranging the equation. This gives us V' = -20 mm.

This means that the final image formed by the combination of these two lenses is 20 mm to the left of the concave lens. Since the image is formed on the opposite side of the lens from the object, it is a virtual image. And since the image is smaller than the object, it is also inverted.

In conclusion, the final image formed by the combination of these two lenses is a virtual, inverted image located 20 mm to the left of the concave lens.