Optical System with Diverging Lens and Concave Mirror

In summary, a diverging thin lens and a concave mirror with equal focal lengths are used to determine the final image after two refractions using Gaussian optics. The final image is located between the lens and mirror, at a distance of 21f/34 from the lens. The formula 1/s + 1/s' = 1/f is used, with the standard sign convention that requires the light rays to come from the left. A ray diagram should be drawn carefully to accurately represent the situation.
  • #1
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Homework Statement



A diverging thin lens and a concave mirror have focal lengths of equal magnitude. An object is placed (3/2)f from the diverging lens and the mirror is placed a distance 3f on the other side of the lens. Using Gaussian optics, determine the final image of the system after two refractions. (Pedrotti-3, 2-22)

Homework Equations



$$\frac{1}{s}+\frac{1}{s'}=\frac{1}{f}$$

The Attempt at a Solution



I'm having trouble getting the final answer. After the first refraction, s' = -3f/5 so it is to the left of the lens. Then after the reflection, s' = 18f/13 so it is to the left of the mirror and to the right of the lens. So now in the last refraction, s = 3f - 18f/13 = 21f/13 but it should be negative since it is to the right of the lens. After using the formula (s = -21f/13, f = -f), I get s' =-21/8 so it is to the right of the lens but the answer should be: Final image between lens and mirror at 21f/34 from lens. If s = +21f/13 then it will work out. Why is it positive instead of negative even though the object is to the right of the lens before the second refraction?
 
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  • #2
The second image is real, and it is to the right of the lens. The sign of the object distance is determined with respect to the incoming light. It is a real object for the other side of the lens. The image is virtual, but for that situation, virtual means to the right of the lens.

Edit: Wrong ray diagram.

lensmirror.JPG

Correct one:

lensmirror2.JPG
 
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  • #3
I think there's a double negative in there because the object is to the right and the lens is diverging or something.
Go through it carefully. Image distances to the right of the lens are positive right? A check-list often helps.

Personally I do this by sketching the ray diagram - it fits on a page.
I can never remember the sign conventions.

[edit] ehild beat me to it ... moral: check with ray diagrams.
 
  • #5
Also, there are rules for drawing the ray diagrams when the light rays come from the left to the right. How are you drawing the ray diagram when the light rays are coming from the right?
 
  • #6
Imagine you keep a magnifying glass in your hand. You can illuminate it with a torch either from left or from right can you not? And then you get a real image on the other side - on the right side in the first case and on the left side in the second case. Why do you think that the light must come from the left always??
The same rules apply in both cases. Use "front" and "behind" instead of left and right.

If it is a converging lens, the rays parallel with the axis of the lens go through the focus behind the lens.
In case of a diverging lens, the parallel rays travel behind the lens as if they emerged from the front-side focus.
 
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  • #7
ehild said:
Imagine you keep a magnifying glass in your hand. You can illuminate it with a torch either from left or from right can you not? And then you get a real image on the other side - on the right side in the first case and on the left side in the second case. Why do you think that the light must come from the left always??

It's just because the formula 1/s + 1/s' = 1/f used with the standard sign convention requires that the light rays must come from the left. Also I think your ray diagram is wrong (if you were drawing it to represent the the final image in the original question)
 
  • #8
There are quite a few standard sign conventions for lenses :) Try to think of a real lens and a real situation.
Yes, my ray diagram is wrong. Sorry. Can you tell please, how to correct it? :)
 

What is an optical system with a diverging lens and concave mirror?

An optical system with a diverging lens and concave mirror is a combination of two optical elements that work together to produce a magnified image of an object. The diverging lens spreads out the light rays, while the concave mirror reflects and converges them, resulting in an enlarged image.

How does an optical system with a diverging lens and concave mirror work?

When an object is placed in front of the diverging lens, the lens refracts the light rays, causing them to diverge. The concave mirror then reflects and converges these diverging rays, creating an enlarged virtual image that appears behind the mirror. The final image is formed by the intersection of the reflected rays.

What are the characteristics of the image formed by an optical system with a diverging lens and concave mirror?

The image formed by this type of optical system is virtual, upright, and magnified. It is located behind the concave mirror and its size depends on the positions of the object, lens, and mirror.

What is the difference between a real and virtual image in an optical system with a diverging lens and concave mirror?

A real image is formed when the light rays actually converge and intersect at a point, while a virtual image is formed when the light rays only appear to intersect due to the use of optical elements. In the case of an optical system with a diverging lens and concave mirror, the image formed is virtual.

What are the practical applications of an optical system with a diverging lens and concave mirror?

This type of optical system is commonly used in magnifying devices such as microscopes and telescopes. It is also used in laser systems, optical projectors, and optical sensors for measuring distances. Additionally, it can be used in vision correction for people with nearsightedness.

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