Question about optics and thin lenses

In summary, the conversation discusses the question of whether a lens with a focal length of 18.18 cm can produce an image on a screen that is 35 meters away. It is solved using the equation 1/f - 1/q = 1/p and it is determined that the object must be placed 18.27 cm from the lens to achieve this. The question then arises if the screen is moved further away, such as 300 meters or 500 meters, would the image still be in focus and visible. It is suggested that whether the image is visible depends on the brightness and sensitivity of the eyes or film.
  • #1
rambo5330
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0

Homework Statement


Hey, I was just doing some questions on thin lenses, and thought of a question I really want answered just can't find anything on in the text or anywhere.

So the question involves a lens with a focal length of f = 18.18 cm

now the question asks if the lens would be able to produce an image on a screen that is 35 meters away. Now I can solve this by
1/f - 1/q = 1/p
stating that if you place the object 18.27 cm from the lens you can get an object at 35 meters that will be magnified and inverted, correct ?
now my question is.. say we make the screen 300 meters away
this equation shows that you will still get an image at 300 meters away by placeing the object 18.19 cm away from the lens? however the object will be hugely magnified, is this correct? I am haveing trouble grasping this.. will the image still be in focus at that length ? it just doesn't seem practical.. even moving it to 500 meters i could still get an image.. when does it end? does light interference eventually wash out the image?


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  • #2
Yes, the equation tells you at what distance the image will be in focus.
Or more usually in a camera how far to move the lens from the film to make an image of an object at a distance X

Whether you could see the object of course depends on how bright it is, and how sensitive your eyes or the film are
 
  • #3


I would like to clarify a few things before addressing your question. Firstly, the equation you have mentioned is called the thin lens equation and is used to calculate the position of an image formed by a thin lens. The variables in this equation are as follows:

- f: focal length of the lens
- p: distance of the object from the lens
- q: distance of the image from the lens

Now, coming to your question, it is correct that if the object is placed at a distance of 18.27 cm from the lens, an inverted and magnified image will be formed at a distance of 35 meters from the lens. This image will be in focus as long as the object is at the calculated distance and the lens is able to refract the light rays properly.

Now, when you move the screen further away, for example, at a distance of 300 meters, the image will still be formed at that distance if the object is placed at a distance of 18.19 cm from the lens. However, the image will be highly magnified and may not be in focus. This is because the lens is designed to form an image at a certain distance, and when the screen is moved too far away, it may not be able to properly refract the light rays to form a clear image.

As for your question about light interference, it is not related to the distance of the image from the lens. Light interference occurs when two or more light waves overlap and either amplify or cancel each other out. This can happen regardless of the distance of the image from the lens.

In conclusion, the thin lens equation can be used to calculate the position of an image formed by a thin lens. However, there are practical limitations to how far the image can be formed and still be in focus. It is important to take into consideration the focal length of the lens and the distance at which the object is placed in order to obtain a clear and focused image.
 

1. What is the difference between a convex and a concave lens?

A convex lens is thicker in the middle and thinner at the edges, causing light rays to converge and create a real image. A concave lens is thinner in the middle and thicker at the edges, causing light rays to diverge and create a virtual image.

2. How does a lens refract light?

A lens refracts light by bending the light rays as they pass through the lens, causing them to converge or diverge depending on the shape of the lens. This bending of light is due to the change in speed as the light passes from one medium (air) to another (glass).

3. What is the focal length of a lens?

The focal length of a lens is the distance between the lens and its focal point. It is the point at which parallel light rays converge or appear to converge after passing through the lens.

4. How does the thickness of a lens affect its optical properties?

The thickness of a lens affects its optical properties by changing the amount of bending or refraction of light as it passes through the lens. Thicker lenses tend to have a shorter focal length, causing greater refraction and a larger image. Thinner lenses have a longer focal length and cause less refraction and a smaller image.

5. What is the difference between a positive and negative lens?

A positive lens, also known as a converging lens, has a thicker middle and causes light rays to converge, creating a real image. A negative lens, also known as a diverging lens, has a thinner middle and causes light rays to diverge, creating a virtual image.

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