Understanding Thin Lens Formula: Image Location & Projection

In summary, the conversation discusses the use of the "thin lens formula" and the placement of images on either side of the lens. It is explained that a negative distance indicates a virtual image, which in this case is located 1.35 m on the patient's side of the lens and cannot be projected onto a screen. The doctor's side of the lens is where a real image would be located, but in this case, the image is virtual and negative. This is why the final calculation is -1.35.
  • #1
luysion
35
0
Hi, in questions involving lenses, when using the "thin lens formulae"
if my di is negative doesn't that mean the image is on the other side of the lens?
in this example however it doesn't seem this holds true..


A doctor examines a patient's skin lesion with a 15 cm focal length magnifying glass. If this lens is held 13.5 cm from the lesion, where is the image and could it be projected onto a screen?

the answer is;
1.35 m on the patient's side of the lens and could not be projected onto a screen.

I get how to get 1.35 but i get -1.35 so shouldn't it mean that it would be on the doctors side?

can somone please explain this to me.

Cheers!
 
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  • #2
luysion said:
A doctor examines a patient's skin lesion with a 15 cm focal length magnifying glass. If this lens is held 13.5 cm from the lesion, where is the image and could it be projected onto a screen?

I get how to get 1.35 but i get -1.35 so shouldn't it mean that it would be on the doctors side?

Hi luysion! :smile:

Distance is positive for real objects and images, and negative for virtual ones.

The light is coming from the patient …

if it was focussing onto a real image, it would be on the doctor's side of the lens (where you could put a screen) …

since it's on the patient's side, it's virtual and negative, which is why you get -1.35. :wink:
 

1. What is the thin lens formula and how is it used?

The thin lens formula is a mathematical equation that describes the relationship between the distance of an object from a lens, the distance of the image from the lens, and the focal length of the lens. It is used to calculate the location and size of an image formed by a thin lens.

2. What is the difference between a real image and a virtual image?

A real image is formed when light rays actually converge at a point and can be projected onto a screen. It is formed by convex lenses and concave mirrors. On the other hand, a virtual image is formed when light rays only appear to converge at a point, but do not actually do so. It cannot be projected onto a screen and is formed by concave lenses and convex mirrors.

3. How does the position of the object and the lens affect the location of the image?

The position of the object and the lens determine the distance of the image from the lens. As the distance of the object from the lens increases, the image distance decreases and vice versa. Additionally, the position of the lens also affects the location of the image. Moving the lens closer to the object will result in a larger image distance, while moving it further away will result in a smaller image distance.

4. What is the significance of the focal length in the thin lens formula?

The focal length is the distance between the lens and the point where parallel light rays converge or appear to converge. It is a crucial component in the thin lens formula as it determines the power of the lens and the properties of the image formed. A shorter focal length results in a more powerful lens and a larger image, while a longer focal length results in a less powerful lens and a smaller image.

5. Can the thin lens formula be used for all types of lenses?

The thin lens formula is only applicable for thin lenses, which are lenses with small thickness compared to their radii of curvature. These include convex and concave lenses, but not thick lenses or lenses with irregular shapes. In these cases, more complex formulas must be used to calculate image location and projection.

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