Light: Goemetric Optics

Dx!2) A biconvex lens is formed by using a piece of plastic(n=170). the radius of the front surface is 20cm and the radius of the back surface is 30cm. what is the focal length of the lens?In summary, the first problem involves finding the distance an object must be positioned from a 50mm focal length lens in order to form a real image magnified by a factor of 3. The correct equation to use is 1/do + 1/di = 1/f, where do is the distance of the object from the lens and di is the distance of the image from the lens. For the second problem, a biconvex lens with
  • #1
Dx
Hi!

I have 2 problems that I need some help with, please.

1) How far from a 50mm focal length lens, such as used in many 35mm cameras, must an object be positioned if it is so to form a real image magnified by a factor of 3?

how do i solve for this? I have tried to solve using lens equation but not the answer I would expect, my answer is 52mm. Is this correct if not how did you go about solving for this?


2) A biconvex lens is formed by using a piece of plastic(n=170). the radius of the front surface is 20cm and the radius of the back surface is 30cm. what is the focal length of the lens?

I know that r/2 = f but should i add both radius's and use that /2? I need some help?
Thanks!
Dx
:wink:
 
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  • #2
Hi Dx,

1) Sorry I can't reproduce your answer. What equation did you use, and what values did you plug in?
Anyway I think it's a strange question since in most cases a camera image will be much smaller than the object.

2) I think r/2=f can't be the correct equation since the refractive index n doesn't appear in it. Plus I doubt that the value n = 170 is correct. You would rather expect n to be in the range between n=1 and n=2. As for a biconvex lens, you should add the inverses of both focal lengths to get the inverse focal length, i.e. 1/f = 1/f1 + 1/f2.
 
  • #3

1) Sorry I can't reproduce your answer. What equation did you use, and what values did you plug in?
Anyway I think it's a strange question since in most cases a camera image will be much smaller than the object.

My equation used was 1/do + 1/di = 1/f, since my f lenth is 1/50mm i tried to solve for do. Unless i am incorrect i believe that do and f are the same distance, right? that's how i solved for it using substitution. But the factor of 3 forlateral magnification, right?
m=+3, so di=m*do = 20cm. Sorry I don't have all my notes but i am trying to remember all my steps, i substituted the do answer to get 1/do - 1/f to get my answer. I am starting to think I am really confused here. Yes! Its a very strange question i think not enough info i though at first but my teacher sasy there's enough to solve. I am going to keep trying anyways...thanks!


2) I think r/2=f can't be the correct equation since the refractive index n doesn't appear in it. Plus I doubt that the value n = 170 is correct. You would rather expect n to be in the range between n=1 and n=2. As for a biconvex lens, you should add the inverses of both focal lengths to get the inverse focal length, i.e. 1/f = 1/f1 + 1/f2.

I think for the particular problem its refractive index n = 1.7 but your correct its normally not. i remeber reading that to get the f = r/2 but i need the inverses of the f to solve for this your saying, sir? I think the formula i found is it!
1/f = (1.7-=1)(1/20+1/30) = 17cm
is 17cm correct?

I would appreciate anymore of your help, I am still lost?
Thanks!
Dx :wink:
 
  • #4
Originally posted by Dx
My equation used was 1/do + 1/di = 1/f, since my f lenth is 1/50mm i tried to solve for do.
That is IMO correct.
i believe that do and f are the same distance
I think this is wrong. Because if you place the object in the focus, there will be no image. I think you should rather use di = 3*do, since you want a magnification of 3.
i need the inverses of the f to solve for this
Yes you need a proper formula for calculating f1 and f2. The value of n must appear in this formula. Then you use the formula I gave above. Remember this is only correct, if lenses are thin.
 
  • #5
1) How far from a 50mm focal length lens, such as used in many 35mm cameras, must an object be positioned if it is so to form a real image magnified by a factor of 3?

I am guessing that it is convex since a there is no way to get a magnified image with a concave.

m=-di/do

di=-3do

1/f=1/do - 1/3do

1/f=3/3do - 1/3do

1/f=2/3do

2f/3=do

2(50mm)/3=do

33.33mm=do, here is your answer
 

1. What is light?

Light is a form of electromagnetic radiation that is visible to the human eye. It is made up of particles called photons, which have properties of both particles and waves.

2. What is geometric optics?

Geometric optics is a branch of optics that studies the behavior of light as it travels in straight lines, without considering the wave nature of light. It deals with the principles of reflection, refraction, and image formation.

3. How is light reflected?

Light is reflected when it bounces off a surface. The angle of incidence, or the angle at which the light hits the surface, is equal to the angle of reflection, or the angle at which the light bounces off the surface. This is known as the law of reflection.

4. What is refraction?

Refraction is the bending of light as it passes through a different medium, such as air to water or vice versa. This occurs because light travels at different speeds in different mediums.

5. How are images formed by light?

Images are formed when light rays from an object pass through a lens and are refracted, or bent, to form an inverted image on a screen or our retina. The size and position of the image depend on the distance between the object and the lens, as well as the focal length of the lens.

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