Finding fmax in Thin Lens Refraction Problem

In summary, the conversation involves a student seeking help with a problem from their previous term's problem sheets while studying for an exam. The problem involves finding the maximum value of f in order to have two possible positions for a lens to form an image on a screen. The student has attempted to solve the problem by substituting in s' = L-s and converting it into a quadratic equation, and is seeking confirmation on their solution. The conversation ends with the student thanking the other person for their help.
  • #1
icedragon
9
0

Homework Statement



Hi all. This is a problem from one of the problem sheets from earlier in the term. I'm just revising for my exam tomorrow and can't solve this one. Here it is:

Consider a light source placed at a fixed distance L from a screen, such that a lens of focal length
f can be placed between the source and the screen. Show that as long as f < fmax there are two
positions where the lens can be placed so that an image is formed on the screen, and find a value for fmax.

Homework Equations



1/s + 1/s' = 1/f

The Attempt at a Solution



I substituted in s' = L-s which gives me:

1/s + 1/(L-s) = 1/f

Given that there are two solutions and the nature of that formula I am guessing I need a quadratic equation and then to solve for s. Not sure how to expand that equation so I can solve for s?

Thanks
 
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  • #2
multiply both sides by s(L-s) and you will get a quadratic equation.
 
  • #3
So far so good. f and L are constants, right? Merely solve for s. Yep, you might a quadratic equation with two solutions.
 
  • #4
Thanks for the help. Got the equation in the form (1/f)s^2 - (L/f)s + L = 0 giving the solutions s= (L±sqrt((L/f)^2 - (4L/f))/2

Does that look good? Lecturer hasn't published answers to this problem sheet for some reason. :(
 
  • #5
your quadratic equation loo0ks good. I assume you are using the quadratic formula to solve for s. Just make sure you use the quad formula correctly. Note that when the lens is close to the object, the image is large, and when the lens is close to the screen, the image is small.
 
  • #6
Yep I used the quadratic formula. Thanks for your help. :)
 

1. What is a thin lens refraction problem?

A thin lens refraction problem refers to a situation where light passes through a thin lens and changes direction due to the varying refractive index of the lens. This results in the formation of an image, and the problem involves determining the properties of the image formed, such as its size, location, and orientation.

2. What is the formula for thin lens refraction?

The formula for thin lens refraction is given by the lens maker's equation: 1/f = (n-1)(1/R1 - 1/R2), where f is the focal length of the lens, n is the refractive index of the lens, and R1 and R2 are the radii of curvature of the two lens surfaces.

3. How do you calculate the position of the image formed by a thin lens?

The position of the image formed by a thin lens can be calculated using the thin lens formula: 1/o + 1/i = 1/f, where o is the object distance, i is the image distance, and f is the focal length of the lens.

4. What is the difference between a convex and concave lens in a thin lens refraction problem?

A convex lens is thicker at the center and thinner at the edges, while a concave lens is thinner at the center and thicker at the edges. In a thin lens refraction problem, a convex lens will converge light rays to form a real image, while a concave lens will diverge light rays and form a virtual image.

5. How does the refractive index of a lens affect the image formed in a thin lens refraction problem?

The refractive index of a lens determines how much the light will bend as it passes through the lens, which in turn affects the position, size, and orientation of the image formed. A higher refractive index will result in a more pronounced bending of light and a different image compared to a lens with a lower refractive index.

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