Angular Magnification: Deriving M_pn & Proving M_pn = M_infty + 1

In summary, we looked at two ways to use a convex lens: forming an image at infinity and forming an image at the near-point. We derived the angular magnification for each case and showed that M_pn = M_\infty + 1.
  • #1
EmmaK
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Homework Statement


Conside two ways to use a convex lens (1) image formed at infinity s'=-[tex]\infty[/tex] and (2)s' at the near-point s'=-p_n. The angular magnifications are M_[tex]\infty[/tex] and M_pn respectively.
Derive the angular magnification M_pn and show that M_pn = M_[tex]\infty[/tex] + 1


Homework Equations


M_angular = [tex]\stackrel{\theta '}{\theta}[/tex]


The Attempt at a Solution



I found the object distance for (2) to be [tex]\stackrel{fp_n}{f+p_n}[/tex]

M_pn =[tex]\stackrel{y/s}{y/p_n}[/tex] =p_n /s = [tex]\stackrel{f+p_n}{f}[/tex]

M_[tex]\infty[/tex] = [tex]\stackrel{y/f}{y/p_n}[/tex] =[tex]\stackrel{p_n}{f}[/tex]

so M_[tex]\infty[/tex] +1 = [tex]\stackrel{p_n + f}{f}[/tex] =M_pn

so it worked out, but I am unsure whether i put in [tex]\theta[/tex]' and[tex]\theta[/tex] correctly for each case, as i didnt really understand what i was doing.
 
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  • #2


First, let's define the variables in the problem:

- s': image distance
- \theta': angular size of the image
- f: focal length of the convex lens
- p_n: near-point distance (distance from the lens to the near-point of the eye)

Now, let's look at the two ways to use a convex lens:

1. Image formed at infinity (s' = -\infty)
In this case, the image distance is at infinity, so the angular size of the image (\theta') can be approximated as \theta' \approx y/s, where y is the size of the image and s is the object distance. Since s' = -\infty, we can say that s \approx f (based on the thin lens equation). Therefore, \theta' \approx y/f.

The angular magnification in this case is given by M_\infty = \theta'/\theta. Since we are assuming that the object is at infinity, \theta = y/p_n (based on the small angle approximation). Therefore, M_\infty = \theta'/\theta = (y/f)/(y/p_n) = p_n/f.

2. Image formed at the near-point (s' = -p_n)
In this case, the object distance is equal to the near-point distance, so s = p_n. Therefore, we can say that \theta' \approx y/s = y/p_n.

The angular magnification in this case is given by M_pn = \theta'/\theta. Since the object is at the near-point, \theta = y/p_n. Therefore, M_pn = \theta'/\theta = (y/p_n)/(y/p_n) = 1.

We can see that M_pn = M_\infty + 1, as both expressions simplify to p_n/f.

I'm not sure what you were trying to do with the equations you included in your solution, but I hope this explanation helps clarify the problem. Let me know if you have any further questions!
 

1. What is angular magnification and why is it important?

Angular magnification is the ratio of the angle subtended by an image to the angle subtended by an object. It is important because it allows us to determine the size and clarity of an image formed by an optical instrument.

2. How do you derive the formula for angular magnification (M_pn)?

The formula for angular magnification (M_pn) can be derived by considering the thin lens equation (1/f = 1/v - 1/u) and the small angle approximation (tanθ ≈ θ). By substituting these equations and simplifying, we arrive at M_pn = v/u, where v is the image distance and u is the object distance.

3. How can you prove that M_pn = M_infty + 1?

We can prove that M_pn = M_infty + 1 by using the thin lens equation and the limit of the magnification at infinity (M_infty = -f/u). By substituting these equations and simplifying, we arrive at M_pn = v/u = -f/u + 1 = M_infty + 1.

4. What is the significance of M_infty in the formula for angular magnification?

M_infty represents the magnification at infinity, which is the limit of the magnification as the object distance approaches infinity. This value is important because it tells us how much the image will be magnified when the object is placed very far away from the lens.

5. Can angular magnification be negative?

Yes, angular magnification can be negative. A negative value indicates that the image formed by the lens is inverted compared to the object. This can happen when the object is placed between the lens and its focal point.

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