Deriving the lens formula for a convex lens-say

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  • #1
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I will be thankful if the following points are clarified.

1. While deriving the lens formula for a convex lens-say, 1/f= 1/v - 1/u where u and v are the object and image distances, the minus sign is obtained after applying sign conventions. But, I don't understand the logic behind taking the value of u again with a minus sign and substituting in the formula while solving problems.For example, if the object distance is given as 45 cm and the image distance as 90 cm we again apply the sign convention, take u as -45 cm and substitute in the above formula and calculate f as 30 cm. In fact, while doing so the above formula becomes 1/f = 1/v + 1/u. What is the meaning behind applying negative sign in a formula which is itself obtained by applying the negative sign already?

2. Again, take the convex lens formula, 1/f = (n-1) (1/R1 -1/R2) where n is the refractive index and R1 and R2 are radii of the two faces of the lens. For a biconvex lens, assuming that R1 = R2, and n= 1.5 roughly for glass,then substituting these values in the above formula, we get the value of f as infinity, which is really absurd. Can any one throw light on the above points?
 

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  • #2
Redbelly98
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1. The formula most people, and most textbooks, use is
1/f = 1/v + 1/u​
with v and/or u positive for a real image or object, and negative for virtual image or object. For the typical case where object and image are both real, both are positive as is f. In your example this becomes simply
1/30 = 1/90 + 1/45​

2. For a biconvex lens, R2 would be negative as it bulges in the opposite direction as R1 -- this is the common sign convention for surface radii. So in your example,
(1.5-1) (1/R1 - 1/(-R1)) = 0.5 * (1/R1 + 1/R1) = 0.5 * 2/R1 = 1/R1​
If you wanted to change the convention and say R is positive for any convex face, then the formula would have (1/R1)+(1/R2) instead.

There is some more info here:
https://www.physicsforums.com/library.php?do=view_item&itemid=148

Scroll down to the "Extended explanation" section to read about sign conventions.
 
  • #3
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1. The formula most people, and most textbooks, use is
1/f = 1/v + 1/u​
with v and/or u positive for a real image or object, and negative for virtual image or object. For the typical case where object and image are both real, both are positive as is f. In your example this becomes simply
1/30 = 1/90 + 1/45​

2. For a biconvex lens, R2 would be negative as it bulges in the opposite direction as R1 -- this is the common sign convention for surface radii. So in your example,
(1.5-1) (1/R1 - 1/(-R1)) = 0.5 * (1/R1 + 1/R1) = 0.5 * 2/R1 = 1/R1​
If you wanted to change the convention and say R is positive for any convex face, then the formula would have (1/R1)+(1/R2) instead.

Thank you for your answer.
The above formula was obtained by considering the distances measured in the opposite direction of the incident ray as negative.
So,again my basic question remain unanswered. "Why should we apply the sign convention two times?"
If you apply the other convention - that is distances for real object and real images as positive, we don't face any problem.
 
  • #4
Redbelly98
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If I understand you and your example from Post #1, then:

u = -45 cm
v = 90 cm
f = 30 cm​

Is it true that 1/f = 1/v - 1/u?

1/30 = 1/90 - 1/(-45) ?
1/30 = 1/90 - (-1/45) ?
0.0333... = 0.0111... - (-0.0222...) ?
0.0333... = 0.0111... + 0.0222... ?
0.0333... = 0.0333... ?​

Yes, it does.

If I have misunderstood your example, please clarify.
 
  • #5
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If I understand you and your example from Post #1, then:

u = -45 cm
v = 90 cm
f = 30 cm​

Is it true that 1/f = 1/v - 1/u?

1/30 = 1/90 - 1/(-45) ?
1/30 = 1/90 - (-1/45) ?
0.0333... = 0.0111... - (-0.0222...) ?
0.0333... = 0.0111... + 0.0222... ?
0.0333... = 0.0333... ?​

Yes, it does.

If I have misunderstood your example, please clarify.
Thank you Redbelly, for your sincere efforts. My doubt is not about solving the above problem. Any way, if I have confused you, I am sorry.
 
  • #6
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this one had me puzzled too. it appears that the sign convention used in the lens equation has an alternative to the one I learned many years ago.

http://www.practicalphysics.org/go/Guidance_122.html;jsessionid=alZLdQlAHb1?topic_id=2&guidance_id=1 [Broken] for details.

Where it might get fun is that I have no idea how widespread the change is.

====

On a personal note, as much as adding "curvature" seems like a nice approach I don't care for it as it sacrifices the symetry inherent in optics where the light path can be reversed and nothing changes beyond the trivial exchange of u and v.
 
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