# Image position and magnification for underwater spherical lens

• lorenz0
It depends on the convention being used. If you use your values, p is positive, meaning the image is to the right of the lens, which is not true.Ok, thanks. (That article is a bit confusing because it crosses over di and do in the diagram.)But it does not give the formula for M that you quote. Since the image is to the left of V, my sketch implies it is upright. I wonder if for the M formula you should be using p=-0.5m. It depends on the convention being used. If you use your values, p is positive, meaning the image is to the right of the lens, which is not true.In summary, using the givenf

#### lorenz0

Homework Statement
A spherical glass lens of radius ##R=20cm## with refraction index ##n_2=1.5## is placed underwater. Find the position of the image placed at ##p=50cm## from the vertex ##V## and the magnification of the image.
Relevant Equations
##\frac{n_1}{p}+\frac{n_2}{1}=\frac{n_2-n_1}{R}##, ##M=\frac{n_1 q}{n_2 p}##
Using the data given and recalling that in this configuration ##R<0## I get: ##\frac{1.33}{0.5}+\frac{1.5}{q}=\frac{1.5-1.33}{-0.2}\Rightarrow q\approx -0.427 m=-42.7 cm## so the image is virtual and is ##42.7\ cm## to the left of vertex ##V##. The magnification is ##M=\frac{n_1 q}{n_2 p}=\frac{1.33\cdot (-0.427)}{1.5\cdot 0.5}\approx -0.757## so the image is shorter (##\approx 75.7\%##) than the original and is upside down.

Is this correct? Thanks.

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• lens.png
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Homework Statement:: A spherical glass lens of radius ##R=20cm## with refraction index ##n_2=1.5## is placed underwater. Find the position of the image placed at ##p=50cm## from the vertex ##V## and the magnification of the image.
Relevant Equations:: ##\frac{n_1}{p}+\frac{n_2}{1}=\frac{n_2-n_1}{R}##, ##M=\frac{n_1 q}{n_2 p}##

Using the data given and recalling that in this configuration ##R<0## I get: ##\frac{1.33}{0.5}+\frac{1.5}{q}=\frac{1.5-1.33}{-0.2}\Rightarrow q\approx -0.427 m=-42.7 cm## so the image is virtual and is ##42.7\ cm## to the left of vertex ##V##. The magnification is ##M=\frac{n_1 q}{n_2 p}=\frac{1.33\cdot (-0.427)}{1.5\cdot 0.5}\approx -0.757## so the image is shorter (##\approx 75.7\%##) than the original and is upside down.

Is this correct? Thanks.
It's always worth doing a sanity check with a sketch.
Draw a small vertical object standing at P and rays from its top.
Where would a ray to V be refracted to, nearer the axis or further away? The tip of the image must lie on that line, perhaps projected back. Does your answer match that?

• lorenz0
It's always worth doing a sanity check with a sketch.
Draw a small vertical object standing at P and rays from its top.
Where would a ray to V be refracted to, nearer the axis or further away? The tip of the image must lie on that line, perhaps projected back. Does your answer match that?
Yes, I think it does. The problem confuses me a bit because I don't know if I can assume that the glass lens will behave like a mirror or not. If it does, looking at a similar configuration here (https://opentextbc.ca/universityphysicsv3openstax/chapter/spherical-mirrors/) I would say yes.

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Yes, I think it does. .
I don‘t think so, but maybe I misunderstand the diagram. It appears to show a concave glass lens of arbitrary thickness. That is not what I would have expected from the text.

• lorenz0
I don‘t think so, but maybe I misunderstand the diagram. It appears to show a concave glass lens of arbitrary thickness. That is not what I would have expected from the text.
What is it that you think I should change in my solution?
I too was confused by the discrepancy between the diagram and the text (the glass is blue and the water is gray...) but in the end I decided to follow the diagram, hence my solution.

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What is it that you think I should change in my solution?
I too was confused by the discrepancy between the diagram and the text (the glass is blue and the water is gray...) but in the end I decided to follow the diagram, hence my solution.
I'm not familiar with the relevant equation you quote. (I assume the 1 should be a q.) Can you provide a link, or at least a clear statement of what all the variables mean and the context in which it applies?

• Delta2
Ok, thanks. (That article is a bit confusing because it crosses over di and do in the diagram.)
But it does not give the formula for M that you quote. Since the image is to the left of V, my sketch implies it is upright. I wonder if for the M formula you should be using p=-0.5m.