# 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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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.