Image position and magnification for underwater spherical lens

In summary: 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 given
  • #1
lorenz0
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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.
 

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  • #2
lorenz0 said:
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?
 
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  • #3
haruspex said:
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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  • #4
lorenz0 said:
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.
 
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  • #5
haruspex said:
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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  • #6
lorenz0 said:
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?
 
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  • #8
lorenz0 said:
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.
 

FAQ: Image position and magnification for underwater spherical lens

What is the relationship between image position and object distance for underwater spherical lenses?

The image position for underwater spherical lenses is directly related to the object distance. As the object distance increases, the image position also increases. This means that the image will appear farther away from the lens.

How does the curvature of the lens affect image magnification underwater?

The curvature of the lens plays a significant role in determining the magnification of an image underwater. A more curved lens will result in a larger magnification of the image, while a flatter lens will produce a smaller magnification.

What is the difference between virtual and real images in underwater photography?

A virtual image is one that appears to be behind the lens and cannot be projected onto a screen. In contrast, a real image can be projected onto a screen and is formed by the actual convergence of light rays. In underwater photography, virtual images are more common due to the refraction of light through water.

How does the refractive index of water affect image position and magnification for underwater spherical lenses?

The refractive index of water plays a crucial role in determining the image position and magnification for underwater spherical lenses. As the refractive index of water increases, the image position moves closer to the lens, and the magnification of the image also increases.

Can image position and magnification be controlled in underwater photography?

Yes, image position and magnification can be controlled in underwater photography by adjusting the distance between the lens and the object, as well as the curvature of the lens. Additionally, using different types of lenses or adding external lenses can also affect the image position and magnification.

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