Image position and magnification for underwater spherical lens

[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.

 

[haruspex]

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.

 

[haruspex]

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.

 

[haruspex]

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?

 

[lorenz0]

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?

https://phys.libretexts.org/Bookshe...n#Equation+20:+Refraction+at+a+convex+surface

 

[haruspex]

https://phys.libretexts.org/Bookshelves/University_Physics/Book:_University_Physics_(OpenStax)/University_Physics_III_-_Optics_and_Modern_Physics_(OpenStax)/02:_Geometric_Optics_and_Image_Formation/2.04:_Images_Formed_by_Refraction#Equation+20:+Refraction+at+a+convex+surface

Ok, thanks. (That article is a bit confusing because it crosses over d_i and d_o 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.