Position of the image of an object placed in water

  • #1
lorenz0
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Homework Statement:
A glass plate of width ##s = 3 cm## with refraction index ##n_2 = 1.5## is in contact with water (##n_1 = 1.33##) on the left and with air on the right. An object is ##p = 10 cm## from the slab. Determine where the image of the object is formed.
Relevant Equations:
##\frac{n_1}{p}+\frac{n_2}{q}=\frac{n_2-n_1}{R}##
I tried using the formula for the refraction of a spherical lens ##\frac{n_1}{p}+\frac{n_2}{q}=\frac{n_2-n_1}{R}## consider each slab as a spherical lens with curvature ##R=\infty## and by doing that I get ##\frac{1.33}{10}+\frac{1.5}{q}=0\Leftrightarrow q\approx -11.3 cm##. Since the piece of glass has a width ##s## I suppose I should count ##11.3 cm## to the left of its rightmost edge, so the object should be ##8.3 cm## to the left of the leftmost edge of the piece of glass i.e. ##1.7 cm## to the right of the original object.

Does this make sense? I had never seen before a problem with "rectangular" lenses and three different materials, so I tried to adapt what I already know to this case. Thanks
 

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Answers and Replies

  • #2
TSny
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I tried using the formula for the refraction of a spherical lens ##\frac{n_1}{p}+\frac{n_2}{q}=\frac{n_2-n_1}{R}##
This is the formula for refraction at a spherical surface. The slab has two surfaces. So, you'll use the formula twice. The image of the first surface serves as the object for the second surface.
 
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  • #3
lorenz0
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This is the formula for refraction at a spherical surface. The slab has two surfaces. So, you'll use the formula twice. The image of the first surface serves as the object for the second surface.
So, if I understand correctly, since from the first application of the equation I get that ##q=-11.3 cm## it means that the first image is formed ##1.3 cm## to the left of the original object, and applying it a second time I get ##\frac{1.5}{11.3+3}+\frac{1}{q}=\frac{1-1.5}{\infty}\Leftrightarrow q\approx -9.53 cm## so the second and final image appears ##9.53 cm## to the left of the rightward surface of the glass which implies, since the glass has a thickness of ##3cm##, that it appears ##3.47 cm## to the right of the original image. Is this correct? Thanks
 
  • #4
rude man
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I think the problem statement needs clarification:
- object is in water or in air?
- object is 10 cm to the left or the right edge of the plate?
 
  • #5
lorenz0
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I think the problem statement needs clarification:
- object is in water or in air?
- object is 10 cm to the left or the right edge of the plate?
Looking at the image describing the problem, the object starts inside the water, ##p=10cm## to the left of the leftmost surface of the glass.
 
  • #6
TSny
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So, if I understand correctly, since from the first application of the equation I get that ##q=-11.3 cm## it means that the first image is formed ##1.3 cm## to the left of the original object, and applying it a second time I get ##\frac{1.5}{11.3+3}+\frac{1}{q}=\frac{1-1.5}{\infty}\Leftrightarrow q\approx -9.53 cm## so the second and final image appears ##9.53 cm## to the left of the rightward surface of the glass which implies, since the glass has a thickness of ##3cm##, that it appears ##3.47 cm## to the right of the original image. Is this correct? Thanks

Your work looks good. It would be better to say that the final image appears ##3.47## cm to the right of the original object, rather than to the right of the original image.
 
  • #7
rude man
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Looking at the image describing the problem, the object starts inside the water, ##p=10cm## to the left of the leftmost surface of the glass.
Check. Thanks.
 
  • #8
rude man
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Using only Snell's law and small-angle approx. I can corroborate the answer. Do we agree it's a virtual image?
 
  • #9
lorenz0
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Using only Snell's law and small-angle approx. I can corroborate the answer. Do we agree it's a virtual image?
Yes, I do agree. Thank you.
 

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