# Position of the image of an object placed in water

• lorenz0
In summary, the problem is that when you use the formula for the refraction of a spherical lens, the piece of glass has two surfaces and you need to use the equation twice. The first equation gives the image to the left of the original object, while the second equation gives the image to the right of the original object.

#### lorenz0

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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lorenz0 said:
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.

lorenz0 and Lnewqban
TSny said:
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

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?

rude man said:
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.

lorenz0 said:
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.

lorenz0
lorenz0 said:
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.

lorenz0
Using only Snell's law and small-angle approx. I can corroborate the answer. Do we agree it's a virtual image?

lorenz0
rude man said:
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.

## What is the position of the image of an object placed in water?

The position of the image of an object placed in water depends on the refractive index of water and the distance between the object and the water's surface. When light travels from a less dense medium (air) to a more dense medium (water), it bends towards the normal line. This causes the image to appear closer to the surface of the water than the actual object.

## Why does the image appear closer to the surface of the water?

As mentioned before, light bends towards the normal line when it travels from air to water. This bending of light is known as refraction. The amount of refraction depends on the difference in refractive indices between the two mediums. Since water has a higher refractive index than air, the light bends more when it enters the water, causing the image to appear closer to the surface.

## How does the distance between the object and the water's surface affect the position of the image?

The distance between the object and the water's surface also plays a role in the position of the image. The closer the object is to the water's surface, the more the light bends when it enters the water, causing the image to appear closer to the surface. On the other hand, if the object is placed further away from the water's surface, the light will not bend as much, resulting in the image appearing further away from the surface.

## Does the shape of the object affect the position of the image in water?

Yes, the shape of the object can affect the position of the image in water. Objects with curved surfaces, such as a spoon or a round coin, will have a distorted image in water due to the refraction of light. The image may appear larger or smaller, depending on the curvature of the object's surface.

## Can the position of the image in water be calculated?

Yes, the position of the image in water can be calculated using the laws of refraction. The formula for calculating the position of the image is: 1/f = (n2 - n1)/r, where f is the focal length, n1 is the refractive index of the first medium (air), n2 is the refractive index of the second medium (water), and r is the distance between the object and the water's surface. This formula can be used to determine the position of the image for different objects and distances from the water's surface.