Calculate the position of the image

[Pseudo Statistic]

A question:
"Suppose that a converging lens with focal length 30 cm in water is placed 20cm above a light source at the bottom of the pool. An image of the light source is formed by the lens.
a) Calculate the position of the image with respect to the bottom of the pool."
For a I did 1/f = 1/p + 1/q But got -60cm:
1/30 = 1/20 + 1/q
That doesn't sound right to me.. can anyone tell me what I'm doing wrong?

"b) If instead of water the pool were filled with a material with a different index of refraction, describe the effect, if any, on the image and its position in each of the following cases.
i) The index of refraction of the material is equal to that of the lens."
I'm going to guess there is no image?

"ii) The index of refraction of the material is greater than that of water but less than that of the lens."
No clue..

Can anyone clarify? Thank you.

 

[Doc Al]

A question:
"Suppose that a converging lens with focal length 30 cm in water is placed 20cm above a light source at the bottom of the pool. An image of the light source is formed by the lens.
a) Calculate the position of the image with respect to the bottom of the pool."
For a I did 1/f = 1/p + 1/q But got -60cm:
1/30 = 1/20 + 1/q
That doesn't sound right to me.. can anyone tell me what I'm doing wrong?

What doesn't sound right? (What's the position of the image with respect to the pool bottom?)

The focal length of a lens depends on the difference between the index of refraction of the lens and the index of refraction of the medium the lens is submerged in. (This should make intuitive sense, since without a difference in index of refraction, there would be no refraction and no lens.) For details, study the Lens Maker's formula. (Look here: http://hyperphysics.phy-astr.gsu.edu/Hbase/geoopt/lenmak.html#c1)

 

[kyle8921]

I can provide a response to the question and clarify any confusion.

Firstly, to calculate the position of the image in this scenario, we can use the thin lens equation: 1/f = 1/p + 1/q, where f is the focal length of the lens, p is the distance from the object to the lens, and q is the distance from the lens to the image.

In this case, the focal length of the lens is given as 30 cm and the distance from the object (light source) to the lens is 20 cm. Plugging these values into the equation, we get: 1/30 = 1/20 + 1/q. Solving for q, we get a value of -60 cm.

However, a negative value for q indicates that the image is formed on the same side of the lens as the object, which is not possible in this scenario. This suggests that there may be an error in the calculations or interpretation of the problem.

Moving on to part b) of the question, let's consider the effects of different indices of refraction on the image and its position in the pool.

i) If the index of refraction of the material is equal to that of the lens, there will still be an image formed by the lens. However, the position of the image may change depending on the thickness of the material and its distance from the lens.

ii) If the index of refraction of the material is greater than that of water but less than that of the lens, there will still be an image formed by the lens. However, the image may appear distorted due to the difference in refractive indices of the materials. The position of the image may also change depending on the thickness of the material and its distance from the lens.

In both cases, it is important to consider the effects of refraction and the properties of the material on the position and characteristics of the image formed by the lens. I hope this clarifies any confusion and helps you understand the problem better.