Refraction - Find how far behind a spherical mirror the fish's reflection appears

  • Thread starter JoeyBob
  • Start date
  • #1
JoeyBob
256
29
Homework Statement:
see atached
Relevant Equations:
1/p+1/q=-2/R
So first I looked at where the image of the fish appeared to be when it went through the water surface.

since we can assume the water is flat, R is infinity, so n1/p=-n2/q. plugging in the values (n1=1.29, n2=1, p=3.5) I get q=-0.3686. So the image of the fish appears at 0.369 above the surface or 3.13 before the mirror.

I make this the new q and use the eqn 1/q+1/p=-2/R. calculating for q I get -0.642 or 0.642 behind the mirror.

but the answer is supposed to be 1.64?
 

Attachments

  • question.PNG
    question.PNG
    9 KB · Views: 39

Answers and Replies

  • #2
haruspex
Science Advisor
Homework Helper
Insights Author
Gold Member
2022 Award
39,240
8,545
first I looked at where the image of the fish appeared to be when it went through the water surface.
The light rays would reflect in the mirror then pass through the water surface, so that is not the natural order in which to solve it. Not sure if it gives the right answer.
so n1/p=-n2/q
The fish and its image would both be below the water surface, so not sure how you get the minus sign.
I get q=-0.3686
You do? The ratio of the distances should be as the ratio of the indexes.
 
  • #3
JoeyBob
256
29
The light rays would reflect in the mirror then pass through the water surface, so that is not the natural order in which to solve it. Not sure if it gives the right answer.
i think this is where I am confused. The light is coming from the fish. but then it needs to come back to the fish for the fish to see itself.

First light comes off fish and hits water. Then light travels through air and hits mirror. Then light goes back through air and hits water. Then goes through water and hits fish for fish to see?

This seems overly complicated. First I would calculate the position of the image after it hit the water/air interphase. Then after it hit the mirror. then after it hit the interphase again?

The sad thing is that i got this question right on the quiz I am looking at, but now I can't remember how I did it...

As for how I get n1/p=-n2/q, its from n1/p+n2/q=(n2-n1)/R

Since R is infinity (since we assume the interphase is flat), n1/p+n2/q=0. One becomes negative if we put them on the other side.
 
  • #4
guv
74
14
When I read the question, it seems to suggest the fish is between the air-water interface and a spherical metallic mirror below. What you wrote suggests the mirror is above the air-water interface which itself is above the fish.
 
  • #5
JoeyBob
256
29
When I read the question, it seems to suggest the fish is between the air-water interface and a spherical metallic mirror below. What you wrote suggests the mirror is above the air-water interface which itself is above the fish.
The way I read "3.5m above" was above the water since why else would refractive indexes even be given?

If we assume its 3.5 above the fish, then 1/q+1/p=-2/R, where p=3.5m and R=6.2m

This does indeed give -1.644 m or 1.644 m behind the mirror. What a dumb question.

Who knows what the right answer would be if the sphere was above the water. Way more complicated then...
 
  • #6
haruspex
Science Advisor
Homework Helper
Insights Author
Gold Member
2022 Award
39,240
8,545
but then it needs to come back to the fish for the fish to see itself.
At first I read the question as concerning the image seen by the fish, but in that case neither the depth of water above the fish nor the refractive index are relevant. Of course, it could be a trick question, but I suggest rather that we are to consider the image as seen by an observer above the water.

Edit: my initial reading was correct.
@JoeyBob, is there a further part to the question, where the depth of water becomes important?
 
Last edited:
  • #7
JoeyBob
256
29
At first I read the question as concerning the image seen by the fish, but in that case neither the depth of water above the fish nor the refractive index are relevant. Of course, it could be a trick question, but I suggest rather that we are to consider the image as seen by an observer above the water.

So let's pretend there was a person right above the water looking down and I wanted to find how deep the reflection of the fish appeared to be to the person.

Is that when I would use n1/p+n2/q=(n2-n1)/R where n2 is 1.29 and p=3.5?
 
  • #8
haruspex
Science Advisor
Homework Helper
Insights Author
Gold Member
2022 Award
39,240
8,545
  • #9
guv
74
14
So let's pretend there was a person right above the water looking down and I wanted to find how deep the reflection of the fish appeared to be to the person.

Is that when I would use n1/p+n2/q=(n2-n1)/R where n2 is 1.29 and p=3.5?
The use of spherical lens equation should be correct where you let R approach infinity. To me this problem is worded poorly. Based on you said, it seems to be asking for the position of the secondary image (from a first virtual image under water) formed in the mirror above water. You probably need to combine the spherical lens equation + mirror equation.
 
  • #10
JoeyBob
256
29
The use of spherical lens equation should be correct where you let R approach infinity. To me this problem is worded poorly. Based on you said, it seems to be asking for the position of the secondary image (from a first virtual image under water) formed in the mirror above water. You probably need to combine the spherical lens equation + mirror equation.
You know what, after thinking about it and reading it a few more times I think I know what it means. The fish 3.5 m above the mirror. The fish is also 3.5 m deep. So the mirror is actually 7m deep. Very poorly worded question.

Since the mirror is underneath, I can find the location of the reflection the fish sees from 1/p+1/q=-2/R.

Then if I wanted to find where a person right above the water saw the reflection, I would use the eqn haruspex linked, n(water)=D(real)/D(apparent) where I am looking for D(apparent) and the D(real) is 7m+(how far behind the mirror the image is)
 
  • #11
Steve4Physics
Homework Helper
Gold Member
2022 Award
1,606
1,454
I believe the question deliberately contains ‘distractors’ – data not required to solve the problem. This is to test/exercise your understanding.

The fish looks directly at the sphere and sees its reflection – a virtual image behind the sphere.
There is no refraction as the rays are always in water. So the refractive index is not needed.
The amount of water above the fish makes no difference.
Only the fish-to-sphere distance and the sphere’s radius are required.

The ray diagram (turned on its side) is just the same as for air, like this:
https://buphy.bu.edu/~duffy/PY106/22b.GIF
so you can simply use your formula “1/p+1/q=-2/R“.

Presumably the fish is a red herring.
 
  • #12
haruspex
Science Advisor
Homework Helper
Insights Author
Gold Member
2022 Award
39,240
8,545
I believe the question deliberately contains ‘distractors’ – data not required to solve the problem. This is to test/exercise your understanding.

The fish looks directly at the sphere and sees its reflection – a virtual image behind the sphere.
There is no refraction as the rays are always in water. So the refractive index is not needed.
The amount of water above the fish makes no difference.
Only the fish-to-sphere distance and the sphere’s radius are required.

The ray diagram (turned on its side) is just the same as for air, like this:
https://buphy.bu.edu/~duffy/PY106/22b.GIF
so you can simply use your formula “1/p+1/q=-2/R“.

Presumably the fish is a red herring.
I just reread the question and you are right... it does say "to the fish". I missed that before.
 

Suggested for: Refraction - Find how far behind a spherical mirror the fish's reflection appears

Replies
5
Views
524
Replies
6
Views
534
  • Last Post
Replies
1
Views
291
Replies
56
Views
972
Replies
43
Views
1K
Replies
2
Views
368
Replies
2
Views
410
  • Last Post
2
Replies
67
Views
2K
Replies
3
Views
368
Replies
7
Views
617
Top