# Homework Help: Lens next to an aquarium

Tags:
1. Sep 26, 2014

### franceboy

1. The problem statement, all variables and given/known data
There is a small shiny object in a big aquarium, which is formed like a cuboid and filled with water. The plan site of a plano-convex lens with a focal length f is put on the wall of the aquarium from outside. The object is located on the optical axis of the lens.
The refraction index of water is 1.33, the index of the lens is 1.50. The wall of the aquarium and the less are very thin. Consider only rays which are located nearly to the optical axis.
a) Determine the position of the image of the object as function of the position of the object. Name in each case, whether it is a virtual or a real, an upright or a turned, an enlarged or shrinked image.
b) Calculate the image distance and the enlargemet if the object distance has a value of 2.5*f.
c)Do the sam as in b) BUT now the lens is put on the wall from inside (again with plan side).

2. Relevant equations
n1*sin(a)=n2*sin(b)

3. The attempt at a solution
F
irst of all, we imagine that th wall does not exist.
My idea was to consider an object as point (-xo , yo). Now we have to look at the ray which is parallel to the axis: It will refracted at the crossing between air and lens and it will cross the optical lens at the point (f , 0). We can describe this ray after refraction with the function f(x)=yo-yo/f*x.
Moreover we have to consider the ray which crosses the lens in the point (0 , 0).
I failed at determining the function g(x) which is the ray after being refracted. The intersection point of f(x) and g(x) should be the image, right?

2. Sep 26, 2014

### Redbelly98

Staff Emeritus
It looks like you are on the right track to solving this.

For the ray going through the point (0,0), you can calculate the angle of that ray (1) in the water, and then (2) in the glass. And from the angle in the glass, you should be able to (3) calculate the angle of the ray in air.

3. Sep 30, 2014

### franceboy

Using sin a=tan a, I got xi= -f*xo/(f*nwater-xo). With this equation I could solve the tasks a) and b). However I was not able to solve c) because in a) we can say that the parallel ray crosses the axis in (f,0) but now this can not be said. I think the function g(x) has the same equation because the perpendicular is always the optical axis.
How can I solve c)?

4. Oct 2, 2014

### Redbelly98

Staff Emeritus
Are you able to figure out the new focal length of the lens when it is in the water? That is, how does placing the lens in water affect the focal length compared to when it is in air?

If your class has covered this aspect of lenses, that could be a good place to start.

If you class has not covered this, see if your textbook does.

p.s. You're correct about g(x). A thin lens does not appreciably affect the path of a ray going through its center.

5. Oct 3, 2014

### franceboy

My textbook covers only the lensmaker equation. With this equation I can calculate R1. But it does not tell us how to deal with the case of a lens between water and air.
Is my solution for a) right?

6. Oct 8, 2014

### franceboy

Is there anyone who can help me?