Calculating Focal Length of a Drop of Water

[thereddevils]

Homework Statement

What is the focal length of a drop of water in the form of a sphere with radius 4mm given that the refractive index of water = 4/3

Homework Equations

The Attempt at a Solution

Here, i think i will need to assume that the light rays are coming from infinity, and using this formula

$$\frac{n_1}{u}+\frac{n_2}{v}=\frac{n_2-n_1}{r}$$

n1/u is 0. Substituting n2=4/3 and n1=1,r=4, i got v=16 and for distant light rays, image distance equals the focal length.

But i am wrong, the answer given is 6mm.

 

[cepheid]

It is possible that the formula you used is not applicable to this situation. When I use the lensmaker's equation (which seems like a natural thing to use), I get an answer that is close but not quite equal to the answer you have been given.

http://en.wikipedia.org/wiki/Lens_(optics)#Lensmaker.27s_equation

If I use the thin lens equation (which is just below the lensmaker's equation in that article), I get the answer that you've been given in your solutions. HOWEVER, I do not think that the thin lens approximation applies here, so really you should be using the lensmaker's equation. It's possible that whoever made your solutions incorrectly applied the thin lens equation to this situation.

EDIT: In applying these formulae, keep in mind that the radius of curvature of the far surface should be negative.

 

[NobodySpecial]

A sphere is known as a ball lens in optics - it;s very useful for coupling light into fibres for example. http://www.edmundoptics.com/technical-support/optics/understanding-ball-lenses/

See http://spie.org/x34513.xml for a derivation

 

[thereddevils]

It is possible that the formula you used is not applicable to this situation. When I use the lensmaker's equation (which seems like a natural thing to use), I get an answer that is close but not quite equal to the answer you have been given.

http://en.wikipedia.org/wiki/Lens_(optics)#Lensmaker.27s_equation

If I use the thin lens equation (which is just below the lensmaker's equation in that article), I get the answer that you've been given in your solutions. HOWEVER, I do not think that the thin lens approximation applies here, so really you should be using the lensmaker's equation. It's possible that whoever made your solutions incorrectly applied the thin lens equation to this situation.

EDIT: In applying these formulae, keep in mind that the radius of curvature of the far surface should be negative.

Thank you.

 

[rwisz]

Your attempt at a solution is correct, but you have made a mistake in your calculation. Let's go through it step by step to find where the error occurred.

First, we need to understand the equation you have used. It is known as the lens maker's equation, which relates the focal length (f), object distance (u), and image distance (v) of a lens or spherical object.

The equation is given as:

\frac{n_1}{u}+\frac{n_2}{v}=\frac{n_2-n_1}{r}

Where n1 and n2 are the refractive indices of the two mediums (in this case, air and water), u is the object distance, v is the image distance, and r is the radius of curvature of the spherical object.

In this case, we are given the radius of the drop of water (r = 4mm) and the refractive index of water (n2 = 4/3). We need to find the focal length (f) of the drop of water.

Using the lens maker's equation, we can rearrange it to solve for f:

f = \frac{1}{\frac{n_2-n_1}{r} - \frac{n_1}{u}}

Substituting the values we have, we get:

f = \frac{1}{\frac{4/3 - 1}{4} - \frac{1}{\infty}}

Since light rays are coming from infinity, u = \infty, which means that \frac{1}{u} = 0.

Substituting this in the equation, we get:

f = \frac{1}{\frac{4/3 - 1}{4} - 0} = \frac{1}{\frac{1/3}{4}} = \frac{1}{\frac{1}{12}} = 12mm

So, the focal length of the drop of water is 12mm, not 16mm as you have calculated. But the given answer is 6mm. Where did we go wrong?

The mistake lies in the assumption that the light rays are coming from infinity. In reality, light rays coming from a distant source will be nearly parallel, but not exactly parallel. This means that the object distance (u) is not infinite, but a finite value. Let's say the light rays are coming from a distance of