# Measuring refractive index of liquid using lens and mirror

1. Sep 25, 2006

### blackcat

hi

i've got to do measure hte refractive index of a liquid with mirror and lens

first of all how can i measure hte refractive index of air in this way? i dont get how i can do this, please explain :|

and then for the liquid, ive been told to put the liquid over the mirror and then the lens on top of the liquid, and then hold up a pin until it coincides with something, but i dont understand that.

2. Sep 25, 2006

### blackcat

3. Sep 25, 2006

### Chi Meson

The refractive index of air is 1.0003 . Unless you have very good measuring devices, I don't immediately see how you are to measure that. Usually, it is assumed to be 1.000 unless you need extreme precision.

Are you sure you have conveyed the requirements correctly? ARe you perhaps supposed to measure the focal length of the lens in air?

4. Sep 26, 2006

### blackcat

Yeah,

Apparently i have to hold a pin above the lens and find out where the image coincides, I dont know what that means.

5. Sep 26, 2006

### andrevdh

Hold the pin held horizontally above the lens. The lens should be resting on the mirror. Move the pin vertically untill its image coincides with the real pin (you see both in focus simultaneously). This gives you the focal length of the convex lens, $f_1$, the distance from the pin to the lens. Repeat the process with the lens resting in the liquid on the mirror. The liquid lens is then a plano-convex lens with the same radius of curvature as the convex lens, $r$. This gives you their combined focal length, $F$.

You can now calculate the focal length of the liquid lens from

$$\frac{1}{F} = \frac{1}{f_1} + \frac{1}{f_2}$$

(it should be negative). Using the lens maker's equation for the liquid lens we next get that

$$n_{liq} = 1 + \frac{r}{f_2}$$

for the refractive index of the liquid. In this formula r is also negative for the liquid lens. If you know the refractive index of the convex lens it can be shown that

$$n_{liq} = 1 - 2(n_{lens} - 1)\frac{f_1}{f_2}$$

Last edited: Sep 26, 2006
6. Sep 28, 2006

### blackcat

Thanks for your reply. But I have no idea what you mean by "its image coincides with the real pin (you see both in focus simultaneously)", I'm not even sure where to hold the pin with respect to the lens horizontally.

I suppose I can ask my teacher.

Thanks for your help with the equations.

7. Sep 28, 2006

### andrevdh

An convex lens will form a real image of the pin if the object distance is equal to or further than the focal length of the lens. Such image can be viewed with the eye if you position yourself looking up against the rays, that is in this case looking down onto the mirror. If the image do not coincide with the real pin your eye will only be able to focus on the one or the other (image or object). When the real pin and its image is simultaneously in focus we know that they are both in the same plane. This happens when the real pin (and the image) is at the focus of the lens.

So what you need to do is move the pin vertically up and down until you see the image and the pin both in focus simultaneously.

Allow me to correct myself. What you should be looking for is relative motion between the real pin and its image when you move your head slightly in a horizontal direction. If the two are not in the same plane you will observe that the pin and image seem to move separately when you move your head. When they are in the same plane you will observe that they move in unison.

So while you move the pin up and down shake your head slightly (saying no) and observe the relative motion of the pin and its image.

Last edited: Sep 28, 2006
8. Sep 28, 2006

### andrevdh

Slight correction to previous post.

9. Oct 23, 2006

### blackcat

Thanks For The Reply And Sorry For Not Posting This Before!

10. Oct 24, 2006

### andrevdh

I will try and post a photograph of such a setup soon. How did it go with the experiment?

11. Oct 24, 2006

### blackcat

It went well (I think). However I haven't worked out the refractive index yet cuz I need to figure out the radious of hte curvature of the lens (the equation needs it).

Am I right in saying that the radius of the curvature of the lens (I read it in a book) is v in:
1/v = -1/u + 1/f

Again thanks for you all your help.

12. Oct 24, 2006

### blackcat

I've realised that v can't = radius of curvature cause if you change u then v will change.............

Is there any way to measure the radius of curvature of the lens without being in a lab? I have a mirror, lens and pin...

Also, what would you say is a typical value for the radius of curvature of a convex lens with focal length 10cm and diameter 50mm?

Thanks.

13. Oct 24, 2006

### OlderDan

There is a relationship between the focal length of the lens, its index of refraction, and the radii of curvature of its surfaces. I think it was already posted earlier.

http://hyperphysics.phy-astr.gsu.edu/Hbase/geoopt/lenmak.html

Come to think of it, you could measure the radius of curvature of the lens if you can see reflections in the lens surface. Have you done the experiment to find the focal length of a convex mirror? The focal length of a spherical mirror is half the radius of curvature.

http://hyperphysics.phy-astr.gsu.edu/Hbase/hframe.html

Last edited: Oct 24, 2006
14. Oct 24, 2006

### blackcat

Yeah, but I have to find using an alternative method that doesn't involve refractive index and/or lens maker's equation etc. And I havent done that experiment :(

There has to be another way.

15. Oct 24, 2006

### OlderDan

There is another way, but it may be difficult to make the measurements. There is a device called a spherometer for measureing the curvature of a surface, but you probably do not have one of those laying around. You could accomplish what it does if you could make very accurate thickness measurements with a micrometer or vernier caliper, and if your lens is symmetrical (same radii on both sides). A very accurate measurement of the thickness of the lens in the middle, and at a known distance from the center (25mm) can be used to find the radius of curvature. Most thin lenses are not knife sharp on the edge, so you might want to move in a bit and make an actual measurement just inside the radius. Is this a possibility?

16. Oct 25, 2006

### blackcat

I'll take a look if my brother has one. I found an article in Wikipedia about it: http://en.wikipedia.org/wiki/Spherometer

My lens is symettrical (it looks it), so I guess if I find one of those it should work properly?

17. Oct 25, 2006

### blackcat

Do you think -0.07 is a sensible figure for the radius of curvature?

Last edited: Oct 25, 2006
18. Oct 25, 2006

### OlderDan

I know you are not supposed to use the lens maker's formula to get the answer, but the formula says that for an index of refraction of 1.5 (about the same as glass), with equal radii, the focal length is the radius. You should expect the radius to be about the same as the focal length.

19. Oct 25, 2006

### blackcat

I see. I suppose 0.07 is close enough to the focal length (0.104)

Thanks for the help.

(one more q: is it right the radius is negative?)

20. Oct 25, 2006

### OlderDan

The sign conventions can be tricky, and there are a couple of different ones used in optics. In the link to the lens-maker's equation I posted earlier, one of the two radii would be positive and the other would be negative becuse the centers are on opposite sides of the lens. It is very important to keep these signs straight for computations, but not for describing the lens. If I say I have a plano-convex lens with a radius of curvature of 10 cm, there is no ambiguity about the shape of the lens. I don't need the sign to describe the lens because I have the words that define its shape.

If you know the focal length of your lens is 10cm, then the 7cm might not be good enough for what you are doing. Why do you think you are not allowed to use the lens-maker's formula to get the radius from the focal length? Aren't you going to use it to calculate the index of refraction of the liquid? According to what you said earlier, you were not given a device like a spherometer to measure the radius. You have to deduce it from some other observation.