What Is the Refractive Index of a Liquid Based on Newton’s Colobars?

[EPhoenix]

Homework Statement

Well it's problem to translate it but I will try...

We measured Newtons colobars, the radius of third dark colobar in reflective light is r3 (without any data). When we filled the space between lens and board, that radius (r3) had 4th dark colobar... determine the refractive index of liquid!

colobar = http://i46.tinypic.com/bfkqxf.png"

Homework Equations

I found this on internet...

Δ=2d+(λ/2)
d= sqr(r)/R

light
kλ=2d + (λ/2)

dark

(k + 0.5)λ= 2d + (λ/2)

The Attempt at a Solution

i don't have a clue really... I don't even know how to translate it...

It's not for me, it's for my brother (high school)... He asked me to try to find solution... But it's not in my field (i'm medical college student)... so...

CAN ANYONE HELP ME...

 

[mark726]

Hello,

I understand that you are trying to determine the refractive index of a liquid based on measurements of Newton's colobars. This can be a challenging problem, but I will try to guide you through the steps.

First, let's define some terms. Newton's colobars refer to the interference pattern created by light passing through a thin film, which in this case is the liquid between the lens and the board. The dark and light colobars refer to the areas of constructive and destructive interference, respectively.

To find the refractive index of the liquid, we need to use the equation Δ=2d+(λ/2), where Δ is the thickness of the liquid, d is the distance between the lens and the board, and λ is the wavelength of the light used. We also have the equation d= sqr(r)/R, where r is the radius of the third dark colobar and R is the radius of the curvature of the lens.

Now, when we fill the space between the lens and the board, the radius of the third dark colobar changes to become the fourth dark colobar. This means that the distance between the lens and the board has increased by one wavelength of light. So we can set up the equation kλ=2d + (λ/2), where k is the number of wavelengths that fit in this new distance.

Combining these equations, we can solve for the refractive index, n, using the equation n=λ/(2d + (λ/2)).

I hope this helps you and your brother in solving the problem. Let me know if you have any further questions. Good luck!