In one line
Filling the gap between the lens and plate with a liquid shrinks the rings by a factor \(\sqrt{\mu}\), so the refractive index is the ratio of \(D_{n+p}^2 - D_n^2\) in air to that in the liquid.
Explained simply
Light travels slower in a liquid, so the optical path in the gap becomes \(\mu\) times larger. The same ring order is reached at a smaller thickness, which happens closer to the centre, so all the rings get smaller. Comparing ring sizes with and without the liquid gives \(\mu\) directly, without needing \(\lambda\) or \(R\).
How it works
- First measure diameters with air: \((D_{n+p}^2 - D_n^2)_{air} = 4p\lambda R\).
- Put a few drops of the liquid between the lens and plate without disturbing the set-up.
- The dark-ring condition becomes \(2\mu t = n\lambda\), so \(D_n^2 = 4n\lambda R/\mu\).
- Measure again: \((D_{n+p}^2 - D_n^2)_{liq} = 4p\lambda R/\mu\).
- Divide: \(\mu = \dfrac{(D_{n+p}^2 - D_n^2)_{air}}{(D_{n+p}^2 - D_n^2)_{liq}}\).
- The liquid's index must be less than that of the glass so the reflection conditions (and the dark centre) stay the same as with air.
Key formula
\(\mu = \dfrac{(D_{n+p}^2 - D_n^2)_{air}}{(D_{n+p}^2 - D_n^2)_{liquid}}\). \(\mu\): refractive index of the liquid (no unit). For the same ring, \(D_{air}/D_{liq} = \sqrt{\mu}\).
Quick example
The 10th dark ring has diameter 0.50 cm in air and 0.435 cm with a liquid. \(\mu = D_{air}^2/D_{liq}^2 = 0.25/0.1892 \approx 1.32\), close to water. Another form: if \(D_{n+p}^2 - D_n^2\) is 0.240 cm2 in air and 0.180 cm2 in liquid, \(\mu = 0.240/0.180 = 1.33\).
Common mistakes
- Inverting the ratio and getting \(\mu < 1\).
- Using diameters instead of squared diameters in the ratio.
- Thinking rings grow bigger with liquid; they shrink.
How MU asks it
Asked as "Explain how Newton's rings can be used to find the refractive index of a liquid" (5 marks) or as a numerical (3 to 5 marks) giving ring diameters in air and liquid. Write the two equations, divide them, and comment that rings contract by \(\sqrt{\mu}\).