Question

Newton's rings are formed in reflected light
of wavelength 6000A with a liquid between
the plane and curved surfaces. The diameter
of 7thdark ring is 0.34 cm and the radius of
curvature of curved surface is 100 cm.
Calculate the refractive index of liquid.

Answer 1

Answer on Question #77790, Physics / Optics

Question. Newton's rings are formed in reflected light of wavelength $6000\,\mathrm{\AA}$ with a liquid between the plane and curved surfaces. The diameter of 7th dark ring is $0.34\,\mathrm{cm}$ and the radius of curvature of curved surface is $100\,\mathrm{cm}$. Calculate the refractive index of liquid.

Given. $\lambda = 6000\,\mathrm{\AA} = 6000 \cdot 10^{-10}\,\mathrm{m};\ m = 7;\ d_m = 0.34\,\mathrm{cm} = 0.34 \cdot 10^{-2}\,\mathrm{m};\ R = 100\,\mathrm{cm} = 1\,\mathrm{m};\ \beta = 0$.

Find. $n - ?$

Solution.

For the dark ring system:

$$2dn \cos \beta + \frac{\lambda}{2} = (2m + 1)\lambda \rightarrow 2dn \cos \beta = m\lambda \rightarrow 2dn = m\lambda \rightarrow n = \frac{m\lambda}{2d}$$

From the figure

$$R^2 = (R - d)^2 + r^2$$ $$R^2 = R^2 - 2Rd + d^2 + r^2$$

Because $d$ is very small, we have that

$$d = \frac{r^2}{2R}$$

Finally

$$n = \frac{4Rm\lambda}{d_m^2} = \frac{4 \cdot 1 \cdot 7 \cdot 6000 \cdot 10^{-10}}{(0.34 \cdot 10^{-2})^2} = 1.45$$

Answer. $n = 1.45$.

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