Question

Newtons ring are formed with reflected light of wavelength 5895A°with a liquid between the plane and curved surface .The diameter of 5 th ring is 0.3cm and radius of curvature is 1 m.determine the refractive index of liquid?

Answer 1

Answer on Question #81220 Physics / Optics

Question. Newtons ring are formed with reflected light of wavelength 5895 Å with a liquid between the plane and curved surface. The diameter of 5 th dark ring is 0.3 cm and radius of curvature is 1 m. Determine the refractive index of liquid?

Given. $\lambda = 5895\mathring{\mathrm{A}} = 5895\cdot 10^{-10}m;m = 5;d_m = 0.3cm = 0.3\cdot 10^{-2}m;R = 1m;\beta = 0.$

Find. $n - ?$

Solution.

For the dark ring system:

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

From the figure

$$\boldsymbol {R} ^ {2} = (\boldsymbol {R} - \boldsymbol {d}) ^ {2} + \boldsymbol {r} ^ {2}$$ $$\boldsymbol {R} ^ {2} = \boldsymbol {R} ^ {2} - 2 \boldsymbol {R} \boldsymbol {d} + \boldsymbol {d} ^ {2} + \boldsymbol {r} ^ {2}$$

Because $\pmb{d}$ is very small, we have that

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

Finally

$$n = \frac {4 R m \lambda}{d _ {m} ^ {2}} = \frac {4 \cdot 1 \cdot 5 \cdot 5 8 9 5 \cdot 1 0 ^ {- 1 0}}{(0 . 3 \cdot 1 0 ^ {- 2}) ^ {2}} = 1. 3 1$$

Answer. $n = 1.31$ .

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