A plano-convex lens of radius is placed on an optically fla

Question

A plano-convex lens of radius $1.0\text{ m}$ is placed on an optically flat glass plate and is illuminated by an extended monochromatic source. Assume that the point of contact is perfect. The diameters of the $10^{\text{th}}$ and $5^{\text{th}}$ dark rings in the reflected light are $4.50 \times 10^{-3}\text{ m}$ and $3.36 \times 10^{-3}\text{ m}$ . Next, the space between the lens and the glass plate is filled with a liquid. The diameter of the $5^{\text{th}}$ ring changes to $3.0 \times 10^{-3}\text{ m}$ . Calculate the refractive index of the liquid when the ring is (i) dark, and (ii) bright, if the wavelength of light is $589\text{ nm}$ .

19 Jan 2026

Answer :

Word Count : 644

The phenomenon described is Newton’s rings, where a plano-convex lens on a flat glass plate produces circular interference fringes due to the air (or liquid) film between them. The radius of the (m)-th ring is given by: [ r_m^2 = m \lambda R \quad \text{(for dark rings in reflected light, air gap)} ] where (R) is the radius of curvature of the lens, (\lambda) the wavelength in the medium, and (m) the ring order. --- Step 1: Wavelength in air from dark rings For the dark rings in air: [ r_{10}^2 = 10 \lambda R, \quad r_5^2 = 5 \lambda R ] Given: [ r_{10} = 4.50 \times 10^{-3}, \text{m}, \quad r_5 = 3.36 \times 10^{-3}, \text{m}, \quad R = 1.0, \text{m} ] Check consistency: [ r_{10}^2 / r_5^2 = \frac{(4.50 \times 10^{-3})^2}{(3.36 \times 10^{-3})^2} = \frac{20.25 \times 10^{-6}}{11.29 \times 10^{-6}} \approx 1.794 ] [ 10/5 = 2 ] Close enough for experimental data. --- Step 2: Wavelength in liquid When _________ ________ _______ ________ _______ ______.
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