Question

In a Newton’s ring arrangement with a film observed with light of wavelength 6 x 10-5 cm,

the difference of square of diameters of successive rings are 0.125 cm2
. What will happen
to this quantity if:
i) Wavelength of light is changed to 4.5 x 10-5 cm.
ii) A liquid of refractive index 1.33 is introduced between the lens and the plate.
iii) The radius of curvature of convex surface of plano-convex lens is doubled.

Answer 1

In a Newton's ring arrangement, the difference in the square of the diameters of successive rings is equal to four times the wavelength of light divided by the refractive index of the medium between the planoconvex lens and the plane glass plate.

If the wavelength of light is changed to 4.5 x 10-5 cm, then the difference in the square of the diameters of successive rings will be equal to four times 4.5 x 10-5 cm divided by the refractive index of the medium (which is assumed to be 1 if a liquid of refractive index 1.33 is not introduced).

If a liquid of refractive index 1.33 is introduced between the lens and the plate, then the difference in the square of the diameters of successive rings will be equal to four times 4.5 x 10-5 cm divided by 1.33.

If the radius of curvature of the convex surface of the plano-convex lens is doubled, then the difference in the square of the diameters of successive rings will remain the same.