A plano-convex glass lens of radius of curvature R = 1.5 m rests on an optically

Question

A plano-convex glass lens of radius of curvature R = 1.5 m rests on an optically flat glass plate. The arrangement is illuminated from above with monochromatic light of 513nm wavelength. The indexes of refraction of the lens and plate are 1.6. The condition for a bright (constructive) interference ring is given below for a thickness t.

t = ( m + 1/2 )( /2 ), m = 0, 1, 2, ...

For t << R, the radius of a fringe is related to t as follows.

Determine the radii of the first and second bright fringe in the reflected light.
......mm (first bright fringe)
........mm (second bright fringe)

Explanation / Answer

here,

Radius of curvature, R = 1.5 m
Wavelength, w = 513*10^-9 m
Refractive index, n = 1.6
index of air, na = 1

Thickness, t = (m + 0.5)*w/2
also radius, r = sqrt(2*R*t)

by equating and substituting, we get,

r^2 = (m+0.5)*w*R

The first fringe corresponds to, m = 0
r^2 = (0 + 0.5)*(513*10^-9)*1.5
r = 6.203*10^-4 m or 0.6203 mm ( 1m = 1000 mm)

The Second fringe corresponds to, m = 1
r^2 = (1 + 0.5)*(513*10^-9)*1.5
r = 1.074*10^-3 m or 1.074 mm

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