Question

The index of refraction of glass can be increased by diffusing in impurities. It is then possible to make a lens of constant thickness. Given a disk of radius a and thickness d, find the radial variation of the index of refraction n(r) which will produce a lens with focal length F. You may assume a thin lens (d << a).

• A. n(r) = n₀ – $\frac { \mathrm { rF } } { 2 \mathrm {~d} ^ { 2 } }$
• B. n(r) = n₀ –$\frac { \mathrm { rd } } { 2 \mathrm {~F} ^ { 2 } }$
• C. n(r) = n₀ – $\frac { \mathrm { r } ^ { 2 } } { 2 \mathrm { dF } }$
• D. n(r) = n₀ –

Answer 1

Answer: (C)

n(r) = n₀ – $\frac { \mathrm { r } ^ { 2 } } { 2 \mathrm { dF } }$

Answer 2

Let the refractive index of the material of the disk be n and the radial distribution of the refractive index of the impurity-diffused disk be represented by n(r), with n(0) = n₀.

Fig.

Incident plane waves entering the lens refract and converge at the focus F as shown in Fig. We have

[n(r) – n₀]d = –+ F,

i.e., n(r) = n₀ – $\frac { \sqrt { \mathrm { F } ^ { 2 } + \mathrm { r } ^ { 2 } } - \mathrm { F } } { \mathrm { d } }$ .

For F >> r, we obtain

n(r) = n₀ –.