(a) Monochromatic light of wavelength 589 nm is incident from air on a water surface. If 𝜇 for water is 1·33, find the wavelength, frequency and speed of the refracted light.
(b) A double convex lens is made of a glass of refractive index 1·55, with both faces of the same radius of curvature. Find the radius of curvature required, if the focal length is 20 cm.

(a) Given, wavelength $\mathrm{\lambda }=589\mathrm{nm}=589\times 10-9\mathrm{m}$ 
Speed of light in air, $\mathrm{c}=3\times {10}^8\mathrm{m}/\mathrm{s}$
Refractive index of water, μ = 1.33
The frequency of light does not depend on the property of the medium in which it is travelling. Hence, the frequency of the refracted ray in water will be equal to the frequency of the incident or reflected light in air.
So Refracted frequency, $\mathrm{v}=\frac{\mathrm{c}}{\mathrm{\lambda }}=5.09\times {10}^{14}\mathrm{Hz}$
Speed of light in water is related to the refractive index of water as:
$\mathrm{v}=\frac{\mathrm{c}}{\mathrm{\mu }}=2.26\times {10}^8\mathrm{m}/\mathrm{s}$
Wavelength of light in water is given by the relation
${\mathrm{\lambda }}_1=\frac{\mathrm{v}}{\mathrm{v}}=444\times {10}^{-9}\mathrm{m}=444\mathrm{nm}$
(b) Given: μ = 1.55,
f = 20cm 
We know that,
$\begin{array}{l}\frac{1}{\mathrm{f}}=(\mathrm{\mu }-1)\left(\frac{1}{{\mathrm{R}}_1}-\frac{1}{{\mathrm{R}}_2}\right)\\ \frac{1}{20}=(1.55-1)\left(\frac{1}{\mathrm{R}}+\frac{1}{\mathrm{R}}\right)\\ \frac{1}{20}=0.55\times \frac{2}{\mathrm{R}}\\ \mathrm{R}=20\times 0.55\times 2\\ \mathrm{R}=22\mathrm{cm}\end{array}$