Question

Light with a wavelength of 493 nm in vacuum travels through a transparent substance.
(a) What is the wavelength, in nanometers, of this light in water, with a refractive index of 1.33?
(b) What is the wavelength, in nanometers, of this light in Plexiglass, with a refractive index of 1.48?
(c) What is the wavelength, in nanometers, of this light in flint glass, with a refractive index of 1.77?

Answer 1

We can find the wavelength of light in the medium from the formula:

$$\lambda=\dfrac{\lambda_0}{n},$$

here, $\lambda_0=493\ nm$ is the wavelength of light in vacuum, $\lambda$ is the wavelength of light in the medium and $n$ is the refractive index of medium.

a) For the case of water, we get:

$$\lambda=\dfrac{493\cdot 10^{-9}\ m}{1.33}=371\cdot 10^{-9}\ m=371\ nm.$$

b) For the case of plexiglass, we get:

$$\lambda=\dfrac{493\cdot 10^{-9}\ m}{1.48}=333\cdot 10^{-9}\ m=333\ nm.$$

c) For the case of flint glass, we get:

$$\lambda=\dfrac{493\cdot 10^{-9}\ m}{1.77}=278.5\cdot 10^{-9}\ m=278.5\ nm.$$

Answer:

a) $\lambda=371\ nm.$

b) $\lambda=333\ nm.$

c) $\lambda=278.5\ nm.$