Question

The wavelength of sodium light in air is 589 nm. (a) Find its frequency in air. (b) Find its wavelength in water (refractive index = 1.33). (c) Find its frequency in water. (d) Find its speed in water.

Answer 1

Given

Wavelength of sodium light in air,

\[\lambda_a  = 589\text{ nm} = 589 \times  {10}^{- 9}   m\]

Refractive index of water, μ_w= 1⋅33

We know that

\[f = \frac{c}{\lambda},\]

\[\text{where c = speed of light} = 3 \times  {10}^8   m/s\]

f  = frequency

λ = wavelength

(a) Frequency in air,

\[f_{\text{air}}  = \frac{c}{\lambda_a}\]

\[f_{\text{air}} = \frac{3 \times {10}^8}{589 \times {10}^{- 9}}\] 

\[= 5 . 09 \times  {10}^{14}\text{ Hz}\]

(b)

Let wavelength of sodium light in water be \[\lambda_w.\]

We know that

\[\frac{\mu_a}{\mu_w} = \frac{\lambda_\omega}{\lambda_a}\]

where `μ_a` is the refractive index of air which is equal to 1 and `λ_w` is the wavelength of sodium light in water.

\[\Rightarrow \frac{1}{1 . 33} = \frac{\lambda_\omega}{589 \times {10}^{- 9}}\]

\[ \Rightarrow \lambda_\omega  = 443\text{ nm}\]

(c) Frequency of light does not change when light travels from one medium to another.

\[\therefore    f_\omega  =  f_a \]

\[= 5 . 09 \times  {10}^{14}   Hz\]

(d) Let the speed of sodium light in water be \[\nu_\omega\] and speed in air, `nu_a = c.`

Using \[\frac{\mu_a}{\mu_\omega} = \frac{\nu_\omega}{\nu_a},\] we get

\[\nu_\omega  = \frac{\mu_a c}{\mu_\omega}\] 

\[= \frac{3 \times {10}^8}{\left( 1 . 33 \right)} = 2 . 25 \times  {10}^8   m/s\]