This question aims to study the **effect of material on the wave speed** when it travels from one material to another.

Whenever **a wave strikes the surface of another material**, a part of it is **bounced back** into the previous medium (called **reflection** phenomenon) and a part of it **enters into** the new medium (called **refraction** phenomenon). During the refraction process, the **frequency of the light waves remains the same, **however the **speed and wavelength change**.

The relationship between the speed (v), wavelength ($ \lambda $) and the frequency f of a wave are given by the following mathematical formula:

\[ f_{ solid } \ = \ \dfrac{ v_{ solid } }{ \lambda_{ solid } } \]

## Expert Answer

Given:

\[ \lambda_{ air } \ = \ 670 \ nm \ = \ 6.7 \times 10^{ -7 } \ m \]

\[ \lambda_{ solid } \ = \ 420 \ nm \ = \ 4.2 \times 10^{ -7 } \ m \]

Let’s **assume** that:

\[ \text{ Speed of Light in Air } \approx v_{ air } \ = \ \text{ Speed of Light in Vacuum } = \ c \ = 3 \times 10^8 m/s \]

**Part(a) – Calculating the frequency of the light waves in the given solid:**

\[ f_{ air } \ = \ \dfrac{ v_{ air } }{ \lambda_{ air } } \]

\[ \Rightarrow f_{ air } \ = \ \dfrac{ 3 \times 10^8 m/s }{ 6.7 \times 10^{ -7 } \ m } \ = \ 4.478 \times 10^{ 14 } \ Hz \]

During the refraction process, the** frequency remains constant**, so:

\[ f_{ solid } \ = \ f_{ air } \ = \ 4.478 \times 10^{ 14 } \ Hz \]

**Part(b) – Calculating the speed of the light waves in the given solid:**

\[ f_{ solid } \ = \ \dfrac{ v_{ solid } }{ \lambda_{ solid } } \]

\[ \Rightarrow v_{ solid } \ = \ f_{ solid } \ \lambda_{ solid } \]

\[ \Rightarrow v_{ solid } \ = \ ( 4.478 \times 10^{ 14 } \ Hz )( 4.2 \times 10^{ -7 } \ m \]

\[ \Rightarrow v_{ solid } \ = \ 1.88 \times 10^8 m/s \]

## Numerical Result

\[ f_{ solid } \ = \ 4.478 \times 10^{ 14 } \ Hz \]

\[ v_{ solid } \ = \ 1.88 \times 10^8 m/s \]

## Example

For the **same conditions given in the above question**, calculate the **speed and frequency** for a solid in which the **wavelength of the light** waves **reduces to 100 nm**.

Given:

\[ \lambda_{ air } \ = \ 670 \ nm \ = \ 6.7 \times 10^{ -7 } \ m \]

\[ \lambda_{ solid } \ = \ 1 \ nm \ = \ 1 \times 10^{ -7 } \ m \]

Using the same **assumption**:

\[ \text{ Speed of Light in Air } \approx v_{ air } \ = \ \text{ Speed of Light in Vacuum } = \ c \ = 3 \times 10^8 m/s \]

Calculating the **frequency of the light waves** in the given solid:

\[ f_{ solid } \ = \ f_{ air } \ = \ \dfrac{ v_{ air } }{ \lambda_{ air } } \]

\[ \Rightarrow f_{ solid } \ = \ \dfrac{ 3 \times 10^8 m/s }{ 6.7 \times 10^{ -7 } \ m } \ = \ 4.478 \times 10^{ 14 } \ Hz \]

Calculating the** speed of the light waves** in the given solid:

\[ v_{ solid } \ = \ f_{ solid } \ \lambda_{ solid } \]

\[ \Rightarrow v_{ solid } \ = \ ( 4.478 \times 10^{ 14 } \ Hz )( 1 \times 10^{ -7 } \ m ) \ = \ 4.478 \times 10^7 m/s \]