Question

Problem One: A beam of red light (656 nm) enters from air into the side of a glass and then into water. wavelength, c. and speed in both the glass and the water. Find the a. the frequencies, b. the

Answer 1

Part a)

`f_w = f_g = 4.57 * 10^(14) Hz`

Part b)

`\lambda_w = 492 nm`

`\lambda_g = 437.3 nm`

Part c)

`v_w = 2.25 * 10^8 m/s`

`v_g = 2.0 * 10^8 m/s`

Step-by-step explanation:

Part a)

frequency of light will not change with change in medium but it will depend on the source only

so here frequency of light will remain same in both water and glass and it will be same as that in air

`f = (v)/(\lambda)`

`f = (3 * 10^8)/(656 * 10^(-9))`

`f = 4.57 * 10^(14) Hz`

Part b)

As we know that the refractive index of water is given as

`\mu_w = 4/3`

so the wavelength in the water medium is given as

`\lambda_w = (\lambda)/(\mu_w)`

`\lambda_w = (656 nm)/(4/3)`

`\lambda_w = 492 nm`

Similarly the refractive index of glass is given as

`\mu_w = 3/2`

so the wavelength in the glass medium is given as

`\lambda_g = (\lambda)/(\mu_g)`

`\lambda_g = (656 nm)/(3/2)`

`\lambda_g = 437.3 nm`

Part c)

Speed of the wave in water is given as

`v_w = (c)/(\mu_w)`

`v_w = (3 * 10^8)/(4/3)`

`v_w = 2.25 * 10^8 m/s`

Speed of the wave in glass is given as

`v_g = (c)/(\mu_g)`

`v_g = (3 * 10^8)/(3/2)`

`v_g = 2 * 10^8 m/s`