Question

A piece of glass with a flat surface is at the bottom of a tank of water. If a ray of light traveling in the glass is incident on the interface with the water at an angle with respect to the normal that is greater than 50.0 ∘ , no light is refracted into the water. For smaller angles of incidence, part of the ray is refracted into the water. If the light has wavelenght 408nm in the glass. what is wavelength of the light water?

Answer 1

Answer:

λ₀ = 542.6 nm

Explanation:

The speed of light is related to the wavelength and the frequency by the equation

      c = λ f

When a beam of light strikes a surface, it makes the electrons in it oscillate, this is a forced oscillation, so the oscillation frequency equals the frequency of the intendent radiation, then the electrons radiate the radiation for a while that absorbed; therefore, the light in both media has the same frequency

As the speed of light changes in each medium, the only way to comply with the equation is for the wavelength of the radiation to change, let's use the definition of the index of refraction

    v = $\lambda_{n}$  f

    n = c / v

    n = λ₀ f / $\lambda_{n}$ f

    n = λ₀ / $\lambda_{n}$

    $\lambda_{n}$ = λ₀ / n

Let's apply this equation to our case

In the glass     λ = 408 nm

In the water is

     $\lambda_{n}$ = λ₀ / n

    λ₀ =n  $\lambda_{n}$

The index of refraction of water-related glass can be obtained by the law of refraction

    Water       glass

   n₁ sin θ₁ = n₂ sin θ₂

For an incident angle of 50º the beam is not refracted in the water, this means that it is parallel to the surface of the glass therefore the angle is 90º, this is called internal total reflection

for angles less than 50º the light that refracts the water that has refractive index 1.33 changes its wavelength

calculate

    $\lambda_{n}$ =  λ₀ / n

     λ₀ = n   $\lambda_{n}$

     λ₀ = 1.33 408

     λ₀ = 542.6 nm