# Does the frequency of light changes when it go through a prism

I know the prism can seperate the light into 7 colours, but my teacher said the frequency of light changes and so as the wavelength..

if he is right.... this is the bit i am not sure about

if the frequency of the light changes when it passes through a prism, does it mean the energy also changes? if it does, does it mean that when the light goes through a prism, it changes into 7 different rays which that we dont regard them as the visible light because they all have different frequency and energy ?

plz correct me

i am not even sure about my basic knowledge

many thanks

Meir Achuz
Homework Helper
Gold Member
The frequency does not change when light enters a prism, but the intensity and energy density of the light wave decreases somewhat due to partial reflection of the wave at the air-glass boundary.

jtbell
Mentor
Ok so basically a red light (wavelength of around 600 nm) entering say a prism whose refractive index is 1.5 becomes violet (wavelength of about 400 nm) in the prism, am I right?

No, the light is still red inside the prism. The frequency of the light (or energy of the associated photons, E = hf) is what fundamentally determines the color that we perceive. We usually relate color to wavelength (in vacuum or practically, in air) because wavelength is easier to measure directly.

fluidistic
Gold Member
No, the light is still red inside the prism. The frequency of the light (or energy of the associated photons, E = hf) is what fundamentally determines the color that we perceive. We usually relate color to wavelength (in vacuum or practically, in air) because wavelength is easier to measure directly.

Ok thank you, amazing.

All the frequencies ( or colors) make up white light. The prism seperates them. When Newton did this he also used a second prism placed so the spectrum coming out of the first prism was mixed back into white light which exited the second prism.

...the associated photons, E = hf

Wait a minute. Using this same logic I get an associated momentum, p=h_bar k; an increase in momentum and an imaginary mass.