can some one tell me how the refractive index changes with the wave length of the light?
The speed of light changes when entering some medium other than vacuum. If the frequency of the light (the number of wave cycles entering a surface per unite time) were to change, it would suggest that the boundary of the medium that the light was entering is destroying waves, which a bunch of very important people decided does not happen (in reality (experiment) it doesn't happen).
Since the speed of the light changes (frequency times wavelength), and the frequency doesn't change, then the wavelength must be changing. The speed on light in a vacuum is c, the speed of light in a material is v=c/n. Thus the wavelength must change as λ/n (λ being the wavelength of the light in a vacuum). Does this answer your question?
The answer depends on the spectral range and material.
In the visible range the index increases when the wavelength decreases (for many common materials). It's larger for blue light than for red light. The difference is in general small (few percent for glass). It is a property of the material so you have to look up the numbers for the material of interest.
Look up "light dispersion" for specific values.
For some materials there are spectral ranges in which the index of refraction increases when the wavelength increases (anomalous dispersion).
You can get more information by googling "Cauchys dispersion formula".
If the index of refraction of a material (glass, lucite, etc.) is dependent on the wavelength (or frequency) of light (dispersion), then the material must have attenuation. This is a direct result of the Kramers-Kronig Dispersion relations. See page 311 in Jackson "Classical Electrodynamics" Second Edition. It is useful to read the sections in Jackson on dispersion and causality.
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