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प्रश्न
Obtain the equation for dispersive power of a medium
संख्यात्मक
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उत्तर
- Consider a beam of white light passes through a prism; It gets dispersed into its constituent colours.
- Let δV, δR are the angles of deviation for violet and red light. Let nV and nR are the refractive indices for the violet and red light respectively.

Angle of deviation for different colours - The refractive index of the material of a prism is given by the equation
n = `(sin ("A + D"/2))/(sin ("A"/2))`
Here A is the angle of the prism and D is the angle of minimum deviation.
n = `(sin ("A + δ"/2))/(sin ("A"/2))`
For small angles of A and δm,
`sin ("A + δ"/2) ≈ ("A + δ"/2)`
`sin ("A"/2) ≈ ("A"/2)`
∴ n = `(("A + δ"/2))/(("A"/2))`
`= ("A + δ")/"A" = 1 + δ/"A"`
Further simplification,
`δ/"A" = "n - 1"`
δ = (n – 1) A
Whn white light enters the prism, the deviation is different for different colours. Thus, the refractive index is also different for different colours.
For violet light, δ =(nv – 1) A
For Red light, δR =(nR – 1) A - As, angle of deviation for violet colour δv is greater than the angle of deviation for red colour, δv the refractive index for violet colour nv is greater than the refractive index for red colour nR.
Subtracting δv from δR we get,
δv – δR = (nv – nR) A - The term (δv – δR) is the angular separation between the two extreme colours (violet and red) in the spectrum is called angular dispersion. Clearly, the angular dispersion produced by a prism depends upon.
- Angle of the prism
- Nature of the material of the prism.
- It is defined as the ratio of the angular dispersion for the extreme colours to the deviation for any mean colour.
Dispersive power (ω),
ω = `"angular dispersion"/"mean deviation" = (δ_"v" - δ_"R")/δ`
Substituting for (δv – δR) and (δ),
ω = `("n"_"v" - "n"_"R")/("n - 1")`
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