To determine the dispersive power of the material of a given prism by the spectrometer
Spectrometer, Prism and Mercury Vapour Lamp
The Dispersive power of the material of the given prism is expressed as
ω = (μ 2 - μ 1)/(μ-1)
Where μ 1 and μ 2 are the refractive indices of two colors
μ = (μ 1 + μ 2)/2
Usually the colors chosen are blue and red so that
ω = (μ b - μ r)/(μ - 1)
Where μ = (μ b + μ r)/2
4. Working with the tangent screw of the telescope, the position of the prism is adjusted so that the blue line is just one point of refracting its path after coming to the point of intersection of the cross wires.
5. The readings of the telescope for the minimum deviation of red line are noted.
6. The telescope is brought in line with the collimator and removing the prism, the direct readings on both verniers are noted.
7. The respective differences give the minimum deviations for blue and red colors.
Their refractive indices are found by
μ b = [Sin (A+D b) / 2] / [Sin (A/2)] & μ r = [Sin (A + Dr) / 2] / Sin (A / 2)
The Dispersive power of the material of the prism, for blue and red colors is found by the relation
ω = (μ b - μ r)/(μ - 1)
The observations of the above experiment are as follows
V 1= MSRo+ (LC') VC
Spectral Line | LHS V1 | RHS V2 | LHS V1' | RHS V2' | Dm(LHS)[V1 - V'1] | Dm(RHS)[V2 - V'2] | AVG Dm | μ = [Sin(A + Dm) / 2] / Sin (A / 2) |
Blue | ||||||||
Green | ||||||||
Yellow | ||||||||
Red |
Dispersive power ω = (μ b - μ r)/(μ - 1)
The Dispersive power of the material of the prism = ...............
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