Dispersion of Glass by the Method of Minimum Deviation

Andy Chmilenko, 20310799 Instructor: Jeff Gardiner Section 1

(Dated: 2:30 pm Tuesday September 24, 2013)

I. PURPOSE

The Purpose of this experiment is to determine the indices of refraction of the glass prism at different wavelengths; thence to determine the dispersion curve for that glass and compare it with Cauchy’s equation.

II. ANALYSIS

A. Determining the Apex Angle

\begin{matrix}α_{L} ± 30′′ & α_{R} ± 30′′ & α ± 42′′ \\ 119^{∘}57′ & 240^{∘}8′ & 60^{∘}5′ \\ 87^{∘}16′ & 207^{∘}25′ & 60^{∘}4′ \\ 71^{∘}9′ & 191^{∘}23′ & 60^{∘}7′ \end{matrix}

TABLE I: Measured angles from reflected collimated light off of the prism from the left (α_{L}) and right(α_{R}) sides, and the calculated apex angle (α)

Sample Calculations for α using Table I row 1

α = \frac{|α_{L}−α_{R}|}{2}
α = \frac{|119^{∘}57′−240^{∘}8′|}{2}
α = \frac{120^{∘}11′}{2}
α = 60^{∘}5′30′′

Sample Calculations for Δα

Δα = \sqrt{(Δα_{L})^{2} + (Δα_{R})^{2}} Δα = \sqrt{(30′′)^{2} + (30′′)^{2}} Δα = \sqrt{(0.25′′)^{2} + (0.25′′)^{2}}

Δα = 42.43′′
Δα = 42′′

Sample Calculations for ᾱ

ᾱ = ^{\frac{∑_{i}^{n} α_{i}}{n}}

ᾱ = \frac{60^{∘}5′+60^{∘}4′+60^{∘}7′}{3}

ᾱ = 60^{∘}5′20′′

The apex angle for the prism, α was measured to be 60^{∘}5′ ± 42′′.

B. Determining the Refractive Index

III. CONCLUSION

Colour  Wave- Setting of  Angle of Refractive
length Minimum Minimum Index, n
in Air Deviation Deviation
(Å) Left Right  δ_{m}
Red - H 6563 132^{∘}1′ 227^{∘}36′ 47^{∘}47′ 1.6
Yellow III - Hg 5790 131^{∘}29′ 228^{∘}11′ 48^{∘}21′  1.6
Yellow I - Hg 5770 131^{∘}28′ 228^{∘}14′ 48^{∘}23′  1.6
Green - Hg 5461 131^{∘}7′ 228^{∘}30′ 48^{∘}41′  1.6
Blue-Green - H 4861 130^{∘}23′ 229^{∘}17′ 49^{∘}27′  1.6
Blue - Hg 4358 129^{∘}22′ 230^{∘}19′ 58^{∘}29′ 1.6
Violet - Hg 4078 128^{∘}34′ 231^{∘}4′ 51^{∘}15′  1.6
Violet - Hg 4047 128^{∘}29′ 231^{∘}6′ 51^{∘}19′  1.6

TABLE II: Measured angles from diffracted collimated light for different wavelengths, produced by a Hg lamp and a H lamp.

Sample Calculations for δ_{m} using Table II row 1

2δ_{m} = |L − R|
δ_{m} = \frac{|L−R|}{2}
δ_{m} = \frac{|132^{∘}1′−227^{∘}36′|}{2}
δ_{m} = \frac{95^{∘}35′}{2}