Chemistry & forensics

Question

The refractive index, nD, of common liquids changes much more significantly with temperature variations than is the case for solids, such as glass. The shape of the dispersion curve for liquids and solids does not change significantly over moderate temperature ranges. The change of refractive index with temperature for a material is commonly symbolized as dn/dT and this temperature coefficient of refractive index is normally a negative number, as might be expected given how optical density is affected by temperature changes. For most transparent solids, dn/dT ≈ -1.0 x 10-6/°C. For Sets 2 and 3 below, you may consider a hypothetical liquid where the dn/dT = -3.7 x 10-4/°C. You have seen rough dispersion curves in lecture class. You should recall that, as a good generalization in all of these, the refractive index increased as shorter wavelength light was used. The higher the dispersion of the material the more steeply the curve increased as the short wavelength end of the visible spectrum was approached. For

most liquids the dispersion is greater than it is for transparent solids. You should consider the liquid and solids specified below as behaving in the manner described above. Sketch three sets of rough dispersion curves satisfying the conditions stated in each of the parts of the problem below. Each set should depict two curves, one for the each of the two materials. Be certain to label the axes and specify units of measure where applicable.

Set 1: Two transparent solids (i.e. two pieces of glass) that have different refractive indices (nD) at temperature T1 but have similar dispersions in the range of visible light.

Label the dispersion curves S1 and S2.

According to the curves you drew, which of the two solids, S1 or S2, has a higher refractive index (nD)?

Set 2: A transparent solid (i.e. glass) and a liquid, each with the same refractive index, nD, at temperature T1. Label the dispersion curves as follows: “S” for the solid and “L” for the liquid. Also label the point at which nD is measured for the two materials.

What does the subscript “D” signify in nD?

Set 3: The same two materials used in Set 2 above, but at a temperature T2 which is 30°C higher than T1. Label the dispersion curves as follows: “S” for the solid and “L” for the liquid.

When the temperature increases, does refractive index increase or decrease? ___________ You must show your calculations for the following two questions:

By how much has the refractive index of the liquid changed from T1 to T2? ___________

By how much has the refractive index of the solid changed from T1 to T2? ___________

Answer 1

Everything here follows from two facts you were given: both curves fall as wavelength increases, and both fall further when the temperature rises - but the liquid falls about 370 times faster than the solid.

Rough sketches are easier to get right if you fix three numbers per curve instead of drawing freehand, so each set below is given as a small table. Plot wavelength in nm on the x-axis (400 to 700) and refractive index n on the y-axis, and draw a smooth curve through the three points, steepening toward the blue end.

Set 1 - two solids, different nD, similar dispersion

                     400 nm     589.3 nm     700 nm
   S1 (upper)         1.540       1.530       1.526
   S2 (lower)         1.520       1.510       1.506
   drop across        0.014       ----        ----   (same for both
   the visible                                        = same dispersion)

Two roughly parallel curves, both decreasing to the right, one sitting entirely above the other. Similar dispersion means similar steepness, so they never cross.

Which solid has the higher nD? S1 - the curve lying above the other. Since the two are parallel, whichever is higher at the sodium line is higher at every visible wavelength.

Set 2 - a solid and a liquid with the same nD at T1

                     400 nm     589.3 nm     700 nm
   S  (solid)         1.530       1.520       1.516
   L  (liquid)        1.545       1.520       1.510
   drop 400 to 700    S: 0.014    L: 0.035    (liquid is steeper)
                                  ^
                        the curves CROSS here - this is the point
                        where nD is measured, and it is the only
                        wavelength at which the two match

Same value at 589.3 nm, so the curves touch there and separate on both sides. Because liquids disperse more, L is steeper: it lies above S at the blue end and below S at the red end. Mark the crossing point and label it as where nD is measured.

What does the subscript D signify? It is the sodium D line, 589.3 nm - the bright yellow doublet of a sodium lamp. Refractive index is quoted at a single stated wavelength because n depends on wavelength, and the sodium D line became the convention since a sodium lamp gives a bright, easily reproduced, essentially monochromatic source.

Set 3 - the same two materials, 30 degrees C hotter

When temperature increases, does refractive index increase or decrease? It decreases. dn/dT is negative for both materials: heating expands them, lowers the number density of scattering centers, and lowers the optical density.

Change for the liquid:

delta n = (dn/dT)(delta T) = (-3.7 x 10^-4 per degree C)(30 degrees C) = -0.0111

Change for the solid:

delta n = (-1.0 x 10^-6 per degree C)(30 degrees C) = -0.000030

                     400 nm     589.3 nm     700 nm
   S  (solid)        1.52997     1.51997     1.51597
   L  (liquid)       1.53390     1.50890     1.49890
   L minus S         +0.0039     -0.0111     -0.0171
                          ^
          the crossing has MOVED to roughly 450 nm; at the
          sodium D line the liquid now sits clearly below

So in your Set 3 sketch, S is essentially unmoved from Set 2 - the shift is in the fourth decimal place and would be invisible at any sensible scale - while L has dropped visibly. Draw S in the same place and L noticeably lower, and note that the crossing point has slid toward the blue.

Why a forensic course is asking this

This is the physical basis of the immersion method for comparing glass fragments. A fragment is suspended in a liquid and heated; at the temperature where the liquid and the glass have the same refractive index, the fragment optically disappears - the Becke line vanishes and the edges cannot be seen. That match temperature is measured rather than the index itself.

The technique works precisely because of the numbers you just calculated. The liquid moves hundreds of times faster than the glass, so temperature is effectively a fine-tuning knob on the liquid alone, and a match can be located to a small fraction of a degree. That converts a hard optical measurement into an easy thermal one, and it is sensitive enough to distinguish glass fragments that look identical.