SOLUTION TO HOMEWORK PROBLEM 14

For speckle displacement on the specklegram: p = lD/d , where p is the pitch or spacing between twinned speckles, l is the wavelength of laser light used, D is distance from specklegram to image plane, and d is measured spacing between successive fringes in the pattern.

a) What is the actual displacement of the surface at the point being interrogated?

In the given measurement l = 633 microns, D is 25 cm, and d is 7.9 mm. Thus the speckle displacement is p = (0.633 microns)(25 cm)/(7.9 mm) = 20 microns. To get the actual physical motion of a location on the original object, the magnification of the optical system imaging the original surface onto the specklegram plate must be calculated. From the given data s is 60 cm and the focal length f = 25 cm.

1/f = 1/s + 1/s'

1/s' = 1/f - 1/s

s' = 42.9 cm

M = -s'/s = -42.9/60 = -0.71 (only want magnitude, so can drop sign)

Since fringes are horizontal, displacement is vertical.

Surface displacement is p/M so Dy = 28 microns in the vertical direction.

b) What is the Young's modulus for the studied specimen?

In the Fourier filtering p = nlf/r where f is focal length of transform lens = 25 cm and r is the radial distance of the filter hole from the optic axis = 1.5 cm. So the fringe lines represent (0.633 microns)(25 cm)/(1.5 cm) = 10 microns motion of speckles in the specklegram. Since starting at the very bottom of the specklegram image of the object, the fringes divided the specimen into exactly 4 equal parts, the displacement of the specklegram was 40 microns. Magnifying this by the M of the original lens system, the compressive displacement of the object was (40 microns)/(0.71) = 56.3 microns. The value of Young's modulus is given by Y = (F L₀)/(A DL) where F is the applied force, A is the cross-sectional area, L_o is the original length, and DL is the change in length. From the given data, F = 240 N, A = 1 × 10^-4 m², L₀ = 0.05 m. Therefore Y = (240)(0.05)/[(1 × 10^-4)(56.3 × 10^-6)] = 2.13 GPa.

c) What would happen if the filtering aperture were located at r = 0.75 cm from the optic axis but still on the vertical axis?

This would double the pitch separation that would show up as fringes on the final image. Thus the fringes would divide the image into halves. Each fringe would represent a displacement of 20 microns on the specklegram. None of the other calculations would change.

d) What if the aperture was 1.5 cm from optic axis but rotated 45 degrees from the vertical?

If the aperture was 1.5 cm from the optic axis but rotated 45 degrees from the vertical, a set of straight fringes angled 45 degrees from the horizontal would be seen. Although none of the object moved in that direction, the components of motion would show up. Take the 45 degree angle as the reference direction and break the vertical motion into components parallel to and perpendicular to the reference direction. Whenever the parallel component was 10 microns or a multiple thereof, it would show up as part of the angled fringes. Since the angle is 45 degrees, this means that whenever the vertical compression at a point on the specklegram is 14.14 microns or a multiple thereof, a fringe angled at 45 degrees will pass through the corresponding point on the image.

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