Chapter 26, Problem 38P

Set Up: The maxima are located by dsinθ= mλ, where d = 1.60×10⁻⁶ m for a CD and d = 0.740×10⁻⁶ m for a DVD.

1. (a) Solve: For a CD we have: $\theta =\arcsin \left(\frac{m\lambda }{d}\right)=\arcsin \left(\frac{m(6.328\times {10}^{-7}\text{m})}{1.60\times {10}^{-6}\text{m}}\right)$= arcsin(0.396m). For m = 1 we have θ₁ = 23.3°. For m = 2, we have θ₂ =52.3°. There are no other maxima.
2. (b) = For a DVD we have: θ = arcsin $\left(\frac{m\lambda }{d}\right)$ = $\arcsin \left(\frac{m(6.328\times {10}^{-7}\text{m})}{0.70\times {10}^{-6}\text{m}}\right)$ = arcsin (0.855m). For m = 1 we have θ₁ = 58.8°. There are no other maxima.

Reflect: the reflective surface produces the same interference pattern as a grading with slit separation d.

38. CDs and DVDs as diffraction gratings. A laser beam of wavelength λ = 632.8 nm shines at normal incidence on the reflective side of a compact disc. (a) The tracks of tiny pits in which information is coded onto the CD are 1.60 μm apart. For what angles of reflection (measured from the normal) will the intensity of light be maximum? (b) On a DVD, the tracks are only 0.740 μm apart. Repeat the calculation of part (a) for the DVD.