An observer viewing the microscope slide at near-normal incidence measures the fringe spacing to be 0.29 mm. Calculate the thickness of the hair.

there are $\frac{{60}}{{0.29}}$ fringes=207;
2×1×t=207×5.9×10^-7;
t=61(μm);

or

$\left( {\tan \theta  = } \right)\frac{1}{{6.0\left( {{\rm{cm}}} \right)}} = \frac{{0.5\lambda }}{{\Delta x}}$;
$t = \frac{{\left[ {0.06\left( {\rm{m}} \right) \times 0.5 \times 5.9 \times {{10}^{ - 7}}\left( {\rm{m}} \right)} \right]}}{{0.00029\left( {\rm{m}} \right)}}$;
t=61(μm);
A phase change of $\frac{1}{2}\lambda$, if seen in working, can be ignored and does not affect the answer.
Award [3] for a bald correct answer.