Date | November 2013 | Marks available | 3 | Reference code | 13N.3.HL.TZ0.11 |
Level | Higher level | Paper | Paper 3 | Time zone | Time zone 0 |
Command term | Calculate | Question number | 11 | Adapted from | N/A |
Question
This question is about wedge fringes.
A glass microscope slide of length 6.0cm is placed on a glass plate and illuminated using a monochromatic source of light of wavelength 590nm. A hair is trapped at one end of the slide forming an air wedge between the glass plate and the slide.
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.
Markscheme
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.