| Date | May 2011 | Marks available | 2 | Reference code | 11M.3.HL.TZ1.15 |
| Level | Higher level | Paper | Paper 3 | Time zone | Time zone 1 |
| Command term | Deduce | Question number | 15 | Adapted from | N/A |
Question
This question is about wedge film interference.
One flat, glass slide is placed at an angle on top of a second identical slide. The slides are in contact along one short edge and are separated at the other edge by a thin piece of paper, as shown below.
A thin wedge of air of variable thickness, t , is trapped between the two slides. The arrangement is viewed normally from above, using light of wavelength 590 nm. The glass plates are coated, so that reflection only takes place at the bottom surface of the top plate and the top surface of the bottom plate.
A series of straight bright and dark fringes, equally separated and parallel to the short edge of the slides, is seen.
Deduce that the thickness of the air wedge t that gives rise to a bright fringe, is given by \(2t = (m + \frac{1}{2})\lambda \).
The length of the air wedge, L, is 8.2 cm. The bright fringes are each separated by a distance of 1.2 mm. Calculate the thickness of the paper.
Markscheme
phase change of π occurs on reflection at one slide but not the other;
constructive interference occurs when path difference between two reflected rays is \(\frac{\lambda }{2},\frac{{3\lambda }}{2},\frac{{5\lambda }}{2}\) etc;
the extra distance travelled is twice the thickness of the air (wedge) hence 2t = [m+½] λ;
number of fringes = \(\frac{{82}}{{1.2}} = 68\);
fringe separation corresponds to a change in thickness of \(\frac{\lambda }{2}\);
thickness of paper\( = 68 \times \frac{{590 \times {{10}^{ - 9}}}}{2} = 2.0 \times {10^{ - 5}}{\rm{m}}\);
