IB DP Physics Mock Exam SL Paper 1B Set 5 - 2025 Syllabus

(a)
To reduce absolute uncertainty, measure the total distance across multiple fringes (for example, 10 fringes) and divide this distance by the number of fringes (minus one).

(b)
From the relation \(s = \frac{\lambda D}{d}\), decreasing \(d\) produces a larger fringe spacing \(s\). As the ruler’s absolute uncertainty remains the same, the fractional uncertainty \(\frac{\Delta s}{s}\) becomes smaller.

(c)
(i) Since \(s = (\lambda D)\frac{1}{d}\), plotting \(s\) against \(\frac{1}{d}\) (or \(d^{-1}\)) gives a straight line through the origin.
(ii) The absolute uncertainty is \(5\%\) of \(611.089\,nm \approx 30.55\,nm\), which rounds to \(\pm 30\,nm\). The wavelength should be rounded to the same precision, giving \(610\,nm\).
Final result: \(\lambda = 610 \pm 30\,nm\).