| Date | May 2016 | Marks available | 4 | Reference code | 16M.3.SL.TZ0.2 |
| Level | Standard level | Paper | Paper 3 | Time zone | Time zone 0 |
| Command term | Determine | Question number | 2 | Adapted from | N/A |
Question
A student measures the refractive index of the glass of a microscope slide.
He uses a travelling microscope to determine the position x1 of a mark on a sheet of paper. He then places the slide over the mark and finds the position x2 of the image of the mark when viewed through the slide. Finally, he uses the microscope to determine the position x3 of the top of the slide.
The table shows the average results of a large number of repeated measurements.
The refractive index of the glass from which the slide is made is given by
\[\frac{{{x_3} - {x_1}}}{{{x_3} - {x_2}}}\].
Determine
(i) the refractive index of the glass to the correct number of significant figures, ignoring any uncertainty.
(ii) the uncertainty of the value calculated in (a)(i).
After the experiment, the student finds that the travelling microscope is badly adjusted so that the measurement of each position is too large by 0.05mm.
(i) State the name of this type of error.
(ii) Outline the effect that the error in (b)(i) will have on the calculated value of the refractive index of the glass.
After correcting the adjustment of the travelling microscope, the student repeats the experiment using a glass block 10 times thicker than the original microscope slide. Explain the change, if any, to the calculated result for the refractive index and its uncertainty.
Markscheme
(i) refractive index = 1.5
Both correct value and 2SF required for [1].
(ii) fractional uncertainty \({x_3} - {x_1} = \frac{{0.04}}{{1.15}} = 0.035\) AND \({x_3} - {x_2} = \frac{{0.04}}{{0.76}} = 0.053\)
sum of fractional uncertainty = 0.088
«uncertainty = their RI × 0.088» = 0.1
Accept correct calculation using maximum and minimum values giving the same answer.
(i) systematic error
Accept “zero error/offset”.
(ii) calculated refractive index is unchanged
because both numerator and denominator are unchanged
Accept calculation of refractive index with 0.05 subtracted to each x value.
numerator and denominator will be 10 times larger so refractive index is unchanged
relative/absolute uncertainty will be smaller
“Constant material” is not enough for MP1.
