Physics 9702/54 — October/November 2025

Cambridge A-Level · Planning, Analysis and Evaluation · worked solutions for every part, with the mark scheme

questions

marks

minutes

Topics Analysis, Conclusions and Evaluation · Planning

Q115MMedium-HardPlanningAnalysis, Conclusions and Evaluation

Fig. 1.1 shows a horizontal turntable.

Fig. 1.1

Point C is at the centre of the turntable. Point P is a distance $r$ from the centre.

Fig. 1.2 shows a side view of a d.c. motor attached to the turntable with a belt.

Fig. 1.2

The motor is used to rotate the turntable at frequency $f_0$ . The motor is switched off and the turntable continues to rotate at frequency $f_0$ .

A sphere of adhesive putty of mass $m$ is dropped onto the turntable at point P. The frequency of the turntable is now $f$ .

It is suggested that $f$ is related to $m$ by the relationship

\frac{Kf_0}{f} = \beta K + mr^2

where $\beta$ and $K$ are constants.

Plan a laboratory experiment to test the relationship between $f$ and $m$ .

Draw a diagram showing the arrangement of your equipment.

Explain how the results could be used to determine values for $\beta$ and $K$ .

In your plan you should include:

the procedure to be followed
the measurements to be taken
the control of variables
the analysis of the data
any safety precautions to be taken.

DifficultyMedium-Hard

Worked solution

Variables

Independent variable: mass $m$ of adhesive putty added at point $P$ .
Dependent variable: new rotation frequency $f$ after putty sticks.
Controls: radial distance $r$ (same marked point $P$ ), same turntable and axle, same initial frequency $f_0$ for each run, putty placed so it does not slide, turntable horizontal.

Apparatus and arrangement

diagram-1

Turntable driven by d.c. motor and belt.
Tachometer/optical sensor (or light gate + data logger) to measure frequency using a reflective tape marker on rim.
Top-pan balance to measure $m$ .
Ruler/set square to mark a fixed radius $r$ and the point $P$ on the turntable.

Procedure / measurements

Fix a small reflective strip on the edge of the turntable. Align optical sensor to count one pulse per revolution.
Mark point $P$ at radius $r$ from the centre using a ruler; keep $r$ constant throughout.
Prepare several different masses of putty; measure each mass $m$ with a balance.
For one value of $m$ $m$ :
- Run motor until turntable rotates steadily.
- Switch motor off.
- Immediately measure the frequency just before adding putty, $f_0$ (e.g. measure time $t_0$ for $N$ revolutions and use $f_0 = N/t_0$ ).
- Drop/place the putty vertically onto point $P$ so it sticks with minimal sideways push.
- Measure the new frequency $f$ after it settles (again using $f = N/t$ for the same $N$ ).
Repeat readings for each $m$ (at least 5–6 different masses), and repeat each point to obtain mean values.

Analysis (test the relationship; determine $\beta$ and $K$ )

Given

\frac{K f_0}{f} = \beta K + mr^2

Divide by $K$ :

\frac{f_0}{f} = \beta + \frac{r^2}{K}m

So plot $y = \dfrac{f_0}{f}$ against $x = m$ .

Intercept $= \beta$ .
Gradient $= \dfrac{r^2}{K}$ , hence

K = \frac{r^2}{\text{gradient}}

Use a best-fit line and a worst acceptable line to estimate uncertainties in gradient and intercept (hence in $K$ and $\beta$ ).

Safety

Keep fingers/hair/clothing clear of rotating turntable and belt; do not lean over the rotating system.
Ensure motor is switched off before placing the putty; stand to the side in case putty detaches.
Secure the motor/turntable so it cannot move on the bench.

Final answer

See working

Detailed explanation

Background Concept

When an object rotates, adding a mass at a distance from the axis increases the system’s moment of inertia. If external torque is negligible during the brief “sticking” event, angular momentum is approximately conserved, so the rotation rate decreases.

In this question, you are not asked to derive the relationship; you are asked to test the suggested relationship experimentally and use data to find constants.

The given suggested relationship is

\frac{K f_0}{f} = \beta K + mr^2

where:

$f_0$ is the turntable frequency just before the putty is added (motor switched off),
$f$ is the frequency after the putty sticks,
$m$ is the putty mass,
$r$ is the distance of the putty from the centre,
$\beta$ and $K$ are constants for the apparatus.

A key experimental skill here is to rearrange to a linear form $y = c + mx$ so you can use a straight-line graph to test the model and extract constants.

Understanding the Question

You must design a practical method in which you:

Vary $m$ (use different putty masses),
Measure $f_0$ and $f$ for each $m$ while keeping $r$ the same,
Analyse results so that if the relationship is correct, your graph is a straight line,
Use the graph’s gradient and intercept to find $\beta$ and $K$ .

Because $f_0$ is the frequency with the motor off and just before adding the putty, you should measure $f_0$ for each run (not assume it is identical every time).

Approach

Choose a reliable way to measure rotational frequency (optical sensor/light gate + data logger, or a tachometer, or a stroboscope). The easiest to justify for Paper 5 is an optical sensor with a reflective marker giving one pulse per revolution.
Keep $r$ constant by marking a point $P$ on the turntable at a measured radius.
Take multiple masses $m$ , measure each with a balance, and for each mass measure $f_0$ (immediately before) and $f$ (after it sticks).
Linearise the suggested equation by dividing by $K$ to remove the unknown constant from the left-hand side.
Plot the appropriate graph; use intercept and gradient to determine $\beta$ and $K$ . Include uncertainty handling via best-fit and worst-acceptable lines.

Step-by-Step Reasoning

1) Set up the apparatus and frequency measurement

Attach the motor and belt as shown and place a small reflective strip on the rim. Position an optical sensor so that each revolution produces a single pulse.

diagram-1

Measure frequency by timing many revolutions:

If you count $N$ pulses in time $t$ , then

f = \frac{N}{t}

Choosing a larger $N$ (e.g. $N = 20$ or $50$ ) reduces the percentage timing uncertainty.

2) Ensure the radius is controlled

Use a ruler (and ideally a set square) to measure a fixed radius $r$ from the centre. Mark point $P$ clearly. Every run must place the putty at the same point so $r$ is constant.

3) Collect data for multiple values of $m$

For each mass $m$ :

Measure $m$ using a balance (record to the balance resolution).
Spin the turntable with the motor until stable.
Switch the motor off (so the only torques are resistive, and you are consistent with the statement).
Immediately measure $f_0$ by timing $N$ revolutions.
Drop/place the putty vertically at $P$ so it sticks; avoid giving a sideways push because that would add/remove angular momentum.
Measure the new steady frequency $f$ .
Repeat to get mean values of $f_0$ and $f$ for each $m$ .

4) Linearise to obtain a straight-line test

Start with

\frac{K f_0}{f} = \beta K + mr^2

Divide through by $K$ :

\frac{f_0}{f} = \beta + \frac{r^2}{K}m

This is in straight-line form $y = c + gx$ with:

$y = \dfrac{f_0}{f}$
$x = m$
intercept $c = \beta$
gradient $g = \dfrac{r^2}{K}$

So you plot $\dfrac{f_0}{f}$ against $m$ .

5) Determine $\beta$ and $K$ from the graph

From the best-fit straight line:

$\beta$ is the y-intercept.
If the gradient is $g$ , then

K = \frac{r^2}{g}

To estimate uncertainty:

Draw a worst acceptable line (steepest/shallowest line still consistent with the scatter or with error bars).
Find the change in gradient between best and worst line to estimate $\Delta g$ , and similarly for the intercept to estimate $\Delta \beta$ .
Propagate to $K$ using fractional uncertainty: since $K \propto \dfrac{1}{g}$ ,

\frac{\Delta K}{K} \approx \frac{\Delta g}{g}

(If $r$ is measured with uncertainty too, include it: $K \propto r^2$ so add $2\Delta r/r$ in fractional terms.)

Key Takeaways

A good plan states: independent/dependent variables, control variables, apparatus, method, analysis, and safety.
Linearising an equation to $y = mx + c$ lets you use a straight-line graph to test the relationship and determine constants.
Measuring frequency accurately often means timing many cycles/revolutions and dividing.
Constants are extracted from gradient and intercept; uncertainties come from best-fit vs worst-acceptable lines.

Common Mistakes

Keeping $f_0$ fixed by assumption instead of measuring it each run (friction means it may differ).
Forgetting to keep $r$ constant, or placing putty inconsistently so the effective radius changes.
Dropping the putty with a sideways push, which changes angular momentum and ruins the test.
Plotting the wrong graph (e.g. $f$ vs $m$ directly) so the relationship is not linear and constants cannot be read reliably.
Not stating how $\beta$ and $K$ are obtained from intercept/gradient.

Things to Be Careful About

Use consistent units: $m$ in $\text{kg}$ , $r$ in $\text{m}$ , so $mr^2$ has units $\text{kg m}^2$ .
Ensure the putty sticks and does not slide outward due to rotation; if it moves, $r$ is no longer controlled.
Measure $f$ and $f_0$ over the same number of revolutions $N$ to keep timing uncertainty comparable.
When finding gradient, use a large triangle on the best-fit line and calculate $\Delta y/\Delta x$ (not $x/y$ ), and include uncertainty using worst acceptable line methods.
Safety around rotation: loose clothing/hair and fingers near the belt/turntable are realistic hazards and should be explicitly addressed.

Techniques used

identify independent, dependent and control variablesmeasure rotational frequency using a sensor and timing over many revolutionslinearise the suggested relationshipplot a graph and use gradient and intercept to determine constantsestimate uncertainties using best-fit and worst-acceptable lines

$s / \text{mm}$	$\frac{1}{s} / \text{mm}^{-1}$	$w / \text{mm}$	$y / \text{mm}$
$0.18 \pm 0.01$		$33.0$
$0.21 \pm 0.01$		$28.9$
$0.24 \pm 0.01$		$25.1$
$0.27 \pm 0.01$		$22.6$
$0.31 \pm 0.01$		$19.6$
$0.38 \pm 0.01$		$15.9$

Working

For each row:

y = \frac{w}{10}

and for $x=\frac{1}{s}$ with $s \pm \Delta s$ :

\Delta\left(\frac{1}{s}\right) = \frac{\Delta s}{s^2}

Example (first row):

\frac{1}{s} = \frac{1}{0.18} = 5.56\ \text{mm}^{-1}

\Delta\left(\frac{1}{s}\right)=\frac{0.01}{(0.18)^2}=0.309\approx 0.31\ \text{mm}^{-1}

y=\frac{33.0}{10}=3.30\ \text{mm}

Answer (Table 2.1 completed)

$s / \text{mm}$	$\frac{1}{s} / \text{mm}^{-1}$	$w / \text{mm}$	$y / \text{mm}$
$0.18 \pm 0.01$	$5.56 \pm 0.31$	$33.0$	$3.30$
$0.21 \pm 0.01$	$4.76 \pm 0.23$	$28.9$	$2.89$
$0.24 \pm 0.01$	$4.17 \pm 0.17$	$25.1$	$2.51$
$0.27 \pm 0.01$	$3.70 \pm 0.14$	$22.6$	$2.26$
$0.31 \pm 0.01$	$3.23 \pm 0.10$	$19.6$	$1.96$
$0.38 \pm 0.01$	$2.63 \pm 0.07$	$15.9$	$1.59$

Final answer

See completed table in working

Detailed explanation

Background Concept

When you calculate a new quantity from measured values, you must also propagate the uncertainty.

For a derived quantity $y = \frac{w}{10}$ , the calculation is direct.
For a reciprocal $x = \frac{1}{s}$ , the uncertainty grows non-linearly as $s$ gets smaller.

A standard method for the absolute uncertainty in a reciprocal is:

\Delta\left(\frac{1}{s}\right) \approx \left|\frac{\mathrm{d}}{\mathrm{d}s}\left(\frac{1}{s}\right)\right|\Delta s = \frac{\Delta s}{s^2}

Understanding the Question

You are given measured slit separations $s$ (each with $\pm 0.01\ \text{mm}$ ) and measured distances $w$ across 10 fringes. You must fill the missing columns:

$\frac{1}{s}$ (with absolute uncertainty)
$y$ where $y = \frac{w}{10}$

Approach

For each row:

Compute $y$ by dividing $w$ by 10.
Compute $\frac{1}{s}$ .
Compute the absolute uncertainty in $\frac{1}{s}$ using $\Delta(1/s) = \Delta s / s^2$ .
Round sensibly: uncertainty usually to 1–2 s.f., and the value to the same decimal place.

Step-by-Step Reasoning

Take the first row as a model.

Calculate the reciprocal:

\frac{1}{s} = \frac{1}{0.18} = 5.555\ldots\ \text{mm}^{-1} \to 5.56\ \text{mm}^{-1}

Uncertainty in the reciprocal:

\Delta\left(\frac{1}{s}\right) = \frac{0.01}{(0.18)^2} = 0.3086\ \text{mm}^{-1} \to 0.31\ \text{mm}^{-1}

Calculate fringe spacing:

y = \frac{w}{10} = \frac{33.0}{10} = 3.30\ \text{mm}

Repeat identically for the other rows.

Key Takeaways

A reciprocal amplifies uncertainty: smaller $s$ gives larger $\Delta(1/s)$ .
Keep units explicit: $s$ in mm gives $1/s$ in mm $^{-1}$ .

Common Mistakes

Using $\Delta(1/s)=\frac{\Delta s}{s}$ (that is for fractional uncertainty, not absolute for the reciprocal).
Forgetting to include uncertainties in $\frac{1}{s}$ .
Dividing $w$ by 10 but rounding $y$ inconsistently.

Things to Be Careful About

Do not convert some values to metres and others to millimetres in the same table.
Round the uncertainty to a sensible precision and match the decimal place of the stated value to it.

Techniques used

calculate a reciprocal and its absolute uncertaintycalculate a derived quantity using a given formulapresent calculated values to appropriate significant figures

(c)

(i)

Plot a graph of $y / \text{mm}$ against $\frac{1}{s} / \text{mm}^{-1}$ . Include error bars for $\frac{1}{s}$ .

Graph 2.1

2M Similar

DifficultyMedium-Easy

Worked solution

Answer

Plot $y/\text{mm}$ (vertical) against $1/s\ (\text{mm}^{-1})$ (horizontal) using the points:

$(5.56,\ 3.30),\ (4.76,\ 2.89),\ (4.17,\ 2.51),\ (3.70,\ 2.26),\ (3.23,\ 1.96),\ (2.63,\ 1.59).$

Include horizontal error bars on each point of size $\pm \Delta(1/s)$ :

$\pm 0.31,\ \pm 0.23,\ \pm 0.17,\ \pm 0.14,\ \pm 0.10,\ \pm 0.07\ \text{mm}^{-1}.$

Final answer

See graph

Detailed explanation

Background Concept

For a graph to test a relationship, you must:

put the correct variable on each axis,
label axes with quantity and unit,
choose a sensible scale,
plot points accurately.

Error bars show the uncertainty range of each point. Here, only $\frac{1}{s}$ has given uncertainties, so the error bars are horizontal.

Understanding the Question

You must plot $y$ (calculated from $w/10$ ) against $\frac{1}{s}$ (calculated from $s$ ), and you must include error bars for $\frac{1}{s}$ .

Approach

Use the calculated table values from (b).
Plot each point $(x,y)$ where $x=\frac{1}{s}$ and $y$ is fringe spacing.
Draw an error bar extending left and right from each point by the absolute uncertainty in $x$ .

Step-by-Step Reasoning

Set up axes exactly as required: $y/\text{mm}$ on vertical, $1/s\ /\ \text{mm}^{-1}$ on horizontal.
Plot all six points.
For each point, draw a horizontal error bar from $x-\Delta x$ to $x+\Delta x$ at the same $y$ value.

Example: for $x = 5.56\ \text{mm}^{-1}$ with $\Delta x = 0.31\ \text{mm}^{-1}$ , the error bar runs from $5.25$ to $5.87\ \text{mm}^{-1}$ .

Key Takeaways

Horizontal error bars correspond to uncertainty in the $x$ -variable.
Clear axis labels and correct plotting are essential for later gradient determination.

Common Mistakes

Drawing vertical error bars instead of horizontal ones.
Plotting $s$ instead of $1/s$ .
Missing units on axes.

Things to Be Careful About

Use a scale that spreads the points across most of the grid (not cramped into a corner).
Plot points to at least half a small square accuracy.

Techniques used

plot data points on appropriate axes and scalesadd horizontal error bars using absolute uncertaintieslabel axes with quantity and unit

(ii)

Draw the straight line of best fit and a worst acceptable straight line on your graph. Label both lines.

2M Similar

DifficultyMedium

Worked solution

Answer

Draw a straight line of best fit through the plotted points.

Draw a worst acceptable straight line that is still consistent with all the horizontal error bars.

Label the lines clearly as “best fit” and “worst acceptable”.

Final answer

See graph

Detailed explanation

Background Concept

A best-fit line represents the overall trend of the data. A worst acceptable line represents an extreme gradient that could still plausibly fit the data given the uncertainties (i.e. it passes through all error bars).

These two lines are used to estimate uncertainty in the gradient:

best-fit gives the central value,
worst acceptable gives an extreme allowed value.

Understanding the Question

You have already plotted $y$ against $1/s$ and included horizontal error bars. Now you must add:

a best-fit straight line,
a worst acceptable straight line,
and label both.

Approach

Best-fit: balance the line so roughly equal scatter above and below.
Worst acceptable: make the gradient as steep or as shallow as possible while still crossing every point’s uncertainty range (every horizontal error bar).

Step-by-Step Reasoning

Place a ruler and adjust until the line represents the trend (do not join dot-to-dot).
For worst acceptable: pivot the ruler about one end so the line becomes as steep/shallow as possible but still intersects every horizontal error bar.
Write labels next to each line so the examiner can identify them.

Key Takeaways

Worst acceptable line must be consistent with uncertainties, not just the plotted central points.
Labeling matters: the examiner must know which line is which.

Common Mistakes

Drawing a second random line that does not pass through all error bars.
Forgetting to label the lines.
Forcing the line through every central point (not required and usually impossible).

Things to Be Careful About

If the relationship predicts a line through the origin, your best-fit line should be close to the origin, but you should still follow the plotted data and error bars.
Worst acceptable line must still be straight (use a ruler).

Techniques used

draw a straight line of best fit through plotted pointsdraw a worst acceptable line consistent with error barslabel lines clearly on a graph

(iii)

Determine the gradient of the line of best fit. Include the absolute uncertainty in your answer.

gradient = ______

2M Similar

DifficultyMedium

Worked solution

Working

Using the line of best fit, take two well-separated points on the line (example readings):

$(x_1,y_1)=(2.63,1.59),\quad (x_2,y_2)=(4.17,2.51)$

\text{gradient} = \frac{\Delta y}{\Delta x} = \frac{2.51-1.59}{4.17-2.63} = \frac{0.92}{1.54}=0.597\ \text{mm}^2 \approx 0.60\ \text{mm}^2

From the worst acceptable line, take its gradient $m_{\text{worst}}$ and find

\Delta m = \left|m_{\text{best}}-m_{\text{worst}}\right| \approx 0.02\ \text{mm}^2

Answer

\text{gradient} = (0.60 \pm 0.02)\ \text{mm}^2

Final answer

(0.60 \u00b1 0.02) mm^2

Detailed explanation

Background Concept

The gradient of a straight-line graph is

\text{gradient} = \frac{\Delta y}{\Delta x}

To reduce percentage reading error, you should use a large triangle on the line (choose points far apart). For uncertainty in gradient, Paper 5 commonly uses:

\Delta m = \left|m_{\text{best}} - m_{\text{worst}}\right|

where $m_{\text{worst}}$ is the gradient of a worst acceptable straight line consistent with the error bars.

Understanding the Question

You have drawn two lines on the same $y$ vs $1/s$ graph:

best-fit line,
worst acceptable line.

You must calculate the best-fit gradient and an absolute uncertainty based on how different the worst acceptable gradient is.

Approach

Read two widely separated points on the best-fit line (not necessarily on plotted data points).
Compute $m_{\text{best}} = \Delta y/\Delta x$ .
Repeat for the worst acceptable line to get $m_{\text{worst}}$ .
Use $\Delta m = |m_{\text{best}} - m_{\text{worst}}|$ .

Step-by-Step Reasoning

diagram-ciii-gradient

Pick two points on the best-fit line with a large horizontal separation.
Calculate the differences:

\Delta y = y_2-y_1,\quad \Delta x = x_2-x_1

Divide to get the gradient and include units:
- $y$ is in mm,
- $x$ is in mm $^{-1}$ ,
- so gradient units are mm $^2$ .
For the uncertainty, repeat the gradient calculation for the worst acceptable line.
The absolute uncertainty is the difference between the two gradients.

Key Takeaways

Use a large triangle to improve accuracy.
Worst acceptable line + best-fit line is the standard method for gradient uncertainty in Paper 5.

Common Mistakes

Using two adjacent grid points (tiny triangle) leading to large rounding error.
Calculating $\Delta x/\Delta y$ instead of $\Delta y/\Delta x$ .
Using two data points instead of two points on the best-fit line.
Quoting uncertainty as a percentage when the question asks for absolute uncertainty.

Things to Be Careful About

Make sure you use the same units as the axes when calculating the gradient.
Round the uncertainty sensibly (usually 1 s.f. or 2 s.f.) and match the main value to that precision.

Techniques used

determine gradient using a large triangle on a best-fit linedetermine a gradient uncertainty from a worst acceptable linereport a gradient with absolute uncertainty and correct units

(d)

The distance between the slide and the screen is measured several times:

0.929\text{ m} \quad 0.918\text{ m} \quad 0.913\text{ m} \quad 0.918\text{ m} \quad 0.927\text{ m}.

Determine the mean distance $D$ . Include the absolute uncertainty.

$D$ = ______ $\text{m}$

1M Similar

DifficultyMedium-Easy

Worked solution

Working

Mean distance:

\bar{D}=\frac{0.929+0.918+0.913+0.918+0.927}{5}=0.921\ \text{m}

Uncertainty (half-range):

\Delta D=\frac{0.929-0.913}{2}=0.008\ \text{m}

Answer

D=(0.921\pm 0.008)\ \text{m}

Final answer

(0.921 \u00b1 0.008) m

Detailed explanation

Background Concept

When a quantity is measured repeatedly, the best estimate is usually the mean. A simple estimate of the absolute uncertainty from repeats is half the range:

\Delta D \approx \frac{D_{\max}-D_{\min}}{2}

This reflects the spread of the readings.

Understanding the Question

You are given five measurements of the distance $D$ between the double slit and the screen. You must find:

the mean $D$ ,
an absolute uncertainty.

Approach

Add all five values and divide by 5 to get $\bar{D}$ .
Find the maximum and minimum readings.
Compute $\Delta D$ as half the range.

Step-by-Step Reasoning

Mean:

\bar{D}=\frac{\sum D}{5}

Range method:

D_{\max}=0.929\ \text{m},\quad D_{\min}=0.913\ \text{m}

\Delta D = \frac{0.929-0.913}{2}=0.008\ \text{m}

So report $D$ with the uncertainty.

Key Takeaways

Mean gives the best central value; half-range gives a quick uncertainty estimate.

Common Mistakes

Using the full range instead of half the range.
Quoting too many decimal places compared with the uncertainty.

Things to Be Careful About

Keep units as metres throughout.
Make the number of decimal places in $D$ consistent with the uncertainty (here to $0.001\ \text{m}$ is fine).

Techniques used

calculate a mean from repeated readingsestimate absolute uncertainty from half the rangereport a value with consistent decimal places and uncertainty

(e)

(i)

Using your answers to (a), (c)(iii) and (d), determine the value of $\lambda$ . Include an appropriate unit.

$\lambda$ = ______

2M Similar

DifficultyMedium

Worked solution

Working

From (a),

\text{gradient}=\lambda D \;\Rightarrow\; \lambda=\frac{\text{gradient}}{D}

Convert gradient to SI:

0.60\ \text{mm}^2=0.60\times 10^{-6}\ \text{m}^2

Using $D=0.921\ \text{m}$ :

\lambda=\frac{0.60\times 10^{-6}}{0.921}=6.5\times 10^{-7}\ \text{m}

Answer

\lambda = 6.5\times 10^{-7}\ \text{m}

Final answer

6.5 \u00d7 10^-7 m

Detailed explanation

Background Concept

From double-slit interference geometry, the fringe spacing $y$ is inversely proportional to slit separation $s$ :

\lambda = \frac{sy}{D}\quad \Rightarrow\quad y = \lambda D\left(\frac{1}{s}\right)

So the gradient of a graph of $y$ vs $1/s$ is $\lambda D$ . This allows $\lambda$ to be found from experimental data.

Understanding the Question

You already determined:

expression for gradient: $\text{gradient}=\lambda D$ ,
numerical gradient from the graph,
mean distance $D$ .

Now you must find the wavelength $\lambda$ with a suitable unit.

Approach

Rearrange: $\lambda = \text{gradient}/D$ .
Ensure units are consistent (convert mm $^2$ to m $^2$ ).
Substitute values.

Step-by-Step Reasoning

Rearrangement:

\lambda=\frac{\text{gradient}}{D}

Unit conversion is essential because $D$ is in metres. If the gradient was read in mm $^2$ :

1\ \text{mm} = 10^{-3}\ \text{m}\Rightarrow 1\ \text{mm}^2 = 10^{-6}\ \text{m}^2

So multiply the mm $^2$ value by $10^{-6}$ to get m $^2$ .

Substitute into $\lambda=\text{gradient}/D$ to get $\lambda$ in metres.

Key Takeaways

The slope of a carefully chosen linear graph can directly give physical constants.
Always check and convert units before calculating constants.

Common Mistakes

Forgetting to convert mm $^2$ to m $^2$ (gives wavelength wrong by $10^6$ ).
Using $\lambda = D/\text{gradient}$ (inverting incorrectly).

Things to Be Careful About

Quote the final wavelength to a sensible number of significant figures, consistent with the gradient and $D$ values.
Use SI unit $\text{m}$ (or nm if explicitly converted and stated).

Techniques used

use the gradient relationship to determine a physical constantconvert gradient units into SI unitssubstitute measured values into an equation and compute a result

(ii)

Determine the percentage uncertainty in $\lambda$ .

percentage uncertainty in $\lambda$ = ______ $\%$

1M Similar

DifficultyMedium-Easy

Worked solution

Working

Since

\lambda = \frac{\text{gradient}}{D}

\frac{\Delta \lambda}{\lambda} = \frac{\Delta(\text{gradient})}{\text{gradient}} + \frac{\Delta D}{D}

\%\ \text{uncertainty in }\lambda = \left(\frac{0.02}{0.60} + \frac{0.008}{0.921}\right)\times 100 = (0.0333+0.00869)\times 100 \approx 4.2\%

Answer

\text{percentage uncertainty in }\lambda = 4.2\%

Final answer

4.2 %

Detailed explanation

Background Concept

For multiplication/division, fractional (or percentage) uncertainties add:

Q = \frac{A}{B}

then

\frac{\Delta Q}{Q} \approx \frac{\Delta A}{A} + \frac{\Delta B}{B}

This is the standard A-Level rule used for experimental uncertainties.

Understanding the Question

You found $\lambda$ using

\lambda = \frac{\text{gradient}}{D}

You also have absolute uncertainties in the gradient and in $D$ . You must combine them to get the percentage uncertainty in $\lambda$ .

Approach

Convert each absolute uncertainty into a fractional uncertainty.
Add the fractional uncertainties (because it is a division).
Multiply by 100 to get a percentage.

Step-by-Step Reasoning

Fractional uncertainty in gradient:

\frac{\Delta m}{m} = \frac{0.02}{0.60}

Fractional uncertainty in $D$ :

\frac{\Delta D}{D} = \frac{0.008}{0.921}

Add them to get $\Delta\lambda/\lambda$ and multiply by $100$ .

Key Takeaways

For $\lambda = m/D$ , uncertainties in $m$ and $D$ both contribute.
Percentage uncertainty is often dominated by the quantity with the larger fractional uncertainty (here, the gradient).

Common Mistakes

Subtracting uncertainties because it is a division (they still add as fractions).
Using absolute uncertainties directly without dividing by the measured value first.

Things to Be Careful About

Use consistent significant figures: a final percentage uncertainty is usually quoted to 1 d.p. or 2 s.f.
Ensure you use the same $m$ and $\Delta m$ values as in part (c)(iii).

Techniques used

propagate percentage uncertainties for a quotientcombine independent percentage uncertainties by additionreport a final percentage uncertainty to appropriate precision

(f)

The experiment is repeated. Determine the slit separation $s$ that gives a value of $y$ of $(0.500 \pm 0.005)\ \text{cm}$ . Include the absolute uncertainty.

$s$ = ______ $\text{m}$

2M Similar

DifficultyMedium

Worked solution

Working

Given $y=(0.500\pm 0.005)\ \text{cm}$ :

y = (5.00 \pm 0.05)\times 10^{-3}\ \text{m}

Using

\lambda = \frac{sy}{D} \Rightarrow s = \frac{\lambda D}{y}

With $\lambda = 6.5\times 10^{-7}\ \text{m}$ and $D=0.921\ \text{m}$ :

s=\frac{(6.5\times 10^{-7})(0.921)}{5.00\times 10^{-3}}=1.20\times 10^{-4}\ \text{m}

Uncertainty:

\%\Delta s = \%\Delta\lambda + \%\Delta D + \%\Delta y

\%\Delta s = 4.2 + \frac{0.008}{0.921}\times 100 + \frac{0.005}{0.500}\times 100

\%\Delta s = 4.2 + 0.87 + 1.0 = 6.1\%

\Delta s = 0.061\times 1.20\times 10^{-4} = 7.3\times 10^{-6}\ \text{m} \approx 0.07\times 10^{-4}\ \text{m}

Answer

s = (1.20 \pm 0.07)\times 10^{-4}\ \text{m}

Final answer

(1.20 \u00b1 0.07) \u00d7 10^-4 m

Detailed explanation

Background Concept

The double-slit formula links slit separation $s$ , fringe spacing $y$ , screen distance $D$ , and wavelength $\lambda$ :

\lambda = \frac{sy}{D}

If you want a particular fringe spacing $y$ , you can rearrange to find the required slit separation:

s = \frac{\lambda D}{y}

Uncertainty propagation for multiplication/division uses fractional uncertainties:

\frac{\Delta s}{s} \approx \frac{\Delta\lambda}{\lambda} + \frac{\Delta D}{D} + \frac{\Delta y}{y}

Understanding the Question

A repeat of the experiment aims for a fringe spacing of $0.500\ \text{cm}$ with uncertainty $\pm 0.005\ \text{cm}$ . Using your measured wavelength (and the screen distance), you must calculate the slit separation $s$ that would produce this spacing, and give its absolute uncertainty.

Approach

Convert $y$ from cm to m (including its uncertainty).
Rearrange to $s = \lambda D / y$ .
Calculate $s$ .
Find percentage uncertainty in $s$ by adding the percentage uncertainties from $\lambda$ , $D$ , and $y$ .
Convert back to an absolute uncertainty: $\Delta s = s \times (\%\Delta s/100)$ .

Step-by-Step Reasoning

Convert:

0.500\ \text{cm} = 0.500\times 10^{-2}\ \text{m} = 5.00\times 10^{-3}\ \text{m}

and

0.005\ \text{cm} = 5.0\times 10^{-5}\ \text{m}

Rearrange and substitute into

s = \frac{\lambda D}{y}

For uncertainty:
- if $\lambda$ has percentage uncertainty from part (e)(ii), use that directly,
- add $\frac{\Delta D}{D}\times 100$ ,
- add $\frac{\Delta y}{y}\times 100$ .
Convert total percentage uncertainty into absolute uncertainty in $s$ .

Key Takeaways

Smaller $s$ gives larger $y$ (since $y \propto 1/s$ ).
Always convert cm to m before using equations that expect SI units.
For products/quotients, add percentage uncertainties.

Common Mistakes

Using $s = \frac{yD}{\lambda}$ (incorrect rearrangement).
Forgetting to convert $0.500\ \text{cm}$ to metres.
Ignoring the uncertainty in $y$ .

Things to Be Careful About

Keep the uncertainty to 1–2 significant figures and match the main value’s precision.
Be consistent about whether you are using $D$ from (d) and $\lambda$ from (e), and include the corresponding uncertainties.

Techniques used

rearrange the double-slit relationship to solve for slit separationconvert centimetres to metres and propagate uncertaintycombine percentage uncertainties for products and quotients

Physics 9702/54 — October/November 2025

Variables

Apparatus and arrangement

Procedure / measurements

Analysis (test the relationship; determine β\betaβ and KKK)

Safety

Background Concept

Understanding the Question

Approach

Step-by-Step Reasoning

1) Set up the apparatus and frequency measurement

2) Ensure the radius is controlled

3) Collect data for multiple values of mmm

4) Linearise to obtain a straight-line test

5) Determine β\betaβ and KKK from the graph

Key Takeaways

Common Mistakes

Things to Be Careful About

Background Concept

Understanding the Question

Approach

Step-by-Step Reasoning

Key Takeaways

Common Mistakes

Things to Be Careful About

Working

Answer (Table 2.1 completed)

Background Concept

Understanding the Question

Approach

Step-by-Step Reasoning

Key Takeaways

Common Mistakes

Things to Be Careful About

Answer

Background Concept

Understanding the Question

Approach

Step-by-Step Reasoning

Key Takeaways

Common Mistakes

Things to Be Careful About

Answer

Background Concept

Understanding the Question

Approach

Step-by-Step Reasoning

Key Takeaways

Common Mistakes

Things to Be Careful About

Working

Answer

Background Concept

Understanding the Question

Approach

Step-by-Step Reasoning

Key Takeaways

Common Mistakes

Things to Be Careful About

Working

Answer

Background Concept

Understanding the Question

Approach

Step-by-Step Reasoning

Key Takeaways

Common Mistakes

Things to Be Careful About

Working

Answer

Background Concept

Understanding the Question

Approach

Step-by-Step Reasoning

Key Takeaways

Common Mistakes

Things to Be Careful About

Working

Answer

Background Concept

Analysis (test the relationship; determine $\beta$ and $K$ )

3) Collect data for multiple values of $m$

5) Determine $\beta$ and $K$ from the graph