Aim
To investigate the relationship between the object distance u and the image distance v for a convex lens and to find its focal length.
Variables
- Manipulated: Object distance, u
- Responding: Image distance, v
- Constant: The same convex lens (same focal length)
Apparatus & materials
- Convex lens
- Lens holder
- Illuminated object (light source with a cross-wire or arrow)
- White screen
- Metre rule (optical bench)
Procedure
- Estimate the focal length by focusing a distant object sharply onto the screen and measuring the lens-to-screen distance.
- Set the illuminated object at an object distance u = 15.0 cm from the lens (greater than the focal length).
- Move the screen until a sharp, inverted image forms, then measure the image distance v.
- Repeat for u = 20.0, 25.0, 30.0 and 35.0 cm.
- Record u and v, and calculate 1/u and 1/v for each setting.
Tabulating results
Record the object distance u and the image distance v in cm, and the derived values 1/u and 1/v in cm⁻¹. Keep the decimal places consistent.
The graph
Plot 1/v (y-axis) against 1/u (x-axis). The straight line cuts each axis at 1/f, so the intercept gives the focal length.
Analysis
The lens formula is 1/f = 1/u + 1/v. The intercept of the 1/v against 1/u graph on either axis equals 1/f, so f = 1 ÷ intercept.
Precautions
- Line up the centres of the object, lens and screen at the same height.
- Adjust the screen for the sharpest possible image before measuring.
- Measure u and v from the centre of the lens.
Light and Optics · Graph skills
Sample results and what they show
These are example readings for a convex lens, with the object distance u and image distance v measured on an optical bench, and the reciprocals worked out.
- u = 15.0 cm, v = 30.0 cm, 1/u = 0.0667 cm⁻¹, 1/v = 0.0333 cm⁻¹
- u = 20.0 cm, v = 20.0 cm, 1/u = 0.0500 cm⁻¹, 1/v = 0.0500 cm⁻¹
- u = 25.0 cm, v = 16.7 cm, 1/u = 0.0400 cm⁻¹, 1/v = 0.0599 cm⁻¹
- u = 30.0 cm, v = 15.0 cm, 1/u = 0.0333 cm⁻¹, 1/v = 0.0667 cm⁻¹
- u = 35.0 cm, v = 14.0 cm, 1/u = 0.0286 cm⁻¹, 1/v = 0.0714 cm⁻¹
As the object is moved further from the lens, the sharp image forms closer to the lens, so v falls as u rises. Each image is real and inverted because the object is always beyond the focal length.
Adding 1/u and 1/v for any row gives the same total, close to 0.100 cm⁻¹, which is 1/f. That constant sum is the message of the experiment and points straight to the focal length of the lens.
Reading the graph and finding the answer
Plot 1/v on the y-axis against 1/u on the x-axis. The points lie on a straight line sloping downwards, and the line cuts both axes at the same value, 1/f.
Read the y-intercept where the best-fit line meets the 1/u = 0 axis; here it is 0.100 cm⁻¹. Then:
- f = 1 ÷ intercept = 1 ÷ 0.100 cm⁻¹ = 10.0 cm
You can check with the x-intercept, which should also be 0.100 cm⁻¹, and a large triangle on the line gives a gradient of about −1, as expected for 1/v = 1/f − 1/u. So the focal length of the lens is 10.0 cm.
Take the intercept from the drawn line, not a single point. The SPM Physics 4531 papers give no formula sheet, so recall the lens formula 1/f = 1/u + 1/v yourself.
Choose axis scales that spread the plotted points across more than half of the grid, and draw the best-fit line so the points are balanced evenly on both sides of it. If one point is clearly off the line, treat it as an anomalous reading, ring it, and leave it out when drawing the line and taking the gradient.
Marks examiners look for
Precautions that keep u and v accurate:
- Line up the centres of the illuminated object, lens and screen at the same height.
- Adjust the screen for the sharpest image before reading each v.
- Measure u and v from the centre of the lens along the bench.
- Work in a darkened room so the image is easy to focus.
For the Paper-3 science process skills, examiners award marks for: naming the manipulated variable (u), responding variable (v) and constant (the same lens); tabulating u, v, 1/u and 1/v with units and consistent decimals; plotting at least five points with labelled axes and a best-fit straight line; and reading the intercept carefully to find f. State the relationship, that 1/v decreases as 1/u increases, and quote the focal length with its unit as the conclusion.
Repeat each reading and take the average to reduce random error, and quote the final answer to a sensible number of significant figures with its correct unit. In the discussion, name one source of error together with a matching improvement, and note that any point off the line was ignored so the conclusion rests on consistent data.
Source: DSKP KSSM Physics Form 4 and 5 (Versi English) (Bahagian Pembangunan Kurikulum (BPK), KPM)