6.4 Thin Lens Formula
The thin lens formula provides a mathematical relationship between object distance (\(u\)), image distance (\(v\)), and focal length (\(f\)) of a lens. Alongside ray diagrams, this formula enables quantitative determination of the position, size, nature, and magnification of images formed by convex and concave lenses.
Learning Objectives
- State the thin lens formula relating object distance (\(u\)), image distance (\(v\)), and focal length (\(f\)).
- State and apply the standard sign conventions for focal length and image distance for converging and diverging lenses.
- Carry out an experiment to investigate the relationship between object distance and image distance, and determine the focal length of a convex lens using graphical analysis.
- Solve numerical problems involving the thin lens formula and linear magnification (\(m = \frac{v}{u} = \frac{h_i}{h_o}\)).
Sign Convention and Experimental Setup for the Thin Lens Formula
See how distances are measured and assigned signs (\(+\) or \(-\)), and how the light source, object, lens, and screen are aligned in the experiment.
Convex Lens: \(f > 0\)
Real image forms on the opposite side of the lens (\(v > 0\)). Virtual image forms on the same side as the object (\(v < 0\)).
Concave Lens: \(f < 0\)
Virtual image forms on the same side as the object (\(v < 0\)).
Short Explanation
Thin Lens Formula
The thin lens formula relates object distance (\(u\)), image distance (\(v\)), and focal length (\(f\)):
Linear Magnification (\(m\))
Linear magnification is the ratio of image distance to object distance, or image height to object height:
Sign Convention
Focal length \(f\) is positive for convex lenses and negative for concave lenses. Image distance \(v\) is positive for real images and negative for virtual images.
Graphical Analysis
Plotting \(\frac{1}{v}\) against \(\frac{1}{u}\) gives a straight line with gradient \(-1\). The intercepts equal \(\frac{1}{f}\).
| Quantity | Positive (\(+\)) | Negative (\(-\)) |
|---|---|---|
| Focal length, \(f\) | Converging lens / convex lens | Diverging lens / concave lens |
| Image distance, \(v\) | Real image (opposite side of object) | Virtual image (same side as object) |
| Object distance, \(u\) | Real object placed in front of the lens | — |
Experiment 6.3: Thin Lens Formula for a Convex Lens
Investigate the relationship between object distance and image distance, and determine the focal length.
Aim
To study the relationship between object distance, \(u\), and image distance, \(v\), for a convex lens, and to determine the focal length of a thin lens using the lens formula.
Variables
Manipulated: Object distance, \(u\)
Responding: Image distance, \(v\)
Constant: Focal length, \(f\)
Apparatus & Materials
Convex lens (\(f = 10.0\text{ cm}\)), lens holder, \(6\text{ V}\) bulb, wooden block, power supply, cardboard with a small triangular shaped hole (object), white screen, metre rule.
Procedure
Set up the apparatus with the bulb behind the triangular hole, followed by the convex lens and white screen, all aligned along a metre rule. Set \(u = 15.0\text{ cm}\), adjust the screen until a sharp image is formed, and measure \(v\). Repeat for \(u = 20.0, 25.0, 30.0, 35.0\text{ cm}\).
Rearranging the thin lens formula gives the linear equation form:
\(\dfrac{1}{v} = -\dfrac{1}{u} + \dfrac{1}{f}\)The gradient is \(-1\). The vertical and horizontal intercepts both equal \(\frac{1}{f}\), so the focal length is \(f = \frac{1}{\text{intercept}}\).
Try to Answer First
Answer in your mind, then press “Check Answer”.
What is the sign of the focal length, \(f\), for a concave lens?
Check Answer
If an object is placed \(15\text{ cm}\) in front of a convex lens with a focal length of \(10\text{ cm}\), what is the image distance, \(v\)?
Check Answer
In an experiment plotting a graph of \(\frac{1}{v}\) against \(\frac{1}{u}\) for a convex lens, what does the vertical intercept represent?
Check Answer
Common Mistakes
- Forgetting to substitute a negative sign for the focal length (\(f\)) when dealing with concave (diverging) lenses.
- Omitting the negative sign for virtual image distance (\(v\)) when solving inverse equations.
- Calculating \(\frac{1}{v}\) correctly (e.g., \(\frac{1}{36}\text{ cm}^{-1}\)) but forgetting to take the reciprocal to state the final image distance (\(v = 36\text{ cm}\)).
- Including negative signs inside the linear magnification formula (\(m = \frac{v}{u}\)), which only uses the magnitudes of the distances.
Concept Misunderstandings
All images produced by a convex lens are real and therefore have positive image distances (\(v > 0\)).
When an object is placed closer than the focal length (\(u < f\)), a convex lens produces an upright, magnified, and virtual image on the same side as the object, giving a negative image distance (\(v < 0\)).
The gradient of a graph of \(\frac{1}{v}\) against \(\frac{1}{u}\) gives the focal length.
The gradient of the graph is constant at \(-1\); the focal length is obtained from the reciprocal of the axes intercepts (\(\frac{1}{f}\)).
Summary
- Thin lens formula: \(\frac{1}{f} = \frac{1}{u} + \frac{1}{v}\).
- Linear magnification: \(m = \frac{v}{u} = \frac{h_i}{h_o}\).
- Sign conventions: \(f\) is positive for convex lenses, negative for concave lenses; \(v\) is positive for real images (opposite side of lens), negative for virtual images (same side as lens).
- Graph of \(\frac{1}{v}\) against \(\frac{1}{u}\) yields a straight line with gradient \(m = -1\) and intercepts equal to \(\frac{1}{f}\).
Short Activity
Objective: Apply the thin lens formula, sign conventions, and graphical analysis.
A. Objective Quiz
1 What is the nature and position of an image if the calculated image distance, \(v\), has a negative value?
2 In Experiment 6.3, a student plots a graph of \(\frac{1}{v}\) against \(\frac{1}{u}\) for a thin convex lens. What does the intercept on the \(\frac{1}{v}\)-axis represent?
B. Fill in the Blanks
3 For a diverging or concave lens, the focal length, \(f\), must always be assigned a sign in the thin lens formula.
4 Linear magnification, \(m\), can be calculated using the ratio of image distance to object distance or image height to .
C. Matching / Drag and Drop
Match each optical parameter or graph feature with its correct description. If using a phone, tap the answer first, then tap the matching box.
