Image formation and magnificationEdexcel A-Level Physics: Revision notes
Section 1
Real and virtual images
A real image is formed where refracted rays actually meet. It can be projected onto a screen and is inverted for a single converging lens.
A virtual image is formed where refracted rays only appear to come from after they have diverged. It cannot be shown on a screen, and for a single lens it is upright.
A converging lens forms a real image when the object is beyond the focal length (u > f) and a virtual image when the object is inside it (u < f). A diverging lens forms only virtual images of real objects.
Section 2
The lens equation
For a thin lens, 1/u + 1/v = 1/f, where u is the object distance, v the image distance and f the focal length, all measured from the lens.
The real-is-positive convention gives signs:
- u is positive for a real object
- v is positive for a real image and negative for a virtual image
- f is positive for a converging lens and negative for a diverging lens
Rearrange as 1/v = 1/f − 1/u. If v comes out negative, the image is virtual.
Take the reciprocal at the end. Calculating 1/v = 20 m⁻¹ gives v = 0.05 m, not 20 m.
Section 3
Magnification
Linear magnification is m = image height ÷ object height = v/u. It has no unit.
- m greater than 1: the image is magnified
- m less than 1: the image is diminished
Magnitudes are usually enough in calculations. If you keep the sign, a negative v gives a negative m for a virtual image, so say which you are giving.
State whether you are giving the magnitude. Use h_image = m × h_object for image sizes.
Section 4
Worked example: real image
A converging lens has f = 0.100 m. An object 3.0 cm tall is 0.150 m from it.
1/v = 1/0.100 − 1/0.150 = 10.0 − 6.67 = 3.33 m⁻¹, so v = 0.300 m (real, so inverted).
m = v/u = 0.300 ÷ 0.150 = 2.0, so the image height is 6.0 cm.
Section 5
Worked example: virtual image
The same lens, with the object 0.050 m from the lens (inside f).
1/v = 10.0 − 20.0 = −10.0 m⁻¹, so v = −0.100 m.
The negative sign shows the image is virtual, 0.100 m from the lens on the object side.
m = 0.100 ÷ 0.050 = 2.0, upright. This is how a magnifying glass works: the object is placed just inside the focal length.
Section 6
Using the equation in practice
A projector places the slide just beyond f, so that a large real image forms far away. A camera has an object far beyond f, so the real image forms just beyond f and is diminished. For closer objects the lens moves away from the sensor, since v increases as u decreases.
A graph of 1/v against 1/u is a straight line with gradient −1 and intercepts of 1/f on both axes, which gives f.
That's the notes covered.
Carry on to the next subtopic.
Exam questions on Image formation and magnification
- A slide projector uses a thin converging lens of focal length 0.100 m. A slide, 24 mm wide, is placed 0.120 m from the lens on the principal axis, and a sharp image is formed on a distant screen.Calculate the width of the image on the screen.2 marks
- A student uses a thin converging lens of focal length 0.060 m as a magnifying glass. She holds an insect specimen 2.0 mm long on the principal axis, 0.045 m from the lens, and views it through the lens from the other side.Calculate the length of the image of the specimen.2 marks
- A camera lens of focal length 0.050 m is used to photograph a person who is 1.80 m tall and standing 4.0 m from the lens. The light-sensitive sensor is placed behind the lens at the position where a sharp image forms.Calculate the distance from the lens to the sensor.3 marks
Written by the Exaim team, led by Shaun Daswani (Head of Upper Secondary, Improve ME Institute; MSc Financial Mathematics, Imperial College London; BSc, UCL) and Jason Daswani (operational lead, Improve ME Institute; LSE).