A student investigates the magnification of the image produced by a thin converging lens using an optical bench setup.
An illuminated shape of height hO=1.8 cmh_{\text{O}} = 1.8\text{ cm}hO=1.8 cm is used as the object. A screen is moved along the bench until a sharp image of the object is formed.
If the magnification m m\,m is 4.00, calculate the height of the image, hIh_{\text{I}}hI, formed on the screen.
The student records the object distance u u\,u and the image distance v v\,v for various positions. The data are shown in the table below:
| u / cmu\text{ / cm}u / cm | v / cmv\text{ / cm}v / cm | mmm |
|---|---|---|
| 15.0 | 59.8 | |
| 18.0 | 36.3 | |
| 24.0 | 24.2 | |
| 30.0 | 19.9 | |
| 36.0 | 18.1 |
Calculate the magnification m m\,m for each value of u u\,u using the equation m=vu\displaystyle m = \frac{v}{u}m=uv. Record the values of m m\,m to 2 decimal places.
When the student plots a graph of u / cmu\text{ / cm}u / cm (y-axis) against mmm (x-axis), they determine the value of the object distance u1 u_1\,u1 at which the magnification m=1.00m = 1.00m=1.00 is exactly 24.1 cm. Use this value of u1 u_1\,u1 to calculate the focal length f f\,f of the lens using the equation:
f=u12 f = \frac{u_1}{2} f=2u1State two practical precautions that should be taken in this experiment to obtain accurate readings.
56 exam-style questions on CIE IGCSE Physics 3.2.3a Converging and diverging lenses. Each one has a worked solution and a mark scheme showing where the marks go.