A student investigates the image formed by a converging lens using an optical bench setup.
At the start of the experiment:
(i) On a scale diagram drawn to a scale of one-quarter (14\frac{1}{4}41) full size, the distance ddd on the page representing the distance from the lens to the screen is measured to be 12.0 cm12.0\text{ cm}12.0 cm. Calculate the actual image distance vvv from the lens.
(ii) Use your value of vvv to complete the first row of Table 1.
Table 1:
object distance u / cmimage distance v / cm16.020.030.024.024.036.018.048.016.0\begin{array}{|c|c|} \hline \text{object distance } u \text{ / cm} & \text{image distance } v \text{ / cm} \\ \hline 16.0 & \\ \hline 20.0 & 30.0 \\ \hline 24.0 & 24.0 \\ \hline 36.0 & 18.0 \\ \hline 48.0 & 16.0 \\ \hline \end{array}object distance u / cm16.020.024.036.048.0image distance v / cm30.024.018.016.0(i) State two ways in which the image of the letter 'P' formed on the screen differs from the original illuminated object when u=16.0 cmu = 16.0\text{ cm}u=16.0 cm.
(ii) The magnification mmm of the image is given by:
m=vu m = \frac{v}{u} m=uvCalculate the magnification mmm when u=16.0 cmu = 16.0\text{ cm}u=16.0 cm.
The focal length fff of the lens can be calculated using the equation:
f=uvu+v f = \frac{uv}{u+v} f=u+vuvUse the values from Table 1 for u=24.0 cmu = 24.0\text{ cm}u=24.0 cm to calculate the focal length fff of the lens.
Using the focal length f=12.0 cmf = 12.0\text{ cm}f=12.0 cm, calculate the expected image distance vvv when the object distance is set to u=30.0 cmu = 30.0\text{ cm}u=30.0 cm.
State one technique used during the experiment to ensure that the image on the screen is focused as sharply as possible.
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.