A student investigates the image formed by a converging lens.
The student sets up the apparatus such that:
The student then continues to move the lens further away from the illuminated object until a second sharp image, which is diminished and inverted, is formed on the screen. The actual object distance is now Y=66.5 cmY = 66.5\text{ cm}Y=66.5 cm.
(i) Describe the appearance of the second sharp image formed on the screen in terms of its orientation and relative size compared to the original object. (ii) Calculate the value of d=Y−Xd = Y - Xd=Y−X. (iii) The focal length f1 f_1\,f1 of the lens can be found using the equation:
f1=D2−d24D f_1 = \frac{D^2 - d^2}{4D} f1=4DD2−d2Calculate the value of f1f_1f1. Give your answer to 3 significant figures.
The student repeats the procedure for an object-to-screen distance D=110.0 cmD = 110.0\text{ cm}D=110.0 cm and records the following results:
Calculate a second value, f2f_2f2, for the focal length of the lens using the same equation. Give your answer to 3 significant figures.
Two quantities can be considered to be equal within the limits of experimental accuracy if their values are within 10% of each other. State whether your values of f1 f_1\,f1 and f2 f_2\,f2 can be considered equal. Support your statement with a calculation.
This experiment is usually performed in a darkened room. Explain how this makes it easier to determine when the image is in sharp focus.
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.