A student investigates the magnification of an image produced by a thin converging lens.
The height of the illuminated object, h0h_0h0, is 2.4 cm2.4\text{ cm}2.4 cm. At a certain distance, the magnification mmm of the image is 3.003.003.00. Calculate the height of the image, hIh_{\text{I}}hI, formed on the screen.
The student records the object distance uuu and the image distance vvv for various positions. Calculate the magnification mmm for each pair of readings in the table below, using the equation:
m=vu m = \frac{v}{u} m=uvGive your answers to 2 decimal places.
| u/cmu / \text{cm}u/cm | v/cmv / \text{cm}v/cm | mmm |
|---|---|---|
| 16.016.016.0 | 48.048.048.0 | |
| 18.018.018.0 | 36.036.036.0 | |
| 30.030.030.0 | 20.020.020.0 | |
| 36.036.036.0 | 18.018.018.0 |
The relationship between uuu and mmm is given by the formula:
u=f(1+1m) u = f \left(1 + \frac{1}{m}\right) u=f(1+m1)where fff is the focal length of the lens. Use this relationship or your data to determine the value of the object distance u1u_1u1 when the magnification m=1.00m = 1.00m=1.00.
Calculate the focal length fff of the lens using your value of u1u_1u1 from Part (c) and the equation:
f=u12 f = \frac{u_1}{2} f=2u1State two experimental precautions the student should take in order to obtain accurate, clear readings of uuu and vvv.
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.