QuestionA converging lens has a focal length of 20 cm. An object 1.5 cm tall is placed 30 cm from the lens.(a) Calculate the power of the lens. (2)
(b) Calculate the distance of the image from the lens. (2)
(c) Calculate the magnification and the height of the image. (2)
(d) State two properties of the image. (2)Answer(a) Convert the focal length to metres: f = 0.20 m.P = 1 / f = 1 / 0.20 = 5.0 D
The power is +5.0 dioptres, positive because the lens is converging. (2)(b) Using the lens equation (real is positive):1/u + 1/v = 1/f
1/v = 1/20 − 1/30 = 3/60 − 2/60 = 1/60
v = 60 cm
The image is 60 cm from the lens, on the opposite side to the object. (2)(c)m = v / u = 60 / 30 = 2.0
Image height = 2.0 × 1.5 cm = 3.0 cm. (2)(d) The image is real (v is positive), inverted and magnified. (2)Examiner tips- Power must use the focal length in metres; 1/20 = 0.05 D is a common wrong answer.
- Magnification has no unit.
- In the lens equation you may keep all distances in cm, as long as every term uses the same unit.
(b) Calculate the distance of the image from the lens. (2)
(c) Calculate the magnification and the height of the image. (2)
(d) State two properties of the image. (2)Answer(a) Convert the focal length to metres: f = 0.20 m.
The power is +5.0 dioptres, positive because the lens is converging. (2)(b) Using the lens equation (real is positive):
The image is 60 cm from the lens, on the opposite side to the object. (2)(c)
Image height = 2.0 × 1.5 cm = 3.0 cm. (2)(d) The image is real (v is positive), inverted and magnified. (2)Examiner tips- Power must use the focal length in metres; 1/20 = 0.05 D is a common wrong answer.
- Magnification has no unit.
- In the lens equation you may keep all distances in cm, as long as every term uses the same unit.