A brief note on Lenses and Their Power . Read it before a test. Try each question at the end before you open its answer.
The big ideas A convex lens is thick in the middle and brings parallel rays together at F 2 F_2 F 2 . A concave lens is thin in the middle and spreads them, as if from F 1 F_1 F 1 . Ray rules: parallel to the axis, then through F 2 F_2 F 2 ; through O O O , straight on; through F 1 F_1 F 1 , then parallel. A convex lens gives real, inverted images (object beyond F 1 F_1 F 1 ) or a larger virtual image (object inside F 1 F_1 F 1 ). A concave lens always gives a smaller, virtual, upright image. Lens formula 1 v − 1 u = 1 f \dfrac{1}{v} - \dfrac{1}{u} = \dfrac{1}{f} v 1 − u 1 = f 1 , and m = v u m = \dfrac{v}{u} m = u v . Power P = 1 f P = \dfrac{1}{f} P = f 1 with f f f in metres, in dioptres. Lenses in contact: the powers add. The facts at a glance Convex lens: object Image Used in Far away At F 2 F_2 F 2 , tiny Burning spot of sunlight Beyond 2 F 1 2F_1 2 F 1 Between F 2 F_2 F 2 and 2 F 2 2F_2 2 F 2 , smaller Camera, the eye At 2 F 1 2F_1 2 F 1 At 2 F 2 2F_2 2 F 2 , same size Between F 1 F_1 F 1 and 2 F 1 2F_1 2 F 1 Beyond 2 F 2 2F_2 2 F 2 , larger Projector Inside F 1 F_1 F 1 Object's side, virtual, larger Magnifying glass
Sign Convex lens Concave lens f f f and P P P positive negative v v v for a real image positive (never real)
A worked example A slide is 35 cm 35\ \text{cm} 35 cm from a convex lens of focal length 20 cm 20\ \text{cm} 20 cm . Where must the screen be, how large is the picture, and what is the power of the lens?
The image. 1 v = 1 20 cm − 1 35 cm = 7 − 4 140 cm = 3 140 cm \dfrac{1}{v} = \dfrac{1}{20\ \text{cm}} - \dfrac{1}{35\ \text{cm}} = \dfrac{7 - 4}{140\ \text{cm}} = \dfrac{3}{140\ \text{cm}} v 1 = 20 cm 1 − 35 cm 1 = 140 cm 7 − 4 = 140 cm 3 , so v = + 46.7 cm v = +46.7\ \text{cm} v = + 46.7 cm . The size. m = 46.7 cm − 35 cm = − 1.33 m = \dfrac{46.7\ \text{cm}}{-35\ \text{cm}} = -1.33 m = − 35 cm 46.7 cm = − 1.33 : inverted and 1.33 1.33 1.33 times larger. The power. P = 1 0.20 m = + 5 D P = \dfrac{1}{0.20\ \text{m}} = +5\ \text{D} P = 0.20 m 1 = + 5 D . The slide is between F 1 F_1 F 1 and 2 F 1 2F_1 2 F 1 , so the picture is real, inverted and larger: a projector.
Practice with new numbers
A program to try The program finds the image of a convex lens and the power of the lens.
Read it, then run it. It repeats the worked example.
1. Move the slide to u = − 15 u = -15 u = − 15 . What kind of image is it now?
2. Change f = 20 to f = -20, a concave lens. What sign does the power have?
Mistakes to avoid Using the mirror formula for a lens. For a lens it is 1 v − 1 u = 1 f \dfrac{1}{v} - \dfrac{1}{u} = \dfrac{1}{f} v 1 − u 1 = f 1 . Putting f f f in centimetres into P = 1 f P = \dfrac{1}{f} P = f 1 . Use metres. Drawing a lens's real image on the object's side. It forms on the far side. Forgetting that a concave lens has a negative f f f and P P P . Check yourself Which kind of lens can form a real image? Stuck? Show a hint Show the answerA spectacle lens has a power of − 5 D -5\ \text{D} − 5 D . What kind of lens is it, and what is its focal length? Stuck? Show a hint Show the answerAn object stands at 2 F 1 2F_1 2 F 1 of a convex lens of focal length 13 cm 13\ \text{cm} 13 cm . Where is the image? Stuck? Show a hint Show the answerLenses of + 3 D +3\ \text{D} + 3 D and + 1.5 D +1.5\ \text{D} + 1.5 D are in contact. What is the focal length of the pair? Stuck? Show a hint Show the answer