1. You should be able to: 2.3.1 Draw ray diagrams for converging and diverging lenses. 2.3.2 Use the equation 1/u+1/v =1/f for converging lenses. 2.3.3 Perform and describe an experiment to measure the focal length of a converging lens. 2.3.4 Recall and use the equation m=v/u. 2.3 Lenses

2. You should be able to: 2.3.5 Describe the use of lenses to correct myopia and hypermetropia. 2.3.6 Perform calculations on the correction of long sight. 2.3.7 Perform calculations involving the lens power of converging lenses 2.3 Lenses

3. One of the most important practical applications of refraction is in lenses. The crucial property of lenses is that they can form an image. There are two types of lens: Convex (converging) or Concave (diverging) Types of Lenses Ray diagrams show what happens to parallel light rays when they pass through the thin glass lens. Each ray is refracted when it meets each surface, this follows Snell’s Law. (See refraction notes).

4. (1) Convex or Converging lens The parallel rays are converged so that they all pass through one point. This point is called the principal focus (or focal point) of the lens. It lies on the principal axis (shown in red). This is the line passing through the centre of the lens and which is perpendicular to the lens.

5. (2) Concave or Diverging lens The diverging lens makes parallel light rays spread out or diverge. They diverge as if they were coming from the one point. This point is called a virtual principal focus, because the rays do not pass through it, but diverge as if they had come from it. Glasses or contact lenses that are used to correct short sight use diverging lenses (more about this later)

6. Information about the images formed by a lens can be obtained by drawing two of the following rays: Ray Diagrams 1.A ray parallel to the principal axis which is refracted through the principal focus F. 2.A ray through the optical centre C which is undeviated for a thin lens. 3.A ray through the principal focus F which is refracted parallel to the principal axis. F F

7. You need to be able to draw these...

8. The focal length, f of a lens is the distance from its middle to the principal focus F. The focal length depends on the curvature of the glass surface, and on the type of glass that is used. A more powerful lens is one that bends the light more and has a shorter focal length. You can use this equation to calculate the power of a lens: Power = 1 2 or P = 1/f Focal length (m) Power is measured in dioptres (D) : the focal length must be in m. Focal Length and Power

9. To calculate the distance of an image from the lens, we use the lens equation: 1 + 1 = 1 2 Object distance, u Image distance, v focal length, f or 1 + 1 = 1 u v f This equation uses a system called ‘real is positive’, meaning: Distances to real objects and images are positive. Distances to virtual images are negative. Object and image distances are measured from the centre of the lens. The lens Equation

10. Method One: The Rough Method

11. Method Two: The Mirror Method

12. Method Three: Using the lens equation

13. By comparing the lens equation with the equation of the straight line, you can find a value for the focal length: 1/u + 1/v = 1/f compare with y = mx + c Where 1/u on y-axis and 1/v on x axis: 1/u = -1/v + 1/f This means the gradient (m) is -1 and the intercept on the y-axis (c) is 1/f u -1 /cm -1 v -1 /cm -1 Gradient = -1 Intercept = 1/f

14. Whenever a lens forms an image, the ratio of distances is the same as ratio of the height. This is the magnification. From similar triangles: O = I 9 u v Magnification = I = v or image height = image distance, v O u Object height object distance, u OR m = v/u Magnification of a lens u O v I

15. Example A 2.0 cm object is placed 40cm from a converging lens of focal length 15cm. Find the position and size of the image (a) 1/u + 1/v = 1/f so 1/40 + 1/v = 1/15 0.025 + 1/v = 0.066 1/v = 0.066 – 0.025 = 0.041 1/v = 0.041 => v = 24 cm The image is real, 24cm from the lens, on the opposite side (b) Magnification = v/u = 24/40 = 0.6 So size of image = 0.6 x 2 = 12cm

16. You should be able to: 2.3.1 Draw ray diagrams for converging and diverging lenses. 2.3.2 Use the equation 1/u+1/v =1/f for converging lenses. 2.3.3 Perform and describe an experiment to measure the focal length of a converging lens. 2.3.4 Recall and use the equation m=v/u. 2.3 Lenses

17. You should be able to: 2.3.5 Describe the use of lenses to correct myopia and hypermetropia. 2.3.6 Perform calculations on the correction of long sight. 2.3.7 Perform calculations involving the lens power of converging lenses 2.3 Lenses

18. Normal Vision Range (i) Near Point The closest point at which an eye can focus comfortably is called the near point (NP). This is the closest position an object can be brought in front of the unaided eye and be clearly seen without noticeable strain of the eye. For a normal eye the near point is 25cm. 25 cm The inverted image is formed on the back of the retina and then transmitted to the brain by the optical nerve.

19. Normal Vision Range (i) Far Point The most distant point the eye can focus is called the far point (FP). This is the position farthest away from the unaided eye at which an object can be seen clearly. For a normal eye the far point is infinity. Our ability to see clearly becomes more and more limited as we get older.

20. Short Sight (Myopia) Near objects are clearly seen but the image of a distant object is formed in front of the retina, causing blurring of distant vision. The far point has moved from infinity to somewhere closer to the eye. The eyeball may be too long or the lens too powerful (so the focal length is too short since P = 1/f ). Myopic eye with distant image formed in front of retina Myopic eye with uncorrected far point (closer to the eye than infinity)

21. Correcting Myopia VI RO The defect is corrected using a diverging (concave) lens. The focal length of the lens is chosen so it diverges the rays from the far object so that they appear to come from a virtual image at the eyes natural far point. The eye can then focus on this natural far point (the virtual image), but see distant objects as if they were at infinity (the real object). ROReal Object VIVirtual Image Myopic eye corrected with diverging (concave) lens

22. Calculating the focal length The focal length is calculated using the lens equation: 1/f = 1/u + 1/v In this case: u = ∞ v = - (person’s far point) Note: v is negative because v is virtual f = ? Example: A person has a far point of 1200mm and needs a lens to correct his myopia. What lens should they be prescribed? u = ∞ v = - 1200 f = ? 1/f = 1/u + 1/v => 1/f = (1/∞) + (-1/1200) since 1/∞ = 0 then F = -1200mm

23. Long Sight (Hypermetropia) Distant objects can be seen clearly, but near objects are blurred. Images of nearby objects are formed behind the retina. The near point of the hypermetropic eye is farther away than 25cm, the near point for a normal eye. The eyeball may be too short or the lens too weak (so the focal length is too long since P = 1/f ). Hypermetropic eye with near image formed behind of retina Hypermetropic eye with uncorrected near point (further than 25cm from the eye)

24. The defect is corrected using a converging (convex) lens, which converges the rays from the near object so that they appear to come from a virtual image at the eyes natural near point. The eye can then focus on this natural near point, but see close objects. Correcting Hypermetropia VI RO Hypermetropic eye corrected with converging (convex) lens

25. Again, you can use the lens equation to calculate the focal length required to correct the hypermetropia. In this case: u = corrected near point v = - person’s near point (NB: virtual and therefore negative) f = ? 1/f = 1/u + 1/v => 1/f = 1/u + (-1/v) Therefore 1/f = 1/u – 1/v Calculating the focal length Example A person with hypermetropia has a near point of 750mm. What is the focal length of the lens they require to correct their eyesight? What is the power of this lens? Hypermetropia => converging (convex) lens required. u = 250mm v = - 750mm f = ? 1/f = 1/u + 1/v => 1/f = 1/250 + (-1/750) Therefore 1/f = 1/250 – 1/750 F = 375mm; Power of lens = 1/f => P = 1/375 = 2.67D

26. When a person is wearing lenses their far point will be changed as well. If an object is placed at position u, the converging lens will produce a virtual image at infinity. The ‘new’ far point Object placed at u Virtual far point at infinity VI RO

27. Finding the ‘new’ far point In order to find the ‘new’ far point, we can again use the lens equation. In this case: u = person’s far point while wearing lenses v = -∞ f = focal length of the lens Example A person is using a lens of focal length 375mm. What is their far point while wearing this lens? u = person’s far point while wearing lenses v = -∞ f = 375mm Therefore: 1/f = 1/u + 1/v so 1/375 = 1/u – 1/∞ since 1/∞ = 0 Then 1/u = 1/375 so u = 375mm.

28. You should be able to: 2.3.1 Draw ray diagrams for converging and diverging lenses. 2.3.2 Use the equation 1/u+1/v =1/f for converging lenses. 2.3.3 Perform and describe an experiment to measure the focal length of a converging lens. 2.3.4 Recall and use the equation m=v/u. 2.3 Lenses

29. You should be able to: 2.3.5 Describe the use of lenses to correct myopia and hypermetropia. 2.3.6 Perform calculations on the correction of long sight. 2.3.7 Perform calculations involving the lens power of converging lenses 2.3 Lenses