13.4 The Lens Equation.

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Presentation transcript:

13.4 The Lens Equation

Determining Characteristics of Images Formed by Lenses Similar to predicting image properties with mirrors, algebraic equations can be used to predict the position and size of the images formed by lenses instead of ray diagrams

Lens Terminology d0 f h0 hi di Where: d0 = distance from object to the optical centre di = distance from the image to the optical centre h0 = height of the object hi = height of the image f = focal length of the lens

The Thin Lens Equation Where: d0 is always positive di is positive for real images (opposite side of the lens as the object) and negative for virtual images (image on the same side as the object) f is positive for converging lenses and negative for diverging lenses.

Example 1: A converging lens has a focal length of 34 cm Example 1: A converging lens has a focal length of 34 cm. A real image is located 45 cm from the lens. Where is the object located? Solve: 1 – 1 = 1 f di do 1 – 1 = 1 34 45 do 0.029 – 0.022 = 1 do 0.007 X do= 1 do= 1 0.007 do= 143 cm Given: f=34cm di = 45cm Required: do = ? Analysis: Paraphrase: The object is located 143 cm from the optical centre.

Example 2: A diverging lens has a focal length of 10 cm Example 2: A diverging lens has a focal length of 10 cm. A candle is located 36 cm from the lens. What type of image will be formed and where will it be located? Solve: 1 – 1 = 1 f do di -10 36 di -0.1 – 0.03 = 1 di -0.13 X di = 1 di = 1 -0.13 di = -7.7 cm Given: f=-10cm do = 36cm Required: di = ? Analysis : Paraphrase: The image is virtual and is located 7.7 cm from the optical centre.

How does your prediction compare to the demo How does your prediction compare to the demo? Try calculating this on your own. Example 3: A converging lens has a focal length of 5.3 cm. A candle is located at 8 cm from the lens. What type of image is formed and where will it be located?

The Magnification Equation The magnification equation allows you to compare the size of the image with the size of the object. Where: M is the magnification of the lens hi is the image’s height ho is the object’s height ho and hi are positive when measured upward and negative when measured downward M is positive for an upright image and negative for an inverted image Magnification has no units

The magnification of the lens is 2.74 for the inverted image produced. Example 4: Magnification of a converging lens A toy height 8.4 cm is balanced in front of a converging lens. An inverted, real image of height 23 cm is noticed on the other side of the lens. What is the magnification of the lens? Given: ho= 8.4 cm hi= -23 cm Required: M= ? Analysis: Paraphrase: The magnification of the lens is 2.74 for the inverted image produced. Solve: M = hi ho M = -23 cm 8.4cm M = -2.74

Example 5: An object is placed 9. 7 cm in front of a converging lens Example 5: An object is placed 9.7 cm in front of a converging lens. An upright, virtual image of magnification 4.5 cm is noticed. Where is the image located? Solve: M = -di do 4.5 = di 9.7 cm di = 43.65 Given: do = 9.7cm M = 4.5 Required: di = ? Analysis: Paraphrase: The image is located 43.65 cm from the optical centre.

Try this Example 6: A converging lens has a focal length of 15.0 cm. An object of height 1.5 cm is placed 32.0 cm from the lens. Calculate the image distance and the image height.

Homework / Classwork Complete your lens equation worksheets