![]() Fig.1 Burning paper with a magnifying glass. Hold a magnifying glass to sunlight over a paper. Focus the light to a point moving the glass up and down. In a few seconds, the paper will be smoldering. Fig.2.Magnifying glass can invert an image. Ali stood outside a window, when Nelly held the lens inside a room, an upside
down image appeared on an opposite wall. They were a bit puzzled at first. After much deliberation they came to the conclusion that the light rays would
have crossed at the lens as shown here. The change of direction is due to a phenomenon called refraction. When a ray enters a medium such as glass or water the direction changes. This is explained in the Light-introduction. What is shown in blue is solid glass. A , B and C are light rays incident on this. You can see that they converge to F. Do you know why? The short yellow lines are the normals to the respective surfaces. At no. 1 the ray A enters the glass and bends towards the normal. At the junction 2, it bends away from the normal as it enters air. The same thing happens for the ray C in opposite directions as the prism is inverted. The ray B passes un-deviated. All the parallel rays to A,B and C will not behave exactly like this. But in a lens the surfaces are curved the rays can be made to go and cross at a point. This point is the focus. We follow the following system in drawing diagrams.
Fig.4 Parallel beam passing through a convex lens and a concave lens. Convex lens converges the light. To the Focus. Concave lens diverges the light. From the Focus. Formation of images In order to get the nature and position of images we need only a few rays. The following rays are usually sufficient. a. One parallel to the principal axis.b. One through the optical centre. c. One passing through the focus
Concave lenses. Lens formula Instead of ray diagrams we may use this formula to calculate the image distance. v = Image distance. u= Object distance, f = Focal length. 1/v + 1/u = 1/f or { 1/di + 1/ do = 1/f } Magnification formula Height of image = Image distance Height of object Object distance. Example
1.0i) Show how the two rays will emerge through the glass prisms. ii) Name the line marked in yellow. 5 marks 2.0 Draw a diagram and describe a method to find the focal length of a double convex lens using sun light. 5x2=10 marks
3.0
Show how the 5 rays will emerge through the lens. 10 marks. 4.0 Show how the 5 rays will emerge through the lens. 10 marks. 5.0 Red and blue spots are 2 light emitting diodes. i) Where will be the 2 sharpest images be? ii) If the lens is moved to F what changes can be observed in the images? 5x3=15 marks. 6.0 i) Draw a ray diagram to show the image of the object shown. ii) Describe the image. 10x2=20 marks. 7.0 The diagram shows an image formed by a double convex lens. Use ray diagrams to find the position of the object. 10 marks. 8.0 An object 2cm high is placed 15 cm away from a convex lens whose focal length is 5cm. i) Calculate the position of the image and describe it . ii) Describe the image. 10 marks. 9.0 What kind of lenses are normally used for the following. .i) Camera, ii) Flash light, iii) Door viewers (peepholes.) iv) Near sightedness, v) Slide projector. 10 marks For solutions click Answer page |
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