2 Learning objectives At the end of this unit you should be able to : 1. Describe the action of converging lens and diverging lens on a beam of light.2. Define the term focal length of a converging lens.3. Draw a ray diagram to illustrate the formation of real and virtual images of an object by a converging lens.
3 Types of converging lens: A converging (or convex) lens is thicker inthe middle than at the edge.(a) biconvex (b) plano-convex(c) concavo-convex
4 Technical Terms:The principal axis of a lens is a line passing through the optical centre, C, of the lens and perpendicular to the plane of the lens.principalaxisOptical centre, C
5 The optical centre, C, of a lens is the point midway between the lens surfaces on its principal axis. Rays passing through the optical centre are not deviated.principalaxisOptical centre, C
6 The principal focus, F, of a thin converging lens is the point on the principal axis, to which an incident beam parallel to the principal axis is made to converge.F: focusFCFFocal length
7 The focal length, f, of a lens is the distance between the optical centre and the principal focus, F.FCFFocal lengthf
8 The focal plane of lens is the vertical plane which passes through the principal focus and perpendicular to the principal axis.Focal Plane
9 The object distance, u, is the distance between the optical centre and the object. The image distance, v, is the distance between the image and the optical centre. Since light can pass through a lens either the left or right side, a lens has two principal foci.Object distanceuImage distancevobjectF Fimage
11 (For construction of ray diagram) Standard rays(For construction of ray diagram)The ray must parallel to the principal axlestep step step 3CFF
12 Construction rules Incident ray through the optical centre, C Incident ray parallel to the principal axisIncident ray directed towards principal focus, F’objectFCF’Image
13 Formation of image by a converging lens: Converging lens of focal length = fObject distance = uImage distance = v ( 6 Cases)Given:(continue on next slide)
14 Case 1:u > 2fNature of image: real, inverted and diminishedUses: In a camera, in your eye at this momentuvobjectimage2F F F2f
15 Case 2: u = 2f u v 2f Nature of image: real, inverted and same size Uses: Photocopieruvobject2F F F Fimage2f
16 Case 3:Nature of image: real, inverted and magnifiedUses: slide projector, film projector, objective lens of microscopef<u<2fuvobjectF Fimagef
17 Case 4:No image is formed. (Image is formed at infinity). Depending on usageUses: spotlights, eyepiece of telescopeu=fuobjectF Ff
18 Case 5: 0<u<f Nature of image: virtual, upright and magnified Uses: magnifying glass, spectacles for correcting long-sightedness0<u<fimageobjectuF Ffv
19 Case 6: Object at infinite position v f Image nature: real, inverted and diminished imageUses: objective lens of telescopev2F F Fimagef
20 Worked example 1:Given: focal length of convex lens, f, object distance, u and its size. Find by graphically the size and the nature of its image produced.objectF FThe image obtained:Real, inverted andmagnified.Image distance, v > fimage
21 Worked example 2:Given: focal length of convex lens, f, image distance, v and its size. Find graphically the size and the position of the object.objectF Fimage
22 Worked example 3:Given: the size and position of distance of an object and its image . Find by graphically the focal length, f, and the position of the lens.focal lengthfobjectFimage
23 Formation of virtual images by a convex lens: (Case 5) When an object is placed within the focal length of a convex lens, the image formed is virtual, upright and magnified.This principal is used in a magnifying glass.
24 Virtual Image:The image formed by this way is a virtual image, please explain?
25 To find the focal length of a convex lens: Approximate methodPlace a screen at the back of a convex lens. Adjust the position of the lens until a clear image of distance object is obtained on the screen. The distance between the lens and the screen gives the focal length of the convex lens.
26 Types of diverging lens: A diverging (or concave) lens is thicker inat the edge than at the middle.
27 Action of diverging lens on a beam of light If the lens is concave, a beam of light passing through the lens is diverged (spread); the lens is thus called a negative or diverging lens. The beam after passing through the lens appears to be emanating from a particular point on the axis in front of the lens; the distance from this point to the lens is also known as the focal length, although it is negative with respect to the focal length of a converging lens.
28 The Thin-Lens Equation and the Magnification Equation Thin Lens Equation: 1/do + 1/di = 1/fMagnification Equation: m = hi/ho = - di/do
29 Summary of sign conventions for lenses Focal lengthf is + for a converging lens.f is – for a diverging lens.Object distancedo is + if the object is to the left of the lens (real object), as is usual.do is - if the object is to the right of the lens (virtual object).Image distancedi is + for an image (real) formed to the right of the lens by a real object.di is - for an image (virtual) formed to the left of the lens by a real objectMagnificationm is + for an image that is upright with respect to the object.m is - for an image that is inverted with respect to the object.
30 Example: A 1. 70m tall person is standing 2. 50m in front of a camera Example: A 1.70m tall person is standing 2.50m in front of a camera. The camera uses a converging lens whose focal length is 0.05m. (a) Find the image distance (the distance between the lens and the film) and determine whether the image is real or virtual. (b) Find the magnification and the height of the image on the film.Solution:(a) To find the image distance di, we use the thin-lens equation with do = 2.50m and f = 0.05m:1/di = 1/f – 1/do = 1/0.05 – 1/2.50 = 19.6 m-1di = 0.051mSince the image distance is a positive number, a real image is formed on the film(b) The magnification follows from the magnification equation:m = - di/do = - (0.051/2.50) =The image is times as large as the object, and it is inverted since m is negative. Since the object height is ho = 1.70m, the image height ishi = mho = ( )(1.70) = m
31 Example: An object placed 7 Example: An object placed 7.10m to the left of a diverging lens whose focal length is f = -5.08cm (a diverging lens has a negative focal length). (a) Find the image distance and determine whether the image is real or virtual. (b) Obtain the magnification.Solution:The thin-lens equation can be used to find the image distance di:1/di = 1/f – 1/do = 1/(-5.08) – 1/7.10 = cm-1di = cmThe image distance is negative, indicating that the image is virtual and located to the left of the lens.(b) Since di and do are known, the magnification equation shows thatm = - di/do = - (-2.96/7.10) = 0.417The image is upright (m is +) and smaller (m < 1) than the object.
32 Application of Converging Lenses A camera consists of converging lens and light sensitive film mounted in a light-tight box. The lens can be moved to or fro so that a real, inverted, diminished and sharp image is focused on the film. The intensity of light that falls onto the film is controlled by a shutter and a variable aperture diaphragm. The shutter controls the length of time that the film is exposed to light. The diaphragm controls the aperture that allows light to pass through.
33 A projector essentially uses converging lenses to produce a real and magnified image. The condenser lenses direct the light through the slide or film to a projection lens. The projection lens is moved to and fro until a real, magnified and sharp image is focused to the screen. Since the image formed on the screen is inverted, the slide or film has to be put upside down. The real image formed on the screen is then the right way up.
34 A magnifying glass produces a virtual, upright and magnified image A magnifying glass produces a virtual, upright and magnified image. The image appears to be larger and more distant than the object but it cannot be projected on a screen.
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