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. Learning objectives
Types of converging lens: A converging (or convex) lens is thicker in the middle than at the edge. (a) biconvex (b) plano-convex (c) concavo-convex
principal axis Optical centre, C 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. Technical Terms:
principal axis Optical centre, C 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.
F F: focus Focal length C 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
F Focal length f C F The focal length, f, of a lens is the distance between the optical centre and the principal focus, F.
Focal Plane The focal plane of lens is the vertical plane which passes through the principal focus and perpendicular to the principal axis.
F object image Object distance u Image distance v 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.
Construction of rays of a converging lens Construction of rays of a converging lens
F C F Standard rays (For construction of ray diagram) step 1 step2 step 3 The ray must parallel to the principal axle
Construction rules §Incident ray through the optical centre, C §Incident ray parallel to the principal axis §Incident ray directed towards principal focus, F’ F’FC object Image
Formation of image by a converging lens: §Converging lens of focal length = f §Object distance = u §Image distance = v ( 6 Cases) Given: (continue on next slide)
2F F F object image u 2f v u > 2f Nature of image: real, inverted and diminished Uses: In a camera, in your eye at this moment Case 1:
2F F F 2F object image u 2f v u = 2f Case 2: Nature of image: real, inverted and same size Uses: Photocopier
F object image u f v f
"name": "F object image u f v f
F object u f u=f No image is formed. (Image is formed at infinity). Depending on usage Uses: spotlights, eyepiece of telescope Case 4:
F object image u f v 0
"name": "F object image u f v 0
2F F F image Object at infinite position Image nature: real, inverted and diminished image Uses: objective lens of telescope v f Case 6:
F object image 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. The image obtained: Real, inverted and magnified. Image distance, v > f
F image object 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.
object image F focal length f 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. Worked example 3:
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.
Virtual Image: The image formed by this way is a virtual image, please explain?
To find the focal length of a convex lens: §Approximate method Place 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.
Types of diverging lens: A diverging (or concave) lens is thicker in at the edge than at the middle.
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.
The Thin-Lens Equation and the Magnification Equation Thin Lens Equation: 1/d o + 1/d i = 1/f Magnification Equation: m = h i /h o = - d i /d o
Summary of sign conventions for lenses Focal length f is + for a converging lens. f is – for a diverging lens. Object distance d o is + if the object is to the left of the lens (real object), as is usual. d o is - if the object is to the right of the lens (virtual object). Image distance d i is + for an image (real) formed to the right of the lens by a real object. d i is - for an image (virtual) formed to the left of the lens by a real object Magnification m 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.
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 d i, we use the thin-lens equation with d o = 2.50m and f = 0.05m: 1/d i = 1/f – 1/d o = 1/0.05 – 1/2.50 = 19.6 m -1 d i = 0.051m Since the image distance is a positive number, a real image is formed on the film (b) The magnification follows from the magnification equation: m = - d i /d o = - (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 h o = 1.70m, the image height is h i = mh o = ( )(1.70) = m
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: (a)The thin-lens equation can be used to find the image distance d i : 1/d i = 1/f – 1/d o = 1/(-5.08) – 1/7.10 = cm -1 d i = cm The image distance is negative, indicating that the image is virtual and located to the left of the lens. (b) Since d i and d o are known, the magnification equation shows that m = - d i /d o = - (-2.96/7.10) = The image is upright (m is +) and smaller (m < 1) than the object.
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.
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.
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.