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Basic Physics

The Lens Equation

R Horan & M LavelleThe aim of this package is to provide a short self assessment programme for students who want to understand image formation by lenses.Copyright c

2004rhoran@plymouth.ac.uk ,mlavelle@plymouth.ac.uk

Last Revision Date: November 9, 2009Version 1.0

Table of Contents

1.

In troduction(Refraction and Lenses)

2.

The Lens Equation

3.

Conca veLenses

4.

Magnication

5.

Final Quiz

Solutions to Exercises

Solutions to QuizzesThe full range of these packages and some instructions, should they be required, can be obtained from our web pageMathematics Support Materials.

Section 1: Introduction (Refraction and Lenses) 3

1. Introduction (Refraction and Lenses)

Lenses userefractionto produce images of objects. Optical instru- ments, such as telescopes or cameras, and your eyes work this way. The surface of the lens in the diagram below is shaped so that spherical wave fronts of light coming from a point,F, called thefocal point, emerge as parallel wave fronts (so-called plane waves).FF Note that the line drawn above through the middle of the lens is called theprincipal axis.

Section 1: Introduction (Refraction and Lenses) 4

Convex (converging) and concave (diverging) lenses are drawn as,, VW To understand image formation we useray diagrams. Here is an example for a con vexlens :FThe image of the top of the object is formed where the light rays cross. In a perfect lens all the rays from a point on the object will meet at one other point - so we only need to draw two rays!

Section 1: Introduction (Refraction and Lenses) 5

For a con vexlens , we draw the ray diagram as follows: Draw a ray from the top of the object straight through the middle of the lens. Its direction is not changed. Draw a ray from the top of the object parallel to the principal axis. It is refracted by the lens to pass through the focal point.F From the diagram we see that the image in this example isinverted. This is also an example of areal imageas the light rays pass through the image's location and may be seen on a screen placed there.

Section 2: The Lens Equation 6

2. The Lens Equation

An image formed by a convex lens is described by thelens equation1 u +1v =1f whereuis the distance of the object from the lens;vis the distance of the image from the lens andfis thefocal length, i.e., the distance of the focus from the lens.F uf vobject imageN.B.other sign conventions are sometimes used in the literature.

Section 2: The Lens Equation 7

Example 1What image is produced by placing an object6 cma way from a convex lens of focal length 3 cm The question states thatu= 6cmand f= 3cm. This can be substi- tuted into the lens equation as follows: 1u +1v =1f )16 +1v =13 1v =13 16 26
16 =16 Sov= 6cm. From theray diagramwe see that this is aninverted, real image.Fu= 6cmv= 6cm

Section 2: The Lens Equation 8

Here are two quizzes similar to the above example. Quiz

If an ob jectis

12 cm a wayfrom a con vexlens of fo callength 4 cm where will the image be? (Draw a ray diagram of how this image is formed to see if your answer is plausible.) (a)v= 6cm;(b)v= 8cm;(c)v= 16cm;(d)v= 3cm: Quiz

If an ob jectis

8 cm a wayfrom a con vexlens of fo callength 2 cm where will the image be? (Draw a ray diagram of how this image is formed to see if your answer is plausible.) (a)v=85 cm;(b)v= 6cm;(c)v= 10cm;(d)v=83 cm:

Section 2: The Lens Equation 9

Example 2What image is produced by placing an object4 cma way from a convex lens of focal length 8 cm The question states thatu= 4cmand f= 8cm. This can be substi- tuted into the lens equation as follows: 1u +1v =1f 14 +1v =18 1v =18 28
1v =18 This implies thatv=8cm. What does thisnegativeresult mean? Theray diagramon the next page claries the nature of the image.

Section 2: The Lens Equation 10

The ray diagram is drawn using the two rules from

p. 5 .Fu= 4cmv=8cmWe see that the image is on the same side of the lens as the object! This is the signicance ofv, the image position, being negative. The image is alsoupright. It is also evident that the light rays onlyappearto pass through the position of the image. No image would be cast on a screen placed at the image location. This is called avirtual image. You can also see a virtual image by looking in a mirror on a wall { your image will appear to be behind the wall!

Section 2: The Lens Equation 11

Exercise 1.From the lens equation calculate the position of the following images produced by a convex lens. Draw a ray diagram in each case and state whether the images are real or virtual , and also if they are uprigh tor in verted (clic kon the greenletters for the solutions). (a)u= 6cm; f= 2cm(b) u= 6cm; f= 4cm (c)u= 10mm; f= 15mm(d) u= 90mm; f= 15mm Quiz Whic hof the follo wingcom binationshas an uprigh timage 12 cm away from a convex lens? (a)u= 4cm; f= 3cm(b) u= 13cm; f= 1cm (c)u= 3cm; f= 4cm(d) u= 6cm; f= 6cm

Section 3: Concave Lenses 12

3. Concave Lenses

Concave lenses always produceupright, virtual images. For aconca ve lens , the lens equation is the same but the value offis nownegativ e.

Ray diagrams for such lenses are drawn using:

a ray from the top of the object through the middle of the lens; a ray from the top of the object parallel to the principal axis which the lens refracts so it seems to come from the focal point.VW F

Section 3: Concave Lenses 13

Example 3If an object is6 cmfrom a conca velens with fo callength f=3cm, what is its position and nature?

From the lens equation

1u +1v =1f 16 +1v =13 1v =26 16 =12

Sov=2cm. This is a virtual, upright image.

Concave lensesalways create

virtual imagesbetweenthe object and the lens.VW 326

Section 3: Concave Lenses 14

Here are some exercises with concave lenses. Becareful with the sign of the focal length in the lens formula! Exercise 2.Calculate the position of the images formed by the fol- lowingconcave lenses. (click on thegreenletters for the solutions). (a) F oran ob jectwith u= 12cmif the fo callength is 4cm (b) F oran ob jectwith u= 6cmif the fo callength is 2cm (c) F oran ob jectwith u= 4cmif the fo callength is 8cm (d) F oran ob jectwith u= 240mmif the fo callength is 8cm In the next section we will see how knowing the positions of the object and image enables us to work out the magnication of the image.

Section 4: Magnication 15

4. Magnication

The magnicationMof an image is the ratio of the height of the image to the height of the object:M=image heightobject height This number is a dimensionless ratio (a length over a length) and does not have any units. To calculateMfor virtual images, consider the diagram below:

From similar triangles:

h iv =hou hivh o=1u hih o=vuh oh i vu

Section 4: Magnication 16

This result also holds for real images:

h o h iuv

From the diagram, using similar triangles:

h ou =hiv and just as before: M=hih o=vu i.e., exactly the same equations as before.

Section 4: Magnication 17

Rule:The magnication

factorMof a lens is always positive and given by:M=jvju Example 4If an object is8 cmfrom a lens an dpro ducesan image 4 cm from the lens, what is the magnication factor, M? M=vu =48 =12 Thus the image will be half the size of the object. Exercise 3.Calculate the magnication factorMor size of image for the following cases. (click on thegreenletters for the solutions). (a)Mifu= 6cmand v= 24cm (b)Mifu= 6cmand v=12cm (c) The image heigh thiof an object with height2 cm, ifu= 3cm andv=9cm

Section 5: Final Quiz 18

5. Final Quiz

Begin Quiz

Cho osethe solut ionsfrom the options giv en.

1.For a convex lens of focal length 3cm, where will the image of an

object 12cm in front of the lens appear? (a)9cm(b) 4 cm( c)6 cm(d) 3 cmquotesdbs_dbs17.pdfusesText_23

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