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2 nd & 3 th N.U.T.S. Workshops Gulu University Naples FEDERICO II University 3 – Imaging (thin lenses and spherical mirrors)

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Presentation on theme: "2 nd & 3 th N.U.T.S. Workshops Gulu University Naples FEDERICO II University 3 – Imaging (thin lenses and spherical mirrors)"— Presentation transcript:

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2 2 nd & 3 th N.U.T.S. Workshops Gulu University Naples FEDERICO II University 3 – Imaging (thin lenses and spherical mirrors)

3 3- Imaging 2nd & 3th NUTS Workshop ( Jan 2010) 2 Mirrors: Importance of the Shape Plane MirrorSpherical Mirror The same scene as seen reflected by:  plane mirror spherical mirror 

4 3- Imaging 2nd & 3th NUTS Workshop ( Jan 2010) 3 Spherical Mirrors Spherical mirrors appear to be cut from a section of a sphere. They can be concave or convex. CONCAVE CONVEX Center of curvature Axis of the mirror or Optical Axis Each obeys the law of reflection but each kind of mirror makes a different kind of image Center or vertex of the mirror

5 3- Imaging 2nd & 3th NUTS Workshop ( Jan 2010) 4 Concave mirror Here the beams from a distant light source such as the sun or a star By the time they arrive here (into a camera or mirror) only the nearby almost parallel beams enter Concave mirror Beams which arrive at the mirror from a close source, like Alex’s nose, are not almost parallel Whenever we speak of incoming parallel beams you can always visualize beams from the sun or a star Beams from a Distant Light Source are Parallel

6 3- Imaging 2nd & 3th NUTS Workshop ( Jan 2010) 5 Type #1 Useful Rays: Parallel to Opt. Axis R F R/2 R F Rule #1: all rays parallel to the axis are reflected such that to cross the focal point F (type #1 rays) Rule #2: incident rays coming towards the centre of curvature are reflected back on themselves (type #2 rays) The focal point F is halfway between the curvature centre C and the centre of the mirror. The focal distance is positive for concave and negative for convex mirror given the sign convention for distance axis (f=R/2 concave, f=-R/2 convex) C C

7 3- Imaging 2nd & 3th NUTS Workshop ( Jan 2010) 6 Type #3 Useful Rays: Headed for F R F R/2 R F Rule #3: incident rays headed for F (type #3 rays) are reflected so that they are parallel to the axis (this rule is just the reverse of Rule #1) C C Rule #4: Any ray striking the centre of the mirror (type #4 rays) reflects symmetrically about the mirror axis

8 3- Imaging 2nd & 3th NUTS Workshop ( Jan 2010) 7 Spherical Mirror Equation F C xOxO xIxI f hOhO hIhI x I >0 →real image in front of the mirror x I <0 →virtual image on the back of the mirror Magnification if M<0 image inverted

9 3- Imaging 2nd & 3th NUTS Workshop ( Jan 2010) 8 Images from Concave Spherical Mirrors (a) Object beyond C, image is real, reduced, and inverted. (b) Object between C and F, image is real, magnified, and inverted. (c) Object inside F, image is virtual, magnified, and upright. Concave Spherical Mirrors Rules:

10 3- Imaging 2nd & 3th NUTS Workshop ( Jan 2010) 9 Convex Spherical Mirrors R V C F f=- R/2 f=-R/2  f <0 virtual image (in the back side of mirror) M>0 ( upright ) and M<1 ( reduced )

11 3- Imaging 2nd & 3th NUTS Workshop ( Jan 2010) 10 Image Using Convex Mirror The image is always: - virtual (x I <0, on the mirror back) - upright (x I 0, M>0) - reduced in size (|x I |< |x O |, M<1)

12 3- Imaging 2nd & 3th NUTS Workshop ( Jan 2010) 11 Concave Mirror Exercise A technician stands 3.85m (from the vertex) in front of concave mirror with focal length = 5.52 m. a) The location of the tech’s image is? b) Is tech’s image virtual? c) What is its magnification ? d) What is the curvature radius of the mirror? The technician moves to 15.00 m from the concave mirror. Answer again the questions a,b and c. (back to the mirror  virtual image) M=- x I /x O =+12.72/3.85=+3.30,upright R =2 f = 11.04 m [ 8.74 m; real; -0.58]

13 3- Imaging 2nd & 3th NUTS Workshop ( Jan 2010) 12 Where is Alex's image when he is between the mirror centre and it's focal point? Let's find the image of Alex’s nose Here is a ray of type 1 from his nose reflecting off the mirror Here is a ray of type 3 from his nose reflecting off the mirror The image of the nose is at the intersection of reflected rays of type 1 and type 3. Why? Can we use a ray of type 2? Here is Alex's image Is it (a) real or (b) virtual? -Magnified or reduced? From which points can your eye see his nose? a)From all points (A, B, C)? b)From A and C, but not B? c)From B and C, but not A? d)From A only? e)From C only Center Focal point Mirror surface Axis A B C More on Spherical Mirror Image

14 3- Imaging 2nd & 3th NUTS Workshop ( Jan 2010) 13 By a concave mirror, your image is vertically expanded like a bathroom magnifying mirror but in one dimension only. Your image looks skinny Mirrors shaped like a cylinder can make you look fat or skinny! By a convex mirror, your image is compressed vertically like rear view mirror but in one dimension only. Your image looks fat Cylindrical Mirrors Virtual image on other side of mirror is stretched vertically Virtual image on other side of mirror is compressed vertically

15 3- Imaging 2nd & 3th NUTS Workshop ( Jan 2010) 14 Thin Lenses A thin lens consists of a piece of glass or plastic, ground so that each of its two refracting surfaces is a segment of either a sphere or a plane. A lens is thin when the radii of curvature (of the spheres) are much bigger than its thickness. Converging lenses are thickest in the middle and have positive focal lengths Diverging lenses are thickest at the edges and have negative focal lengths

16 3- Imaging 2nd & 3th NUTS Workshop ( Jan 2010) 15 Converging Lens and Its Focal Length, f The parallel beams pass through the lens and converge at the focal point. The parallel rays can come from the left or right of the lens. A thin lens has two focal points equidistance from the lens, corresponding to parallel rays from the left and from the right. A thin lens is one in which the distance between the surface of the lens and the center of the lens is negligible compared to f.

17 3- Imaging 2nd & 3th NUTS Workshop ( Jan 2010) 16 Diverging Lens and Its Focal Length, f The parallel beams diverge after passing through the diverging lens. The focal point is the point where the beams appear to have originated.

18 3- Imaging 2nd & 3th NUTS Workshop ( Jan 2010) 17 Imaging by a Lens f xOxO xIxI Three principal beams (rays) can help you to locate the image: 1.A beam parallel to the lens’ axis on the front side passes through the focal point on the back side. 2.A beam passing through the lens’ center does not get refracted. 3.A beam passing through the focal point on the front side emerges parallel to the lens’ axis on the back side.

19 3- Imaging 2nd & 3th NUTS Workshop ( Jan 2010) 18 Thin Lenses Equation xOxO xIxI Optical axis f > 0 converging lensf < 0 diverging lens

20 3- Imaging 2nd & 3th NUTS Workshop ( Jan 2010) 19 3- Imaging 2nd & 3th NUTS Workshop ( Jan 2010) 19 Thin Lenses Equation: Sign Convention X I > 0 real image, in the opposite “space” of the object with respect to the lens xOxO xIxI xPxP f xIxI X I < 0 virtual image, in the same “space” of the object with respect to the lens

21 3- Imaging 2nd & 3th NUTS Workshop ( Jan 2010) 20 Lens magnification xOxO xIxI Optical axis The two shaded triangles, blue and gold, are right-angled and similar. Therefore, for the lens magnification, M, we have: M>0 image upright M<0 image inverted

22 3- Imaging 2nd & 3th NUTS Workshop ( Jan 2010) 21 Ray Diagram for Converging Lens #1 If the object is further away from the lens than the front focal point, x O > f  the image is real, inverted, and located behind the lens If the object is located at more than 2f, x O > 2f  the image is reduced. If the object is located at f < x O < 2f the image is magnified.

23 3- Imaging 2nd & 3th NUTS Workshop ( Jan 2010) 22 Ray Diagram for Converging Lens #2 If the object is closer to the lens than the front focal point, x O <f  the image is virtual, upright, magnified, and in front of the lens The image can only be viewed from behind the lens!

24 3- Imaging 2nd & 3th NUTS Workshop ( Jan 2010) 23 xOxO f xIxI Ray Diagram for Diverging Lens The situation is pretty much the same no matter where the object is located. The image is virtual and located in the object space. The image is upright and is always reduced in size compared with the object.

25 3- Imaging 2nd & 3th NUTS Workshop ( Jan 2010) 24 O I O I Lens and Apertures What happens to the image if we put in an aperture? The image becames less bright !

26 3- Imaging 2nd & 3th NUTS Workshop ( Jan 2010) 25 Masking Half a Lens O I O I What happens to the image if we mask the top half of the lens ? The image becames less bright !

27 3- Imaging 2nd & 3th NUTS Workshop ( Jan 2010) 26 Given the Object and its Image, Find the Lens! O I O I

28 3- Imaging 2nd & 3th NUTS Workshop ( Jan 2010) 27 Given the Object and its Image, Find the Lens! #2 O I O I

29 3- Imaging 2nd & 3th NUTS Workshop ( Jan 2010) 28 Image Formation with a Lens #1 A light bulb is +56 cm from a converging lens. The image appears on a screen +31cm on the other side of the lens. (a) What is the focal length of the lens? (b) What is the magnification factor of the image? Is the image inverted?YES!

30 3- Imaging 2nd & 3th NUTS Workshop ( Jan 2010) 29 Image Formation with a Lens #2 A lens has a focal length of +35 cm. (a) Find the type and height of the image when a 2.2 cm high object is placed at f +10 cm from the lens. The image is real, behind the lens, and inverted. (b) Describe the image you got

31 3- Imaging 2nd & 3th NUTS Workshop ( Jan 2010) 30 Image Formation with a Lens #3 A lens has a focal length of +35 cm. (a) Find the position and height of the image when a 2.2 cm high object is placed at f -10 cm from the lens. The image is virtual, in front of the lens, and upright. (b) Describe the image you got

32 3- Imaging 2nd & 3th NUTS Workshop ( Jan 2010) 31 3- Imaging 2nd & 3th NUTS Workshop ( Jan 2010) 31 Beams scattered off translucent screen particles at each point Translucent Screen at the Location of Real Image 1.To see the Alex image your eyes have to catch beams refracted by the lens (you have to stay behind the lens) 2.A translucent screen is put at the location of the image 3.Beams from each image point are now scattered by the screen in all directions 4.Now we can see Alex image from the front, the side and the back of the screen

33 3- Imaging 2nd & 3th NUTS Workshop ( Jan 2010) 32 3- Imaging 2nd & 3th NUTS Workshop ( Jan 2010) 32 Negative focal length f means the lens is diverging. Otherwise it is converging. Negative image distance x I from image to lens means the image is at the left of the lens, on same side as object and virtual (beams coming from it never really meet) Otherwise image is real Negative object distance x O means the object seen by the lens is on the “wrong” side of the lens (at the right of lens) Negative magnification M means the image is upside down (inverted) with respect to the object. Meaning of Negative Numbers in the Lens Equation

34 3- Imaging 2nd & 3th NUTS Workshop ( Jan 2010) 33 3- Imaging 2nd & 3th NUTS Workshop ( Jan 2010) 33 Definition of Power, P, in terms of f is Meaning of P P is a measure of the beams bending power of the lens Large P means the lens bends beams more than if P was small The eyeglass or contact lens prescription is usually given in diopters (P) The power of a converging lens is always positive because f is a positive number for a converging lens The converging lens always bends light beams towards the axis behind the lens The power of a diverging (concave) lens is always negative because f is negative for a diverging lens The diverging lens always bends light beams away from the axis behind the lens The “Power” of a Lens

35 3- Imaging 2nd & 3th NUTS Workshop ( Jan 2010) 34 3- Imaging 2nd & 3th NUTS Workshop ( Jan 2010) 34 Power for a Systems of Two Thin Lenses For two lenses, of power P 1 and P 2, in contact to each other, the total power of the lenses system is: P total = P 1 +P 2 or in terms of the two lenses focal f 1 and f 2 :

36 3- Imaging 2nd & 3th NUTS Workshop ( Jan 2010) 35 3- Imaging 2nd & 3th NUTS Workshop ( Jan 2010) 35 Home Work #1 A candle is +36 cm from a concave mirror with a focal length +15 cm. (a) Where is the image? (b) What is the magnification? (c) Is the image real or virtual, upright or inverted? Home Work #2 The image of an object in a +27 cm focal length concave mirror is upright and magnified by a factor of 3. Where is the object? A magnifying lens has a focal length of +30 cm. How far from the page should you hold the lens in order to see the print enlarged 3X? Home Work #3


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