Light Reflection PDF

Summary

This document discusses the nature of light, including particle and wave theories. It also covers the reflection of light from plane and spherical mirrors, along with related concepts like optical energy, interference, and diffraction.

Full Transcript

LIGHT : REFLECTION of these waves. They can travel in vacuum CONTENTS also. The speed of these waves in air or in  The Nature of...

LIGHT : REFLECTION of these waves. They can travel in vacuum CONTENTS also. The speed of these waves in air or in  The Nature of light vacuum is maximum i.e., 3 × 108 m/s. Photoelectric effect was not explained with the  Reflection of light help of wave theory, so Plank gave a new theory  Laws of reflection of light which was known as quantum theory of light.  Nature of image (iii) Quantum theory of light : When light falls on the surface of metals like  Reflection from the plane mirror caesium, potassium etc., electrons are given  Reflection from spherical mirrors out. These electrons are called 'photo- electrons' and phenomenon is called 'photo-  Rules for image formation by ray electric effect'. diagram method This was explained by Einstein. According to  Image formation by spherical plank light consisted of packets or quanta's of mirrors in different cases energy called photons. The rest mass of  Numerical method in spherical mirror photon is zero. Each quanta carries energy  Summary of images by spherical E = h. mirror h Planck's constant = 6.6 × 10–34 J-s. Frequency of light Some phenomenons like interference of light, diffraction of light are explained with the help THE NATURE OF LIGHT of wave theory but wave theory was failed to Light is a form of energy (optical energy) explain the photo electric effect of light. It which helps us in seeing objects by its was explained with the help of quantum theory. So, light has dual nature. presence. (i) Wave nature (ii) Particle nature (A) THEORIES ABOUT NATURE OF LIGHT : (i) Particle nature of light (Newton's  (B) SOURCES OF LIGHT. corpuscular theory) : The objects which emit (give) light are called According to Newton light travels in space luminous objects. It may be natural or man- with a great speed as a stream of very small made. Sun is a natural source of light and particles called corpuscles. electric lamp, and oil lamp, etc. are man- made source of light. This theory was failed to explain interference The Non-luminous objects do not emit light. of light and diffraction of light. So wave However, such objects become visible due to the theory of light was discovered. reflection of the light falling on them. Moon does (ii) Wave nature of light : Light waves are electromagnetic waves so not emit light. It becomes visible due to the there is no need of medium for the propagation reflection of the sunlight falling on it. (C) PROPAGATION OF LIGHT too. Scientists have assigned a value of 299, Light travels along straight lines in a medium 792, 458 m/s to the speed of light in vacuum. or in vacuum. The path of light changes only  According to current scientific theories, no when there is an object in its path or where material particle can travel at a speed greater the medium changes. Apart from vacuum and gases, light can travel through some liquids than that of light in vacuum. and solids as well.  Transparent medium : A medium in which light can travel freely over large distances is REFLECTION OF LIGHT called a transparent medium.  Definition. When light rays are incident on Examples: Water, glycerin, glass and clear an opaque polished surface (medium), these plastics are transparent. are returned back in the same medium. This phenomenon of returning of ray of light  Opaque : A medium in which light cannot in the same medium, is called reflection of travel is called opaque. light. Examples : Wood, metals, bricks, etc., are opaque. Definition of some associated terms :  Translucent : A medium in which light can travel some distance, but its intensity reduces P N Q rapidly. Such materials are called translucent. i r Examples : Oil X Y O (D) THE CHARACTERISTICS OF LIGHT  Reflecting surface : The surface from which  Light is an electromagnetic wave. the light is reflected, is called the reflecting Light travels in a straight line. surface. In diagram, XY is the reflecting  Light is a transverse wave, and does not need surface. any medium to travel. Light can travel  Point of incidence : The point on the through vaccum. Its speed through vaccum is reflecting surface at which a ray of light 3 × 108 m/s. strikes, is called the point of incidence. In  The velocity of light changes when it travels diagram, O is the point of incidence. from one medium to another.  Normal : A perpendicular drawn on the  The wavelength () of light changes when it reflecting surface at the point of incidence, is goes from one medium to another. called the normal. In diagram, ON is the  The frequency (f) of the light wave remains normal. the same in all media.  Incident ray : The ray of light which strikes Light gets reflected back from polished the reflecting surface at the point of incidence surfaces, such as mirrors, polished metal is called the incident ray. In diagram, PO is surfaces, etc. the incident ray. Light undergoes refraction (bending) when it  Reflected ray : The ray of light reflected travels from one transparent medium to from the reflecting surface from the point of another. incidence, is called the reflected ray. In  Light does not need a material medium to diagram, OQ is the reflected ray. travel, that is, it can travel through a vacuum  Angle of incidence : The angle that the incident ray makes with the normal, is called Incident ray the angle of incidence. It is represented by the symbol i. In diagram, angle PON is the angle Reflected ray of incidence.  Angle of reflection : The angle that the Normal Incidence reflected ray makes with the normal, is called the angle of reflection. It is represented by the (ii) Laws of reflection are also obeyed when light symbol r. In diagram, QON is the angle of is reflected from the spherical or curved surfaces as shown in figure (a) and (b) reflection. Plane of incidence : The plane in which the N N I R I R normal and the incident ray lie, is called the plane of incidence. In diagram, the plane of i r i r the bookpage, is the plane of incidence. Plane of reflection : The plane in which the (a) (b) normal and the reflected ray lie, is called the plane of reflection. In diagram, the plane of Reflection from curved surface the book page, is the plane of reflection. (iii) Regular and Irregular Reflection : Regular Reflection – The phenomenon due LAWS OF REFLECTION OF LIGHT to which a parallel beam of light travelling through a certain medium, on striking some  First law : The incident ray, the reflected ray smooth polished surface, bounces off from it, and the normal at the point of incidence, all as parallel beam, in some other fixed lie in the same plane. direction is called Regular reflection.  Second law : The angle of reflection (r) is always equal to the angle of incidence (i). i.e., r = i (For normal incidence, i = 0, r = 0. The ray is reflected back along normal). (i) A ray of light striking the surface normally retraces its path. Regular reflection When a ray of light strikes a surface Regular reflection takes place from the normally, then angle of incidence is zero i.e., objects like looking glass, still water, oil, i = 0. According to the law of reflection, highly polished metals, etc. r = i, r = 0 i.e. the reflected ray is also Regular reflection is useful in the formation of perpendicular to the surface. Thus, an images, e.g., we can see our face in a mirror incident ray normal to the surface (i.e. only on account of regular reflection. However, perpendicular to the surface) retraces its path it causes a very strong glare in our eyes. as shown in figure. Irregular reflection or Diffused reflection : REFLECTION FROM THE PLANE MIRROR Relation between the distances of the object and the image from the plane mirror is that they are equal. To verify this, consider the geometrical construction shown in figure. Rays OP and OD, starting from the object O, fall on the mirror. The ray OP is perpendicular to the Irregular or diffused reflection mirror and hence, reflects back along PO. The The phenomenon due to which a parallel incident ray OD and the reflected ray DE beam of light, travelling through some make equal angles with the normal DG. The medium, gets reflected in various possible two reflected rays when produced backwards directions, on striking some rough surface is called irregular reflection or diffused meet at I, producing a virtual image there. reflection. E The reflection which takes places from D ground, walls, trees, suspended particles in G air, and a variety of other objects, which are O I not very smooth, is irregular reflection. P Irregular reflection helps in spreading light energy over a vast region and also decreases Now, EDG = DIO (DG || IO), its intensity. Thus, it helps in the general EDG = GDO (law of reflection), illumination of places and helps us to see things around us. and  Note : Laws of reflection are always valid no GDO = DOI (DG || IO). matter whether reflection is regular or Hence, DIO = DOI irregular.  OD = DI NATURE OF IMAGE Now OP2 = OD2 – DP2, and  Definition : Incident rays starting from a PI2 = DI2 – DP2 point object, and reflected from a mirror, either actually meet at or appear to come from From (i), since OD = DI, OP2 = PI2 or OP = PI. a point. The other point is called the image of So, in the case of a plane mirror, the image is the point object. formed as far behind the mirror at the same Real Image Virtual Image 1. A real image is formed 1. A virtual image is distance as the object is in front of it. when two or more formed when two or  SOME IMPORTANT RESULTS ABOUT reflected rays meet at more rays appear to REFLECTION FROM PLANE SURFACES a point in front of the be coming from a point mirror. behind the mirror.  Lateral inversion : When you see your 2. A real image can be 2. A virtual image cannot image in a vertical plane mirror such as that obtained on a screen. be obtained on a screen. fixed to an almirah, the head in the image is up and the feet are down, the same way as 3. A real image is inverted 3. A virtual image is erect with respect to the with respect to the you actually stand on the floor. Such an object. object. image is called an erect image. However, if you move your right hand, it will appear as if the left hand of your image is moving. If you  Deviation :  is defined as the angle between keep a printed page in front of a plane mirror, directions of incident ray and emergent ray. the image of the letters appear erect but So if light is incident at an angle of incidence inverted laterally, or sideways. Such an i, inversion is called lateral inversion.  = 180º – (i + r) = (180º – 2i) [as i = r] L R R L  r i Object Image Plane mirror So if light is incident at angle of 30º,    Relative motion of object and image :    = (180º – 2 × 30º) = 120º and for normal Case I : incidence i = 0º,  = 180º If an object moves towards (or away from) a plane mirror at speed v Characteristics of the image formed by a plane mirror : (i) The image formed by a plane mirror is virtual. (ii) The image formed by a plane mirror is erect. The image will also approach (or recede) at (iii) The size of the image formed by a plane speed v mirror is same as that of the size of the object. If object is 10 cm high, then the image of this The speed of image relative to object will be object will also be 10 cm high. v – (–v) = 2v. (iv) The image formed by a plane mirror is at the Case II : same distance behind the mirror as the object is If the mirror is moved towards or (away in front of it. Suppose, an object is placed at 5 cm from) the object with speed 'v' in front of a plane mirror then its image will be at The image will move towards (or away from) 5 cm behind the plane mirror. the object with a speed '2v'. (v) The image formed by a plane mirror is laterally inverted, i.e., the right side of the object appears as  Multiple Reflection the left side of its image and vice-versa. Number of images formed by combination of plane mirrors depends upon angle between mirrors.  90°  If there are two plane mirrors inclined to each other at an angle 90° , the number of images of a  Lateral Inversion point object formed are 3. REFLECTION FROM SPHERICAL MIRROR Radius of curvature : The distance between the pole and the centre of curvature of the  INTRODUCTION : There are two types of mirror, is called the radius of curvature of the spherical mirrors: mirror. It is equal to the radius of the (i) Concave mirror : spherical shell of which the mirror is a A section. In diagram, PC is the radius of curvature of the mirror. It is represented by Principal the symbol R. axis  Focal length : The distance between the pole F P and principal focus of the mirror, is called the focal length of the mirror. In diagram, PF is the focal length of the mirror. It is represented B by the symbol f. (ii) Convex mirror : R A f  for convex 2 Principal R axis f=  for concave 2 P F C Principal section : A section of the spherical mirror cut by a plane passing through its B centre of curvature and the pole of the mirror, is called a principal section of the mirror. It   SOME TERMS ASSOCIATED WITH contains the principal axis. In diagram, APB SPHERICAL MIRRORS. is the principal section of the mirror cut by  Aperture. The diameter of the circular rim of the plane of the book page. the mirror. In diagram AB is the aperture of the mirror.  Pole : The centre of the spherical surface of RULES FOR IMAGE FORMATION BY RAY DIAGRAM METHOD the mirror is called the pole of the mirror. It lies on the surface. In diagram, P is the pole  RULES FOR IMAGE FORMATION FROM of the mirror. CONCAVE MIRROR  Centre of curvature : The centre of the (a)When the light ray incident parallel to the spherical shell, of which the mirror is a principal axis. section, is called centre of curvature of the A ray light parallel to mirror. It lies outside the surface. Every point the principal axis on mirror surface lies at same distance from it. In diagram, C is the centre of curvature of Principal axis P the mirror. C F  Principal axis : The straight line passing through the pole and the centre of curvature of the mirror, is called principal axis of the OR mirror. When the light ray incident towards focus.  Principal focus : It is a point on the principal Reflected ray goes parallel axis of the mirror, such that the rays incident to the principal axis on the mirror parallel to the principal axis after reflection, actually meet at this point (in P case of a concave mirror) or appear to come C F from it (in case of a convex mirror). In diagram, F is the principal focus of the mirror. (b)When the light ray incident towards centre of  SIGN CONVENTION curvature. (a) Description : It is a convention which fixes A ray of light passing the signs of different distances measured. The through the centre of sign convention to be followed is the New curvature Cartesian sign convention. It gives the following rules : C F P 1. All distances are measured from the pole of the mirror. 2. The distances measured in the same direction as the direction of incident light from pole are taken as positive. (c)When the light ray incident on the pole of the 3. The distances measured in the direction mirror. opposite to the direction on incident light Incident ray from pole are taken as negative. 4. Distances measured upward and C Fi perpendicular to the principal axis, are taken r P as positive. Reflected ray 5. Distances measured downward and perpendicular to the principal axis, are taken  RULES FOR IMAGE FORMATION FROM as negative. CONVEX MIRROR (a)When the light ray incident parallel to the IMAGE FORMATION BY SPHERICAL principal axis. MIRROR IN DIFFERENT CASES Reflected ray Introduction : From mirror formula, we find that Incident ray for a mirror of a fixed focal length f, as object P F distance u changes, image distance also Principal axis changes. (A) BY CONCAVE MIRROR :  OR (1) Object at Infinity   When the light ray incident parallel to the A point object lying on the principal axis. principal axis. Rays come parallel to the principal axis and after reflection from the mirror actually meet Incident ray at the focus F. Reflected ray The image is formed at F. It is real and point P F sized (fig.) Principal axis (b)When the light ray incident on the mirror C P F directing towards centre of curvature. Rays traveling towards C behind the mirror Fig. Concave mirror : point object at infinity, 90° image at focus. (2) Object Beyond Centre of Curvature P F C Real object AB has its image AB formed between focus and centre of curvature. The image is real-inverted and diminished. B B A A' C F P P A C F Parallel rays B' to infinity Concave mirror : object beyond centre of Concave mirror : object at focus image at curvature, image between focus and centre of infinity. curvature. (6) Object between Focus and Pole (3) Object at Centre of Curvature Real object AB has its image AB formed behind the mirror. The image is virtual-erect Real object AB, has its image AB formed at and enlarged. centre of curvature. B' The image is real-inverted and has same size as the object. (fig.). B B C FA P A' A C A' P F Fig. Concave mirror : Object between pole and focus, image behind the mirror. B' (B) BY CONVEX MIRROR : Concave mirror : object at centre of (1) Object at infinity curvature, image at centre of curvature A point object lying on the principal axis. (4) Object between Centre of Curvature and Focus Rays come parallel to the principal axis and Real object AB has its image AB formed after reflection from the mirror, appear to beyond centre of curvature. diverge from focus F behind the mirror. The image is real-inverted and enlarged The image is formed at F. (bigger in size than the object). (Fig.) The image is virtual and point sized. [fig.] B A' P P C CA F F B' Fig. Convex mirror : point object at infinity, Concave mirror : object between centre of virtual image at focus. curvature and focus, image beyond centre of curvature. (2) Object at anywhere on principle axis (5) Object at Focus Real object AB has its image formed at infinity. The image is imaginary inverted (reflected rays go downward) and must have very large O IF C size. Image is virtual & point sized NUMERICAL METHOD IN SPHERICAL Between Behind Enlarged Virtual MIRROR P and F the and mirror erect (A) Mirror formula at infinite at focus highly virtual  Definition : The equation relating the object diminished point size convex mirror distance (u) the image distance (v) and the mirror focal length (f) is called the mirror anywhere between diminished virtual formula. on pole & erect 1 1 1 principal focus   axis v u f   Assumptions made :  SOLVED EXAMPLES  (i) The mirror has a small aperture. Ex.1 An object is placed in front of a plane mirror. (ii) The object lies close to principal axis of the If the mirror is moved away from the object mirror. through a distance x, by how much distance (iii) The incident rays make small angles with the mirror surface or the principal axis. will the image move? Sol. Suppose the object O was initially at a (B) linear magnification For spherical mirrors distance d from the plane mirror M as shown  Definition : The ratio of the size of the in fig. The image formed at O’ is at a distance image, as formed by reflection from the mirror to the size of the object, is called linear d behind the mirror. Now, the mirror is magnification produced by the mirror. It is shifted by a distance x to M’ such that the represented by the symbol m. distance of the object from M’ becomes d + x. v height of image The image now formed at O” which is also at m  u height of object a distance d + x from M’. (C) Power of mirror 1 M M' Power of a mirror [in Diopters] = f (in metre) O O' O'' SUMMARY OF IMAGES BY SPHERICAL d MIRROR d So, OM = MO' = d Nature Position Position Size of OM' = M'O" = d + x of of object of Image Image Thus, OO" = OM' + M'O" = 2(d + x)...(1) Image when OO' = OM + MO' = 2d...(2) At infinity At focus Highly Real and F diminished inverted   O'O" = OO" – OO' = 2(d + x) – 2d Beyond C Between Diminished Real and = 2x Concave mirror F inverted Thus, the image is shifted from O' to O" by a and C distance 2x. At C At C Same size Real and inverted Ex.2 An insect is at a distance of 1.5m from a Between Beyond C Enlarged Real and plane mirror. Calculate the following? F and C inverted (i) Distance at which the image of the insect is formed. At F At infinity Highly Real and enlarged inverted (ii) distance between the insect and its image. Sol. (i) The distance of insect from the mirror 15 cm = 1.5 m The distance of insect from the mirror A is also equal to 1.5 m. The image is B' B F formed at 1.5 m behind the mirror. (ii) The distance between the insect and A' image = 1.5 + 1.5 = 3m 30 cm Sol. Here u = –15 cm and = –30 cm Ex.3 A concave mirror is made up by cutting a Size of the object, h = 2 cm portion of a hollow glass sphere of radius 30 h' v Magnification, m = m =  cm. Calculate the focal length of the mirror. h u Sol. The radius of curvature of the mirror = 30 cm h' (30) or =– =2 Thus, the focal length of the mirror h (15) or h' = – 2 × h = – 2 × 3 30cm = = 15cm = – 6 cm 2 So the height of the image is 6 cm. The minus Ex.4 An object is placed at a distance of 15 cm sign shows that it is on the lower side of the from a concave mirror of focal length 10 cm. principal axis, i.e. the image is inverted. Find the position of the image. 15 cm Ex.6 A 1.4 cm long object is placed perpendicular to the principal axis of a convex mirror of A focal length 15 cm at a distance of 10 cm B' from it. Calculate the following : B F (i) location of the image A' 10 cm (ii) height of the image 30 cm (iii) nature of the image Sol. We have u = –15 cm and f = –10 cm A 1 1 1 A' Using the relation,, + = we get v u f B10 cm 6 cm B' F C 1 1 1 + = v  15  10 1 1 1 1 15cm or = – =– v 15 10 30 Sol. (i) For a convex mirror, focal length is positive. or v = –30 cm Therefore, f = +15 cm and u = –10 cm So the image will be formed 30 cm from the 1 1 1 mirror. Since has a negative sign, the image Using the relation, + = , we get v u f is formed to the left of the mirror, i.e. in front of the mirror as shown in fig. 1 1 1 + = v  10 15 Ex.5 A 3 cm long object is placed perpendicular to 1 1 1 5 1 the principal axis of a concave mirror. The or = + = = v 15 10 30 6 distance of the object from the mirror is 15 or = 6 cm cm, and its image is formed 30 cm from the mirror on the same side of the mirror as the Since is positive, the image is formed to the object. Calculate the height of the image right of the mirror at a distance 6 cm from it. formed. (ii) Magnification, Sol. Object distance, u = –30 h' v Image distance, = –60 or m= = – h u (real image is formed on the same side) h' 6 1 1 1 = = + 0.6 Now, using the mirror formula, + = h  10 v u f or h' = + 0.6 × h0 = 0.6 × 1.4 we get = 0.84 cm. 1 1 1 + = Thus, the height of the image is 0.84 cm.  60  30 f (iii) Since h' is positive, the image will be on the 1 3 1 same side of the principal axis as the object. or =– =– f 60 20 Hence, the image is virtual, erect and diminished. or f = – 20 cm Focal length of the mirror = 20 cm Ex.7 An object is placed at a distance of 40 cm h' v Magnification m= =– from a convex mirror of focal length 30 cm. h u Find the position of image and its nature. h' (60) or =– A 3 (30) A' or h' = 3 × (–2) = –6 cm B 40 cm B' F C The height of the image is 6 cm. The negative sign shows that the image is inverted. 30cm Ex.9 A 1 cm high object is placed at 20 cm in front Sol. Here, object distance, u = –40 cm of a concave mirror of focal length 15 cm. Focal length of convex mirror, f = +30 cm Find the position and nature of the image. 1 1 1 Sol. u = –20 cm, f = –15 cm, h0 = 1 cm Now, using mirror formula, + = we v u f 1 1 1 get Using mirror formula, + = we get v u f 1 1 1 1 1 1 + = + = v  40 30 v  20  15 1 1 1 7 or = + = 1 1 1 1 v 40 30 120 or =– + =– v 15 2v 60 120 or =  = – 60 cm 7 The image is formed 60 cm from the mirror. The positive sign shows that the image is Since, the signs of u and are the same, the formed on the right, i.e. behind the mirror. object and image are formed on the same side Now, magnification, of the mirror. Therefore, the image is real. v 120 3 Now magnification, m = – =– =+ u 7  (  40) 7 h' v 60 cm m= = = = –3 Since, the magnification is positive, the image h u  20 cm is erect. Thus, the image is formed 17.1 cm  h’ = –3h = – 3 × 1 cm = – 3cm behind the mirror. The image is virtual, erect and diminished. The negative sign shows that the image is inverted. Thus, the image is real, inverted and Ex.8 A 3 cm high object is placed at a distance of of size 3 cm and formed 60 cm in front of the 30 cm from a concave mirror. A real image is mirror. formed 60 cm from the mirror. Calculate the focal length of the mirror and the size of the image. Ex.10 An object 4 cm high is placed 25 cm in front h' 60 / 7 60 3 of a concave mirror of focal length 15 cm. At or =– =+ = 5 (20) 7  20 7 what distance from the mirror should a screen be placed in order to obtain a sharp image? 3 15 or h' = 5 × = cm Find the nature and size of image. 7 7 Sol. Here, u = –25 cm, f = –15 cm, h = + 4 cm 1 1 1 The height of the image is 2.1 cm. Positive Using the mirror formula, + = , we get v u f sign shows that the image is erect. 1 1 1 or + = v  25  15 Ex.12 A convex mirror used on a automobile has 3 m radius of curvature. If a bus is located 5 m 1 1 1 1 1 2 or = – =– + =– from this mirror, find the position, nature and v  15  25 15 25 75 size of image. 75 or = = –37.5 cm 2 Sol. Here, u = –5 m, r = +3m Thus, the screen must be placed 37.5 cm from r 3 the mirror on the same side as the object.  f= =+ 1.5 m 2 7 h' v Now, magnification, m = =  1 1 1 h u Using the relation, + = , we get v u f h' (37.5)  = – 1.5 1 1 1 4.0 cm (25) or + = v  5 1.5 or h' = – 1.5 × 4 = –6 cm Negative sign shows that the image is 1 1 1 or = + = + 1.15 m inverted. Hence, the image is real, inverted v 1.5 5 and of size 6 cm. The image is 1.15 m behind the mirror. h' v 1.15 Ex.11 An object 5 cm high is placed at a distance of Magnification, m = =– =– = + 0.23 h u (5) 20 cm from a convex mirror of radius of curvature 30 cm. Find the position, nature and Thus, the image is virtual, erect and smaller size of image. in size than the object. Sol. Here, u = –20 cm, h = 5 cm Radius of curvature, r = +30 cm IMPORTANT POINTS TO BE REMEMBER r 30  Laws of reflection :  Focal length, f = =+ = + 15 cm 2 2 (a) Angle of incidence is equal to the angle of 1 1 1 reflection. Using the mirror formula, + = , we get v u f (b) The incident ray, the reflected ray and the 1 1 1 + = normal all lie in the same plane. v  20  15  Mirror : A smooth, highly polished reflecting 1 1 1 7 or = + = surface is called a mirror. There are two types v 15 20 60 of mirrors : (a) plane mirror (b) curved 60 or  = cm mirrors 7 Curved mirrors are classified as spherical The image is formed 8.5 cm from the mirror. The positive sign shows that the image is mirrors or parabolic mirrors depending upon formed on the other side or behind the mirror. the curvature of the mirror. So the image is virtual.   Concave mirror : A spherical mirror whose h' v inner hollow surface is the reflecting surface Magnification, m = = – h u is called concave mirror. A concave mirror screen is called a virtual image. A virtual forms a real, inverted image of an object if image is formed only when the reflected or the object is placed at any place outside F. the refracted rays appear to come from a However, when the object lies between F and point. P, the image formed is erect and virtual.  Sign convention for spherical mirrors :  Convex mirror : A spherical mirror whose According to the sign convention for mirror, outer bulging at surface is the reflecting the focal length of a concave mirror is surface is called convex mirror. The image negative and that of a convex mirror is formed by a convex mirror is always erect, positive. virtual and smaller than the object whatever  Mirrors formula : The relationship, may be the position of the object. A convex 1 1 1 mirror is used as a side-mirror (driver’s   is called the mirror formula. f v u mirror) on automobiles.  Aperture of a mirror : The effective width  Magnification : The ratio of the size of the of a spherical mirror from which reflection image to that of the object is called can take place is called its aperture. magnification. For a mirror, magnification (M) is given by,  Pole : The centre of a curved mirror is called its pole. v h M=– = i u ho  Centre of curvature : The centre of the hollow sphere of which the spherical mirror is 1 power (in diopters)  a part is called its centre of curvature. The f ( metre) centre of curvature of a concave mirror is in  front of it, while that of a convex mirror is CONCLUSIONS FROM THE SIGN CONVENTION behind it. For spherical mirror : Radius of curvature : Radius of the hollow sphere of which the mirror is a part is called Distance of the object, u is negative its radius of curvature. Distance of the real image, v is negative  Principal axis : A straight line passing Distance of the virtual image v is positive through the centre of curvature and pole of a Focal length, f is negative for spherical mirror is called its principal axis. concave & f is + ve for convex Focus : A point on the principal axis of a Radius of curvature, R is negative mirror at which the rays coming from infinity for concave & meet or appear to meet is called its focus. R is + ve for Focus is denoted by F. convex Focal length : The distance between the pole Height of the object, O is positive of a spherical mirror and the focus is called Height of the inverted (real) image, I is negative the focal of a spherical mirror. Height of the erect (virtual) image, I is positive  Real image : The image which can be obtained on a screen is called a real image. A real image is formed only when the reflected or refracted rays actually meet at a point. Virtual image : The image which can be seen into a mirror but cannot be obtained on a EXERCISE # 1 A. Very Short Answer Type Questions Q.13 We known that plane and convex mirrors produce virtual images of objects. Can they Q.1 A ray of light is incident on a plane mirror, i produce real images under some being the angle of incidence. What is the circumstances ? Explain deviation suffered by the ray of light? Q.14 The wall of a room is covered with perfect plane mirror. Two movie films are made, one Q.2 A plane mirror reflects a pencil of light to recording the movement of a man and the form a real image. What is the nature of the other of his mirror image. From viewing the pencil of light incident on the mirror? films later, can an outsider tell which is Q.3 Define principal axis of a spherical mirror. which? Q.15 A concave mirror is held in water. What Q.4 What is the focal length of a plane mirror? would be the change in the focal length of the Q.5 Two perpendicular plane mirror forms mirror?.............. number of images of a point source Q.16 What is the difference between the virtual of light. images produced by (i) plane mirror, Q.6 What is the magnification produced by a (ii) concave mirror, (iii) convex mirror? plane mirror? Q.17 Show that if a ray of light is reflected Q.7 Which mirror would you use for shaving? successively from two mirrors inclined at an angle , the deviation of the ray does not Q.8 Suppose x and y are distances of object and depend upon the angle of incidence. image respectively from a mirror. What shall Q.18 Use the mirror equation to deduce that an 1 be the shape of the graph between and object placed between f and 2f of a concave x mirror produces a real image beyond 2f. 1 for a concave mirror ? y Q.19 Show that a convex mirror always produces a virtual image independent of the location of B.  Short Answer Type Questions the object. Q.9 An object is placed between two plane Q.20 Prove that the virtual image produced by a parallel mirrors. Why do the distant images convex mirror is always diminished in size get fainter and fainter? and is located between the focus and the pole. Q.10 Why mirrors used in search light are Q.21 Show analytically that an object placed parabolic and not concave spherical? between the pole and focus of a concave mirror produces a virtual and enlarged image. Q.11 You read a newspaper because of the light that it reflects. Then why do you not see even Q.22 We know that a virtual image cannot be a faint image of yourself in the newspaper? obtained on a screen. But when we see a virtual image, we are obviously bringing it on Q.12 If you were driving a car, what type of mirror the retina (may be regarded as a screen) of the would you prefer to use for observing traffic eye. Point out the contradiction, if any. at your back and why? Q.23 Why a concave mirror of small aperture Q.30 Find formulae for magnification produced in forms a sharper image? the following cases : (i) concave mirror, when image formed is real (ii) concave mirror, Q.24 What do you understand by the term when image formed is virtual (ii) convex ‘parallax’? mirror. Q.25 How can you distinguish between three different mirrors just by looking at them? Q.31 Draw a ray diagram to show the formation of image of an object placed between the pole Q.26 What is the effect of size of mirror on the and centre of curvature of a concave mirror. nature of image ? Derive the formula connecting object distance Q.27 Is irregular reflection follows the laws of (u), image distance () and focal length (f) for reflections or not ? this particular case for the given concave mirror. State clearly the assumptions and sign conventions used. C.  Long Answer Type Questions Q.32 Express magnification produced by a Q.28 Prove that the radius of curvature of a spherical mirror in terms of (i) u and f(ii)  spherical mirror is equal to twice the focal and f. length of the mirror. Q.29 Derive mirror formula for a concave mirror when image formed is (i) real (ii) virtual Also give the sign convention used. EXERCISE # 2 Q.8 Air is not visible because it- Single Correct Answer type Questions (A) is nearly a perfectly transparent Q.1 A child walks towards a fixed plane mirror at (B) neither absorbs nor reflects light a speed of 5 km h–1. The velocity of the (C) transmits whole of light image with respect to mirror is - (D) all of the above are correct (A) 5 km h–1 (B) –5 km h–1 (C) 10 km h–1 (D) –10 km h–1 Q.9 According to laws of reflection of light - (A) Angle of incidence is equal to the angle Q.2 The letter that does not show lateral of reflection inversion- (B) Angle of incidence is less than the angle (A) Z (B) M (C) O (D) W of reflection Q.3 In a plane mirror, an object is 0.5 m in front (C) Angle of incidence is greater than the of the mirror. The distance between object angle of reflection and image is - (D) None of these (A) 0.5 m (B) 1 m (C) 0.25 m (D) 0.75 m Q.10 Which of the following correctly represents graphical relation between angle of incidence Q.4 An object 0.5 m tall is in front of a plane (i) and angle of reflection (r) ? mirror at a distance of 0.2 m. The size of the y y image formed is- i i (A) 0.2 m (B) 0.5 m (C) 0.1 m (D) 1 m (A) (B) O x O x Q.5 A plane mirror is approaching you at 10 cm r r s–1. Your image shall approach you with a y y speed of- i i (A) + 10 cm s–1 (B) – 10 cm s–1 (C) (D) (C) + 20 cm s–1 (D) – 20 cm s–1 O x O x r r Q.6 The path along which light travels in a Q.11 A light ray falls on a mirror and deviates by homogeneous medium is called a- 60° then the angle of reflection will be (A) beam of light (B) ray of light (A) 30° (B) 90° (C) pencil of light (D) none of these (C) 60° (D) 180° Q.12 A ray of light is incident on a plane mirror at an Q.7 A thin layer of water is transparent but a very angle . If the angle between the incident and thick layer of water is- reflected rays is 80°, what is the value of . (A) translucent (B) opaque (A) 40° (B) 50° (C) most transparent (D) none of these (C) 45° (D) 55° Q.13 Light shows - Q.18 The magnification produced by a concave (A) Random propagation mirror - (B) Curvilinear propagation (A) is always more than one (C) Rectilinear propagation (B) is always less than one (D) None of these (C) is always equal to one (D) may be less than or greater than one Q.14 The image of the moon is formed by a concave mirror whose radius of curvature is Q.19 Choose the correct relation between u, v and 4.8 m at a time when distance from the moon R- is 2.4 × 108 m. if the diameter is of the image 2uv 2 (A) R  (B) R  is 2.2 cm, the diameter of the moon is- uv uv (A) 1.1 × 106 m (B) 2.2 × 106 m 2( u  v ) (C) R  (D) none of these (C) 2.2 × 108 m (D) 2.2 × 1010 m (uv) Q.15 The focal length of a concave mirror is f and the distance of the object from the principal Q.20 The image formed by a concave mirror is focus is a. The magnitude of magnification real, inverted and of the same size as that of obtained will be- the object. The position of the object should (A) (f + a)/f (B) f /a be- (C) f/ a (D) f2/a2 (A) Beyond C (B) Between C and F (C) At C (D) At F Q.16 The magnification of an object placed 10 cm from a convex mirror of radius of curvature Q.21 A boy is standing in front of a plane mirror at 20 cm will be- a distance of 3m form it. What is the distance (A) 0.2 (B) 0.5 between the boy and his image ? (C) 1 (D) infinity (A) 3 m (B) 4.5 m (C) 6 m (D) none of these Q.17 The image formed by a concave mirror is observed to be virtual, erect and larger than the object. then the position of the object should be- (A) between the focus and the centre of curvature (B) at the centre of curvature (C) beyond the centre of curvature (D) between the pole of the mirror and the focus ANSWER KEY EXERCISE-1 1.  – 2i 2. virtual 5. 3 1 1 1 6. 1 7. plane mirror 8.    linear v u f 1 9. as I  10. They reflect light parallel 15. No change r2 26. Image becomes brighter 27. Yes EXERCISE-2 Ques 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 Ans B A B B C B A D A D C B C B B Ques 16 17 18 19 20 21 Ans B D D A C C

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