Reflection of Light Chapter 31 PDF

O I 60 Reflection of Light 1 E3 When a ray of light after incidenting on a boundary separating two media comes back into the same media, then this phenomenon, is called reflection of light. Normal i r   After reflection velocity, wavelength and frequency of light remains same but D YG intensity de...

O I 60 Reflection of Light 1 E3 When a ray of light after incidenting on a boundary separating two media comes back into the same media, then this phenomenon, is called reflection of light. Normal i r   After reflection velocity, wavelength and frequency of light remains same but D YG intensity decreases.  ∠i = ∠r After reflection, velocity, wave length and frequency of light remains same but intensity decreases There is a phase change of  if reflection takes place from denser medium U Boundary Note :   ID Incident ray Reflected ray If light ray incident normally on a surface, after reflection it retraces the path. Real and virtual images U If light rays, after reflection or refraction, actually meets at a point then real image is formed and if they appears to meet virtual image is formed. ST I (Real image) O (Real object) O (Real image) Real image I (Virtual object) (Real object) I (Virtual image) (Virtual image) (Virtual image) O (Virtual object) 2 Reflection of Light Plane Mirror. The image formed by a plane mirror is virtual, erect, laterally inverted, equal in size that of x E3 x 60 the object and at a distance equal to the distance of the object in front of the mirror. r  = (180 – 2i) Note :    Final path 2  Original path   = (360 – 2) U i ID (1) Deviation : Deviation produced by a plane mirror and by two inclined plane mirrors. D YG If two plane mirrors are inclined to each other at 90 o, the emergent ray is anti-parallel to incident ray, if it suffers one reflection from each. Whatever be the angle to incidence. IR IR ST U (2) Rotation : If a plane mirror is rotated in the plane of incidence through angle , by keeping the incident ray fixed, the reflected ray turned through an angle 2. RR   RR  (3) Images by two inclined plane mirrors : When two plane mirrors are inclined to each other at an angle , then number of images (n) formed of an object which is kept between them. 360  360  n  1  ; If  even integer     (i) (ii) If 360   odd integer then there are two possibilities (a) Object is placed symmetrically asymmetrically (b) Object Object /2 /2 Object   is placed Reflection of Light 3  360  n  1    360  60 n Note :  If θ = 0o i.e. mirrors are parallel to each other so n   i.e. infinite images will be If θ = 90o, n    If θ = 72o, n  E3 formed. 360 1  3 90 360  1  4 (If nothing is said object is supposed to be symmetrically 72 ID placed). (4) Other important informations (i) When the object moves with speed u towards (or away) from the plane mirror then image also moves toward (or away) with speed u. But relative speed of image w.r.t. object is 2u. D YG U (ii) When mirror moves towards the stationary object with speed u, the image will move with speed 2u. I O u I O 2u Rest u u Mirror at rest Mirror is moving U (iii) A man of height h requires a mirror of length at least equal to h/2, to see his own complete image. (iv) To see complete wall behind himself a person requires a plane mirror of at least one third the height of wall. It should be noted that person is standing in the middle of the room. ST H E h H E M' h 2 M' M' h 3 E h M' B L d d Concepts  The reflection from a denser medium causes an additional phase change of  or path change of /2 while reflection from rarer medium doesn't cause any phase change.  We observe number of images in a thick plane mirror, out of them only second is brightest. Incident light (100%) 10% 80% 9% Brightest image To find the location of an object from an inclined plane mirror, you have to see the perpendicular distance of the object from the mirror. ID E3  60 4 Reflection of Light Example s A plane mirror makes an angle of 30o with horizontal. If a vertical ray strikes the mirror, find the angle between mirror and reflected ray U Example: 1 (a) 30o (c) 60o (d) 90o Since angle between mirror and normal is 90 o and reflected ray (RR) makes an angle of 30o with the normal so required angle D YG Solution : (c) (b) 45o will be   60. o (a) 60o (b) 120o By using   (360  2 )  ST Solution : (d) Example: 3 30o  = 60o 30o (c) 180o (d) 240o   360  2  60  240 o A person is in a room whose ceiling and two adjacent walls are mirrors. How many images are formed (a) 5 Solution : (c) RR Two vertical plane mirrors are inclined at an angle of 60 o with each other. A ray of light travelling horizontally is reflected first from one mirror and then from the other. The resultant deviation is U Example: 2 IR 30o [AFMC 2002] (b) 6 (c) 7 (d) 8 The walls will act as two mirrors inclined to each other at 90 o and so sill form 360 1  3 90 images of the person. Now these images with object (Person) will act as objects for the ceiling mirror and so ceiling will form 4 images as shown. Therefore total number of images formed = 3 + 4 = 7 I1 I1 O I2 I2 O I3 Three images by I3 Four images by ceiling Note :  The person will see only six images of himself (I1 , I 2 , I 3 , I1' , I 2' , I 3' ) A ray of light makes an angle of 10o with the horizontal above it and strikes a plane mirror which is inclined at an angle  to the horizontal. The angle  for which the reflected ray becomes vertical is (a) 40o Solution : (a) E3 Example: 4 60 Reflection of Light 5 (b) 50o (c) 80o From figure Vertical RR     10  90 IR ID     40 o  10o Horizontal line Plane mirror  A ray of light incident on the first mirror parallel to the second and is reflected from the U Example: 5 (d) 100o second mirror parallel to first mirror. The angle between two mirrors is (a) 30o From geometry of figure (c) 75o D YG Solution : (b) (b) 60o (d) 90o       180 o      60 o Example: 6    A point object is placed mid-way between two plane mirrors distance 'a' apart. The plane mirror forms an infinite number of images due to multiple reflection. The distance between the nth order image formed in the two mirrors is U (a) na (b) 2na (d) n2 a (c) na/2 ST Solution : (b) III order image I3' M M' II order image I2' I order image I1' 3a/2 5a/2 a/2 O a/2 a/2 a I order image II order image I1 I2 a/2 III order image I3 3a/2 5a/2 From above figure it can be proved that seperation between nth order image formed in the two mirrors = 2na Example: 7 Two plane mirrors P and Q are aligned parallel to each other, as shown in the figure. A light ray is incident at an angle of  at a point just inside one end of A. The plane of incidence coincides with the plane of the figure. The maximum number of times the ray undergoes reflections (including the first one) before it emerges out is 6 Reflection of Light (a) l d tan  (b) d l tan  l d  (c) ld tan  Solution : (a) 60 (d) None of these Suppose n = Total number of reflection light ray undergoes before exist out. x = Horizontal distance travelled by light ray in one reflection.  x d l d tan    A plane mirror and a person are moving towards each other with same velocity v. Then the velocity of the image is (a) v (b) 2v ID Example: 8 n also E3 So nx = l l x tan   d (c) 3v (d) 4v If mirror would be at rest, then velocity of image should be 2v. but due to the motion of mirror, velocity of image will be 2v + v = 3v. Example: 9 A ray reflected successively from two plane mirrors inclined at a certain angle undergoes a deviation of 300o. The number of images observable are U Solution : (c) Solution : (b) (b) 11 D YG (a) 10 By using (c) 12 (d) 13   (360  2 )  300  360  2    30 o. Hence number of images  360  1  11 30 Tricky example: 1 U A small plane mirror placed at the centre of a spherical screen of radius R. A beam of light is falling on the mirror. If the mirror makes n revolution. per second, the speed of light on the screen after reflection from the mirror will be (b) 2nR ST (a) 4nR (c) nR 2 (d) nR 4 Solution : (a) When plane mirror rotates through an angle , the reflected ray rotates through an angle 2. So spot on the screen will make 2n revolution per second  Speed of light on screen v  R  2 (2n)R  4nR Tricky example: 2 A watch shows time as 3 : 25 when seen through a mirror, time appeared will be [RPMT 1997; JIPMER 2001, 2002] (a) 8 : 35 (b) 9 : 35 (c) 7 : 35 (d) 8 : 25 Reflection of Light 7 Solution : (a) For solving this type of problems remember Actual time = 11 : 60 – given time So here Actual time = 11 : 60 – 3 : 25 = 8 : 35 60 Tricky example: 3 When a plane mirror is placed horizontally on a level ground at a distance of 60 m from the foot of a tower, the top of the tower and its image in the mirror subtend an angle of 90o at the eye. The height of the tower will be (b) 60 m Solution : (b) Form the figure it is clear that (c) 90 m h  tan 45 o 60  h = 60 m (d) 120 m E3 (a) 30 m ID Tower h 60 m Image D YG U 45o 45o Curved Mirror. ST U It is a part of a transparent hollow sphere whose one surface is polished. C C P P F F Principle axis Converges the light rays Diverges the light rays (1) Some definitions : (i) Pole (P) : Mid point of the mirror (ii) Centre of curvature (C) : a part. Centre of the sphere of which the mirror is (iii) Radius of curvature (R) : curvature. Distance between pole and centre of 8 Reflection of Light (Rconcave = –ve , Rconvex = +ve , Rplane = ) : (v) Focus (F) A line passing through P and C. : object is at  (vi) Focal length (f) : An image point on principle axis for which Distance between P and F. (vii) Relation between f and R f : 60 (iv) Principle axis R (fconcare = –ve , fconvex = + ve , fplane 2 =) : mirror (ix) Aperture : Effective diameter of light reflecting area. Intensity of image  Area  (Aperture)2 : A plane passing from focus and perpendicular to principle axis. ID (x) Focal plane The converging or diverging ability of E3 (viii) Power (2) Rules of image formation and sign convention : Rule (ii) Rule (iii) F D YG U Rule (i) F F F C C (3) Sign conventions : (i) All distances are measured from the pole. (ii) Distances measured in the direction of incident rays are U taken as positive while in the direction opposite of incident rays are taken negative. ST (iii) Distances above the principle axis are taken positive and below the principle axis are taken negative. Note Incident ray + + – Mirror or Lens – Principle axis : Same sign convention are also valid for lenses. Use following sign while solving the problem : Concave mirror Real image (u ≥ f) Virtual image (u< f) Convex mirror Reflection of Light 9 Distance of object u – u – u – Distance of image v – v + v + Focal length f – f – f Height of object O + O+ O + Height of image I – I + I Radius of curvature R – R – Magnification m – m+ + 60 + R + m + (a) Concave At infinity Location of the image  P F C Note :  In virtual inverted Real inverted Away from centre Between f and 2f m < 1, diminished Real inverted of curvature (u > 2f) i.e. At centre of At Real inverted U f < v < 2f centre of m = 1, same size curvature u = 2f curvature i.e. v = 2f as that object Between centre of curvature and focus : Away from centre curvature the of m > 1, magnified Real inverted F < u < 2f v > 2f At focus i.e. u = f At infinity i.e. v = ∞ m = ∞, magnified Real inverted Between pole and focus u < f v>u m > 1 magnified Virtual erect At infinity i.e. u = ∞ At focus i.e., v = f m < 1, diminished Virtual erect Between pole and focus m < 1, diminished Virtual erect ST (b) Convex Erect 1, D YG P Real m