Solution 3S.<\/strong><\/span><\/p>\n(a) Plane mirror: Plane mirror is a highly polished and smooth reflecting surface made from a clear plane glass sheet, usually thin and silvered with suitable reflecting abrasive (for example, mercury) on one side. Once this pasting is done, then the glass becomes opaque but due to the reflecting property of the abrasive, the plane glass sheet becomes a plane glass reflector or a plane glass mirror.
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\n(b) Incident ray: The light ray striking a reflecting surface is called the incident ray.<\/p>\n
(c) Reflected ray: The light ray obtained after reflection from the surface, in the same medium in which the incident ray is travelling, is called the reflected ray.<\/p>\n
(d) Angle of incidence: The angle which the incident ray makes with the normal at the point of incidence is called the angle of incidence. It is denoted by the letter i.<\/p>\n
(e) Angle of reflection: The angle which the reflected ray makes with the normal at the point of incidence is called the angle of reflection. It is denoted by the letter r.
\n<\/p>\n
Solution 4S.<\/strong><\/span><\/p>\n
\nRegular reflection occurs when a beam of light falls on a smooth and polished surface and irregular reflection occurs when a beam of light falls on a rough surface. Since the surface is uneven, from different points light rays get reflected in different directions and give rise to irregular reflection.<\/p>\n
Solution 5S.<\/strong><\/span><\/p>\nReflection of light from a plane mirror is regular reflection and reflection of light from plane sheet of paper is irregular reflection of light.<\/p>\n
Solution 6S.<\/strong><\/span><\/p>\nLaws of reflection:<\/p>\n
\n- The angle of incidence is equal to the angle of reflection.<\/li>\n
- The incident ray, the reflected ray and the normal at the point of incidence, lie in the same plane.<\/li>\n<\/ol>\n
Solution 7S.<\/strong><\/span><\/p>\nLaws of reflection:<\/strong><\/p>\n\n- The angle of incidence is equal to the angle of reflection.<\/li>\n
- The incident ray, the reflected ray and the normal at the point of incidence, lie in the same plane.<\/li>\n<\/ol>\n
Experiment to verify the laws of reflection:<\/u><\/p>\n
Fix a white sheet of paper on a drawing board and draw a line MM1\u00a0<\/sub>as shown in figure. On this line, take a point O nearly at the middle of it and draw a line OA such that\u00a0\u2220MOA is less than 90o<\/sup>. Then draw a normal ON on line MM1<\/sub>\u00a0at the point O, and place a small plane mirror vertical by means of a stand with its silvered surface along MM1<\/sub>.<\/p>\n<\/p>\n
Next fix two pins P and Q at some distance (\u22485 cm) apart vertically on line OA, on the board. Keeping eye on the other side of normal (but on the same side of mirror), see clearly images P’ and Q’ of the pins P and Q. Next fix a pin R such that it is in line with the images of pins P and Q as observed in the mirror. Next, fix one more pin S such that the pin S is in line with the pin R as well as images P’ and Q’ of pins P and Q.<\/p>\n
Draw small circles on paper around the positions of pins as shown in figure. Remove the pins and draw a line OB joining the pin points S and R, which meets the surface of mirror at O. The angles AON and BON are measured and recorded.<\/p>\n
The experiment is then repeated for the angle of incidence\u00a0\u2220AON equal to 40o<\/sup>, 50o<\/sup>, 60o<\/sup>.
\nFrom results, it is observed that angle of incidence is equal to the angle of reflection. This verifies the first law of reflection.<\/p>\nThe experiment has been performed on a flat drawing board, with mirror normal to the plane of board on which white sheet of paper is being fixed. Since the lower tips of all the pins also lie on the same plane (i.e., the plane of paper), it proves the second law of reflection.<\/p>\n
Solution 8S.<\/strong><\/span><\/p>\n<\/p>\n
Solution 9S.<\/strong><\/span><\/p>\n<\/p>\n
Solution 10S.<\/strong><\/span><\/p>\n(a) Angle of incidence = 90o\u00a0<\/sup>– 30o<\/sup>\u00a0= 60o
\n<\/sup>(b) Angle between the incident ray and reflected ray = Angle of incidence + Angle of reflection
\nAngle of reflection = Angle of incidence = 60o
\n<\/sup>Therefore, Angle between the incident ray and reflected ray = 60o<\/sup>\u00a0+ 60o<\/sup>\u00a0= 120o<\/sup><\/p>\n<\/p>\n
Solution 11S.<\/strong><\/span><\/p>\n<\/p>\n
Solution 12S.<\/strong><\/span><\/p>\n<\/p>\n
Solution 13S.<\/strong><\/span><\/p>\n(a) Three characteristics of image formed by plane mirror:<\/p>\n
\n- Image formed in erect (upright)<\/li>\n
- Image formed is virtual<\/li>\n
- Image formed is of the same size as the object<\/li>\n<\/ol>\n
(b) The image is situated at the same perpendicular distance behind the mirror as the object in front of it.<\/p>\n
Solution 14S.<\/strong><\/span><\/p>\n\n\n\nReal Image<\/strong><\/td>\nVirtual image<\/strong><\/td>\n<\/tr>\n\n1. A real image is formed due to actual intersection of the reflected rays.<\/td>\n | 1. A virtual image is formed when the reflected rays meet if they are produced backwards.<\/td>\n<\/tr>\n | \n2. A real image can be obtained on a screen.<\/td>\n | 2. A virtual image cannot be obtained on a screen.<\/td>\n<\/tr>\n | \n3. A real image is inverted with respect to the object.<\/td>\n | 3. A virtual image is erect with respect to the object.<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n Solution 15S.<\/strong><\/span><\/p>\nThe interchange of the left and right sides in the image of an object in a plane mirror is called lateral inversion. \n \nFigure above shows the image formation of a letter P in a plane mirror. \n \nSolution 16S.<\/strong><\/span><\/p>\nThe letters on the front of a ambulance are written laterally inverted, so that the driver of the vehicle moving ahead of the ambulance reads these words laterally inverted as AMBULANCE, in his rear view mirror, and gices side to pass the ambulance first.<\/p>\n Solution 17S.<\/strong><\/span><\/p>\nDue to lateral inversion, , it becomes difficult to read the image of the text of a page formed due to reflection by a plane mirror.<\/p>\n Solution 1M.<\/strong><\/span><\/p>\ni = r<\/p>\n Solution 2M.<\/strong><\/span><\/p>\nErect and of same size<\/p>\n Solution 3M.<\/strong><\/span><\/p>\nvirtual with lateral inversion<\/p>\n Solution 1N.<\/strong><\/span><\/p>\nAngle of incidence (i) + Angle of reflection(r) = 90o\u00a0<\/sup> \nBut, as per the laws of reflection, i = r \nTherefore, 2 i = 90o \n<\/sup>Or, i = r = 45o<\/sup><\/p>\nSolution 2N.<\/strong><\/span><\/p>\nDistance between man and his image = 6m \nDistance between man and mirror + distance between mirror and image = 6m \nBut, Distance between man and mirror (object distance) = distance between mirror and image (image distance) \nTherefore, distance of man from mirror = 6\/2 = 3m<\/p>\n Solution 3N.<\/strong><\/span><\/p>\n(a) Image of the insect is formed 1m behind the mirror. \n(b) Distance between the insect and his image = 1 + 1 = 2 m<\/p>\n Solution 4N.<\/strong><\/span><\/p>\nInitially, distance of the object from the mirror = 60 cm. \nTherefore, image is formed at a distance 60 cm from the mirror, behind it. \nThus, initial distance between the object and image = 60 + 60 = 120 cm \nIf the mirror is moved 25 cm away from the object, \nThe new distance of the object from the mirror = 60 + 25 = 85 cm \nThe new image is now at a distance 85 cm from the mirror behind it. \nThus, new distance of the image from the object = 85 + 85 = 170 cm \nTaking the position of the object as reference point, the distance between the two positions of the image = new distance of image from the object – initial distance of the image from the object \n= (170 – 120) cm = 50 cm \nThus, the image shifts 50 cm away.<\/p>\n Solution 5N.<\/strong><\/span><\/p>\nDistance between man and chart = 3m \nDistance between man and mirror = 2m \nTherefore, distance between chart and mirror = 5 m \nNow, final image is formed on the mirror, which is at a distance of 2 m from the man, therefore, the chart as seen by patient is (5m + 2m =) 7m away.<\/p>\n Exercise 7(B)<\/strong><\/span><\/p>\nSolution 1S.<\/strong><\/span><\/p>\nIf two mirrors make an angle\u00a0\u03b8\u00a0with each other and object is placed in between the two mirrors, the number of images formed is n or (n – 1) depending upon n = 360o<\/sup>\u00a0\/\u00a0\u03b8o<\/sup>\u00a0is odd or even.<\/p>\n(a) If n = 360o<\/sup>\u00a0\/\u00a0\u03b8o<\/sup>\u00a0is odd, \n(i) The number of images formed is n, when the object is placed asymmetrically between the mirrors. \n(ii) The number of images formed is n-1, when the object is placed symmetrically between the mirrors.<\/p>\n(b) If n = 360o<\/sup>\u00a0\/\u00a0\u03b8o<\/sup>\u00a0is even, the number of images is always n-1.<\/p>\nSolution 2S.<\/strong><\/span><\/p>\n<\/p>\n Solution 3S.<\/strong><\/span><\/p>\nFor two mirrors kept perpendicular to each other, three images are formed for an object kept in between them.<\/p>\n <\/p>\n Solution 4S.<\/strong><\/span><\/p>\nFor two mirrors kept parallel to each other, an infinite number of images are formed for an object kept in between them.<\/p>\n Solution 5S.<\/strong><\/span><\/p>\nTwo uses of plane mirror:<\/p>\n \n- In barber’s shop for seeing the hairs at the back of head, two mirrors facing each other are fixed on opposite walls at the front and back of the viewer.<\/li>\n
- In solar heating devices such as a solar cooker, solar water heater, etc., a plane mirror is used to reflect the incident light rays from sun on the substance to be heated.<\/li>\n<\/ol>\n
Solution 1M.<\/strong><\/span><\/p>\n5<\/p>\n Solution 2M.<\/strong><\/span><\/p>\nIn a barber’s shop, two plane mirrors are placed parallel to each other.<\/p>\n Solution 1N.<\/strong><\/span><\/p>\n(a) Angle between the mirrors,\u00a0\u03b8 = 90o \n<\/sup>Now, n = 360o<\/sup>\u00a0\/\u00a0\u03b8o<\/sup>\u00a0= 360o<\/sup>\u00a0\/ 90o\u00a0<\/sup>= 4, which is even. \nHence number of images formed will be (n-1); i.e., 4-1 = 3 images<\/p>\n(b) Angle between the mirrors,\u00a0\u03b8\u00a0= 60o \n<\/sup>Now, n = 360o<\/sup>\u00a0\/\u00a0\u03b8o<\/sup>\u00a0= 360o<\/sup>\u00a0\/ 60o\u00a0<\/sup>= 6, which is even. \nHence number of images formed will be (n-1); i.e., 6-1 = 5 images<\/p>\nSolution 2N.<\/strong><\/span><\/p>\nAngle between the mirrors,\u00a0\u03b8\u00a0= 50o \n<\/sup>Now, n = 360o<\/sup>\u00a0\/\u00a0\u03b8o<\/sup>\u00a0= 360o<\/sup>\u00a0\/ 50o\u00a0<\/sup>= 7.2\u00a0\u00a07, which is odd. \n(i) When placed asymmetrically, number of images formed will be n, i.e. 7. \n(ii) When placed symmetrically, number of images formed will be (n-1); i.e. 7-1 = 6 images<\/p>\nExercise 7(C)<\/strong><\/span><\/p>\nSolution 1S.<\/strong><\/span><\/p>\nA reflecting surface which is a part of a sphere is called a spherical mirror.<\/p>\n Solution 2S.<\/strong><\/span><\/p>\nTwo kinds of spherical mirrors are concave and convex.<\/p>\n Distinction between concave and convex mirror: A concave mirror’s bulging surface is silvered and reflection takes place from the hollow surface but a convex mirror’s inner surface is silvered and reflection takes place from the bulging surface.<\/p>\n Solution 3S.<\/strong><\/span><\/p>\nPole:<\/strong> The geometric centre of the spherical surface of mirror is called the pole of mirror. \nPrincipal axis:<\/strong> It is the straight line joining the pole of the mirror to its centre of curvature. \nCentre of curvature:<\/strong> The centre of curvature of a mirror is the centre of the sphere of which the mirror is a part.<\/p>\nSolution 4S.<\/strong><\/span><\/p>\n<\/p>\n Solution 5S.<\/strong><\/span><\/p>\n<\/p>\n Solution 6S.<\/strong><\/span><\/p>\nFocus of a concave mirror: The focus of a concave mirror is a point on the principal axis through which the light rays incident parallel to principal axis, pass after reflection from the mirror. \nFocal length of a concave mirror: The distance of the focus from the pole of the concave mirror is called its focal length. \n<\/p>\n Solution 7S.<\/strong><\/span><\/p>\nFocus of a convex mirror: The focus of a convex mirror is a point on the principal axis from which, the light rays incident parallel to principal axis, appear to come, after reflection from the mirror. \nFocal length of a convex mirror: The distance of the focus from the pole of the convex mirror is called its focal length. \n<\/p>\n Solution 8S.<\/strong><\/span><\/p>\nIncident ray is directed towards the centre of curvature because the ray is normal to the spherical mirror, so \u2220i = \u2220r = O.<\/p>\n Solution 9S.<\/strong><\/span><\/p>\n<\/p>\n Solution 10S.<\/strong><\/span><\/p>\n<\/p>\n Solution 11S.<\/strong><\/span><\/p>\n<\/p>\n Solution 12S.<\/strong><\/span><\/p>\nTwo convenient rays that are chosen to construct the image by a spherical mirror for a given object:<\/p>\n \n- A ray passing through the centre of curvature:<\/strong> A ray of light passing through the centre of curvature of a concave mirror or a ray directed in the direction of centre of curvature of a convex mirror is reflected back along the same path after reflection.
\n<\/li>\n- A ray parallel to the principal axis:<\/strong> A ray of light parallel to the principal axis, after reflection pass through the principal focus in case of a concave mirror or appears to diverge from it in case of convex mirror.
\n<\/li>\n<\/ol>\nSolution 13S.<\/strong><\/span><\/p>\n<\/p>\n Solution 14S.<\/strong><\/span><\/p>\n<\/p>\n Solution 16S.<\/strong><\/span><\/p>\n \nThe image formed is virtual, erect and magnified.<\/p>\n
Solution 17S.<\/strong><\/span><\/p>\n \nThe image formed is real, inverted and magnified.<\/p>\n
Solution 18S.<\/strong><\/span><\/p>\n \nThe image formed is virtual, erect and diminished.<\/p>\n
Solution 19S.<\/strong><\/span><\/p>\nConvex mirror always produces erect and virtual images. The images formed are diminished, i.e. the size of the image is shorter than the size of the object.<\/p>\n Solution 20S.<\/strong><\/span><\/p>\n(a) If the object is placed between the pole and focus of a concave mirror, the image formed is magnified and erect. \n(b) The image is virtual.<\/p>\n Solution 21S.<\/strong><\/span><\/p>\n | | |