Solution of Optics-Modern-Physics-2E-Sol--DC-Pandey

Solutions of Optics & Modern Physics Lesson 26th to 30th By DC Pandey 26 Reflection of Light Introductory Exercise 26.1 where c is the speed of light m e0 in vacuum hence unit of is m/s m e0 w = 1.5 ¥ 1011 rad/s Since c = Hence -7 11 By = ¥ 10 T sin [500x + 1.5 ¥ 10 t ] Comparing this equation with the standard wave eqution By = B0 sin [ kx + w t ] 2p k = 500 m -1 fi k = l 2p m fi l= k 2p p m= metre fi l= 500 250 fi fi pn = 1.5 ¥ 1011 1.5 n= ¥ 1011 Hz 2p Speed of the wave v = w 1.5 ¥ 1011 = k 500 = ¥ 108 m/s Let E0 be the amplitude of electric field Then E0 = cB0 = ¥ 108 ¥ ¥ 10-7 = 60 V/m Since wave is propagating along x-axis and B along y-axis, hence E must be along z-axis fi E = 60 V/m sin [500x + 1.5 ¥ 1011 t ] Introductory Exercise 26.2 Total deviation produced From figure q = 90∞ - i fi i i 180 °–2 i N1 = 180∞ 90° 90° d = 360∞ - [ i + 90 - i] 90° q q v0 = m/s for plane mirror vi = m/s 180°–2q N2 d = 180∞ - i + 180∞ - 2q d = 360∞ - 2( i + q) Hence rays and are parallel (antiparallel) Velocity of approach = v0 + vi = m/s In figure, AB is mirror, G is ground, CD is pole and M is the man The minimum height to see the image of top of pole is = EN q L A N 2m K N1 M C E tan f = q pole =4m 2m D f 8m L' f f 2m B C' m f = 45∞ So, NK = ¥ tan 45∞ = m Hence in minimum height = m + m = 10 m In DAC¢ C tan q = C = EK + KN = + KN Now in DNKB, NK = tan f fi NK = KB tan f KB =2 In DL¢ LA we get, LL¢ = tan q LA LL¢ fi =2 LL¢ = m fi = tan f BC¢ = =1 CC¢ Maximum height = CA + LL¢ = + = 16 m In DBC¢ C we get, Introductory Exercise 26.3 Here f = - 10 cm (concave mirror) (a) u = - 25 cm Using mirror formula, 1 + = v u f fi fi fi 1 1 = - =+ v f u 10 25 -5 + = v 50 50 v== - 167 cm Hence image is real, inverted and less height of the object (b) Since u = - 10 cm, Hence object is situated on focus of the image formed at • (c) u - 5, f = - 10 1 1 = - =+ v f u 10 fi -1 + = v 10 fi v = 10 cm Hence, image is virtual, erect and two time of the object Here u = - m, f = - m, we have, 1 (a) = v f u fi 1 = -2 + v fi v = - 0.6 m As ball moves towards focus the image moves towards -• and image is real as the distance decreases by focal length image become virtual which moves from + • to zero (b) The image of the ball coincide with ball, when u = - R = - m Using h = ut + fi t= gt Since the incident rays and reflected rays are parallel to each other therefore mirror is plane mirror 2¥ 2h = g 9.8 Let us solve the first case : = 0.639 s Similarly again images match at t = 0.78 s Since image is magnified, hence the mirror is concave -v -v Here, m= fi =5 u u fi v = -5u A Since image is formed at a distance m from mirror …(ii) From Eqs.(i) and (ii), we get - (5 + x) = - 5x fi 4x = fi x = 1.25 Hence mirror is placed at 1.25 m on right side of the object by mirror formula 1 + = , v u f we have 1 =f 6.25 1.25 fi f = - 6.25 , Hence R = f fi R = - a a a A' q 2q F q C …(i) Let distance between mirror and object is x v = - (5 + x) q M q B 40 cm 20 cm By applying the geometry we can prove that, 40 PA¢ = v = cm Further, in triangles ABP and PA¢ B¢ we have, AB A¢ B¢ = 40 ( 40/ 3) \ Thus mirror is concave mirror of radius of curvature 2.08 m A¢ B¢ = AB = cm 3 Similary, we can solve other parts also Simply apply : 1 = = v u f I -v for lateral magnification If = o u magnitfication is positive, image will be virtual If magnification is negative, image will be real and m = 6.25 = - 2.08 m 20 cm AIEEE Corner ¢ Subjective Questions (Level 1) Hence the images distance are nb, where n = 1, 2, K Ans Here v = 39.2 cm, hence v = - 39.2 cm and magnification m = fi hi = ho = 4.85 Hence image is formed at 39.2 cm behind the mirror and height of image is = 4.85 cm Suppose mirror is rotated at angle q about its axis perpendicular to both the incident ray and normal as shown in figure y From figure, angle of incident = 15∞ N 0° i Mirror x q Horizontal (a) 15° 15° y IV Let reflected ray makes an angle q with the horizontal, then i–2q I A R' i–q q i–q q + 15∞ + 15∞ = 90∞ fi q = 60∞ i ° 15 R N Incident ray 50 Reflacted ray 90° I C x q (b) O' 1o cm 30 cm 50 cm B 1o cm O'' 30 cm O''' 50 cm O''' 70 cm In figure (b) I remain unchanged N and R shift to N¢ and R¢ From figure (a) angle of rotation = i, D 40 cm Since mirror are parallel to each other • image are formed the distance of five closet to object are 20 cm, 60 cm, 80 cm, 100 cm and 140 cm From figure (b) it is i - 2q Thus, reflected ray has been rotated by angle 2q I is incident ray –i = 30∞ = –r The distance of the object from images are 2b, b, 6b etc 30° C 30° A 1.6m A 2b 20 cm R O''' b O'' b O' b O''' I A' 4b 4b B x P From D PA¢ A, we get D B B' x = tan 30∞ fi x = 20 tan 30∞ 20 160 cm AB No of reflection = = x 20 cm ¥ tan 30∞ = ª 14 Hence the reflected ray reach other end after 14 reflections The deviation produced by mirror M1 is = 180∞ - a M1 Z' I1 a 180°–2a R2 90°–a a R1 f f 90°–f 180°–2q q A C and the deviation produced by mirror M2 is = 180 - Hence total deviation = 180 - 2a + 180 - 2f = 360 - ( a + f) Using mirror formula, fi 1 = v f u fi - 165 + 11 1 =+ = v 11 16.5 16.5 ¥ 11 fi v=- Hence size of image is hi = ¥ h0 = ¥ = 12 mm R Here u = - 12 cm, f = + = + 10 cm Using mirror formula 1 + = v u f we get 1 1 = - = + v f u 10 12 90 - a + q + 90 - f = 180 Object height h0 = mm u = - 16.5 cm (a) The ray diagram is shown in figure B A' A fi The image is formed on right side of the 60 vertex at a distance cm the image is 11 virtual and erect the absolute magnification v is given by |m|=Ω Ω Ω uΩ fi f B' u = 16.5 cm 6+5 60 60 cm = 5.46 cm v= 11 = a + f=q Hence deviation produces = 180 - 2q R 22 Here f = - = = - 11 cm 2 16.5 ¥ 11 = - 33 cm 5.5 Hence the image is formed at 33 cm from the pole (vertex) of mirror on the object side the image is real, inverted and magnified The absolute magnification v 33 |m|=Ω Ω = =2 Ω uΩ 16.5 In D ABC we get, fi 1 + = v u f Q Ω Ω 60 Ω= |m|=Ω Ω 11 ¥ ( - 12) Ω 11 m T, xy = l( N )2 y For one half life N = N0 2 fi ( xy) T lN 20 xy ÊN ˆ = lÁ ˜ = = 4 Ë ¯ Hence correct options are (a), (b) and (d) The correct options are (a), (b), (c) and (d) A nucleus in excited state emits a high energy photon called as g-ray The reaction is X * ỉỈ X + g Hence by gamma radiation atomic number and mass number are not changed Since after emission of one a atomic number reduced by (2 a4 ) and after 2b atomic number is increased by (2) Hence correct options are (a), (b) and (c) Here half lives are T and 2T and N x = N , N y = N after 4T for first substance = half lives and after 4T the second substance = Half lines N N x = N ÊÁ ˆ˜ = 16 Ë ¯ fi N N y = N ÊÁ ˆ˜ = Ë2¯ N /16 N x= x = = Ny N /4 fi Let their activity are Rx and R y Rx = lx N x and R y = l y N y 0.693 Rx lx N x y= = ¥ = T ¥ R y l y N y 0.693 2T Rx =y= Ry fi fi fi Hence correct options are (b) and (c) Since nuclear forces are vary short range charge independent, no electromagnetic and they exchange (n Ỉ p or p Ỉ n) Hence the correct options are (a), (b), (c) and (d) Q R = R0 A1/ r= A ¥ 1.67 ¥ 10-27 kg M = 4/3 pR pR03 A fi r is independent of A But r = 1.67 ¥ 10-27 kg ¥ 3.14 ¥ (1.3 ¥ 10-15 ) 3 = 1.8 ¥ 1017 kg/m Hence correct options are (b) and (c) 93 ¢ Match the Columns Here N = x, let l is the decay constant fi N = N e - lt = x e - lt dN dN = lx e- lt but dt dt =y t=0 y = lx e = lx y l= x fi fi In reaction P + P Æ Q energy is released Q Binding energy increases when two or more lighter nucleus combine to form haviour nucleus Hence correct match is (a) ỉỈ p Similerly for reaction Hence P + P + RỈR (a) ỉỈ s Correct match is Half life T1/2 = log log x = = log y l y x Hence (b) ỉỈ p We have activity R = lN = l xe- lt at t = l R = lx -l ¥ l e = l x e -1 = lx e but lx = y y R= e fi for reaction P + R = 2Q from graph BE per nucleon increases Hence energy is released Correct match is (c) ỉỈ p For the reaction P+Q=R We will check energy process if BE per nucleon is given Hence data is not sufficient correct match is (d) ỉỈ s Hence (c) ỉỈ r Number of nuclei after time t = l N =xe (b) ỉỈ p -l ¥ l = x e Hence (d) ỉỈ s Thus correct match is Since A and B are radioactive nuclei of ( A + B) decreases with time Hence correct match is (a) ỉỈ q Q A is converted into B and B is converted into C and decay rate of A Ỉ B and B Ỉ C are not known Hence correct match is (b) ỉỈ s (b) ỉỈ r Since at time passes A is converted to B and B is converted to C Hence nuclei of ( B + C) increases Correct match is (c) ỉỈ r (c) ỉỈ p (d) ỉỈ s Similerly the correct match for (d) is (a) ỉỈ s 94 (d) ỉỈ s After emission of a particle mass no decreased by but after emission of 1b particle atomic number will increase or decrease by Hence for (a) (a) ỉỈ p Since BE per nucleon of heavy nuclei is about 7.2 MeV Hence (b) ỉỈ s (b) ỉỈ p, r X-ray photon have wavelength about Å the energy of this wavelength is of order of 10 keV (c) ỉỈ s (c) ỉỈ r (d) ỉỈ q, r Q Visible light energy of order of ª eV Hence (a) ỉỈ p, s For (d) ỉỈ q ... cm F D Length of mirror = AB = 0.75 m The ray diagram is shown in above figure H is the Head of the boy and F is the feet It also shows the paths of the rays that leaves the head of the man enter... m = sin r3 m Let focal length of one part is f ¢ 1 then = ( n - 1) È - ˘ ÍỴ R • ˙˚ f¢ Let refractive index of the liquid s nl fi …(ii) 14 Let radius of curvature of the lens is R 1 R then = (... focal length of = 1 = + F 10 10 fi F = cm Here u = - 7.5 cm , F = cm 1 1 7.5 - fi + - fi = v 7.5 v 7.5 ¥ fi v = + 15 cm height of image 15 v Hence|m|= fi = height of object 7.5 u Height of image = ¥