Page 1 26. Reflection of Light Introductory Exercise 26.2 1. Total deviation produced d q = °  + °  180 2 180 2 i d q = °  + 360 2( ) i From figure q = °  90 i Þ d = °  +  360 2 90 [ ] i i = ° 180 Hence rays 1 and 2 are parallel (anti parallel). 2. v 0 2 = m/s for plane mirror v i = 2 m/s. Velocity of approach = + v v i 0 = 4 m/s. 3. 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 Introductory Exercise 26.1 1. Since c = 1 0 0 m e where c is the speed of light in vacuum hence unit of 1 0 0 m e is m/s. 2. Hence B x t y = ´ + ´  2 10 500 10 7 11 T 1.5 sin [ ] Comparing this equation with the standard wave eqution B B kx t y = + 0 sin [ ] w k =  500 1 m Þ k = 2p l Þ l p = 2 k m Þ l p = 2 500 m = p 250 metre w = ´ 1.5 10 11 rad/s Þ 2 10 11 pn = ´ 1.5 Þ n = ´ 1.5 2 10 11 p Hz Speed of the wave v k = = ´ w 1.5 10 500 11 = ´ 3 10 8 m/s Let E 0 be the amplitude of electric field. Then E cB 0 0 8 7 3 10 2 10 = = ´ ´ ´  = 60 V/m Since wave is propagating along xaxis and B along yaxis, hence E must be along zaxis Þ E = 60 V/m sin [ ] 500 10 11 x t + ´ 1.5 90° 90° 90° i q q i 180°–2i N 1 1 2 180°–2q N 2 Page 2 26. Reflection of Light Introductory Exercise 26.2 1. Total deviation produced d q = °  + °  180 2 180 2 i d q = °  + 360 2( ) i From figure q = °  90 i Þ d = °  +  360 2 90 [ ] i i = ° 180 Hence rays 1 and 2 are parallel (anti parallel). 2. v 0 2 = m/s for plane mirror v i = 2 m/s. Velocity of approach = + v v i 0 = 4 m/s. 3. 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 Introductory Exercise 26.1 1. Since c = 1 0 0 m e where c is the speed of light in vacuum hence unit of 1 0 0 m e is m/s. 2. Hence B x t y = ´ + ´  2 10 500 10 7 11 T 1.5 sin [ ] Comparing this equation with the standard wave eqution B B kx t y = + 0 sin [ ] w k =  500 1 m Þ k = 2p l Þ l p = 2 k m Þ l p = 2 500 m = p 250 metre w = ´ 1.5 10 11 rad/s Þ 2 10 11 pn = ´ 1.5 Þ n = ´ 1.5 2 10 11 p Hz Speed of the wave v k = = ´ w 1.5 10 500 11 = ´ 3 10 8 m/s Let E 0 be the amplitude of electric field. Then E cB 0 0 8 7 3 10 2 10 = = ´ ´ ´  = 60 V/m Since wave is propagating along xaxis and B along yaxis, hence E must be along zaxis Þ E = 60 V/m sin [ ] 500 10 11 x t + ´ 1.5 90° 90° 90° i q q i 180°–2i N 1 1 2 180°–2q N 2 = + = + EK KN KN 6 Now in DNKB, NK KB = f tan Þ NK KB = tan f = 4 tan f In DBC C ¢ we get, tan f = ¢ ¢ = = BC CC 2 2 1 f = ° 45 So, NK = ´ ° = 4 45 4 tan m Hence in minimum height = + = 6 4 10 m m m In DAC C ¢ tan q = = 4 2 2 In DL LA ¢ we get, LL LA ¢ = tan q Þ LL¢ = 4 2 Þ LL¢ = 8 m Maximum height = + ¢ = + = CA LL 8 8 16 m Introductory Exercise 26.3 1. Here f =  10 cm (concave mirror) (a) u =  25 cm Using mirror formula, 1 1 1 v u f + = Þ 1 1 1 v f u =  =  + 1 10 1 25 Þ 1 5 2 50 v =  + Þ v =  =  50 3 167 . 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 1 1 10 1 5 v f u =  =  + Þ 1 1 2 10 v =  + Þ v = 10 cm Hence, image is virtual, erect and two time of the object. 2. Here u =  3 m, f =  1 2 m, we have, (a) 1 1 1 v f u =  Þ 1 2 1 3 v =  + Þ 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 =  =  1 m 2 B A C K N L' L 8 m q q f f f C' C D 2 m pole = 4 m M N 1 2 m 6 m 2 m E Page 3 26. Reflection of Light Introductory Exercise 26.2 1. Total deviation produced d q = °  + °  180 2 180 2 i d q = °  + 360 2( ) i From figure q = °  90 i Þ d = °  +  360 2 90 [ ] i i = ° 180 Hence rays 1 and 2 are parallel (anti parallel). 2. v 0 2 = m/s for plane mirror v i = 2 m/s. Velocity of approach = + v v i 0 = 4 m/s. 3. 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 Introductory Exercise 26.1 1. Since c = 1 0 0 m e where c is the speed of light in vacuum hence unit of 1 0 0 m e is m/s. 2. Hence B x t y = ´ + ´  2 10 500 10 7 11 T 1.5 sin [ ] Comparing this equation with the standard wave eqution B B kx t y = + 0 sin [ ] w k =  500 1 m Þ k = 2p l Þ l p = 2 k m Þ l p = 2 500 m = p 250 metre w = ´ 1.5 10 11 rad/s Þ 2 10 11 pn = ´ 1.5 Þ n = ´ 1.5 2 10 11 p Hz Speed of the wave v k = = ´ w 1.5 10 500 11 = ´ 3 10 8 m/s Let E 0 be the amplitude of electric field. Then E cB 0 0 8 7 3 10 2 10 = = ´ ´ ´  = 60 V/m Since wave is propagating along xaxis and B along yaxis, hence E must be along zaxis Þ E = 60 V/m sin [ ] 500 10 11 x t + ´ 1.5 90° 90° 90° i q q i 180°–2i N 1 1 2 180°–2q N 2 = + = + EK KN KN 6 Now in DNKB, NK KB = f tan Þ NK KB = tan f = 4 tan f In DBC C ¢ we get, tan f = ¢ ¢ = = BC CC 2 2 1 f = ° 45 So, NK = ´ ° = 4 45 4 tan m Hence in minimum height = + = 6 4 10 m m m In DAC C ¢ tan q = = 4 2 2 In DL LA ¢ we get, LL LA ¢ = tan q Þ LL¢ = 4 2 Þ LL¢ = 8 m Maximum height = + ¢ = + = CA LL 8 8 16 m Introductory Exercise 26.3 1. Here f =  10 cm (concave mirror) (a) u =  25 cm Using mirror formula, 1 1 1 v u f + = Þ 1 1 1 v f u =  =  + 1 10 1 25 Þ 1 5 2 50 v =  + Þ v =  =  50 3 167 . 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 1 1 10 1 5 v f u =  =  + Þ 1 1 2 10 v =  + Þ v = 10 cm Hence, image is virtual, erect and two time of the object. 2. Here u =  3 m, f =  1 2 m, we have, (a) 1 1 1 v f u =  Þ 1 2 1 3 v =  + Þ 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 =  =  1 m 2 B A C K N L' L 8 m q q f f f C' C D 2 m pole = 4 m M N 1 2 m 6 m 2 m E Using h ut gt = + 1 2 2 Þ t h g = = ´ 2 2 2 9.8 = 0.639 s Similarly again images match at t = 0.78 s. 3. Since image is magnified, hence the mirror is concave. Here, m v u =  Þ  = v u 5 Þ v u =  5 …(i) Let distance between mirror and object is x. Since image is formed at a distance 5 m from mirror v x =  + ( ) 5 …(ii) From Eqs.(i) and (ii), we get  + =  ( ) 5 5 x x Þ 4 5 x = Þ x = 1.25 Hence mirror is placed at 1.25 m on right side of the object by mirror formula 1 1 1 v u f + = , we have 1 1 1 f =   6.25 1.25 Þ f =  6.25 6 , Hence R f = 2 Þ R =  =  6.25 2.08 3 m Thus mirror is concave mirror of radius of curvature 2.08 m. 4. Since the incident rays and reflected rays are parallel to each other therefore mirror is plane mirror. 5. Let us solve the first case : By applying the geometry we can prove that, PA v ¢ = = 40 3 cm Further, in triangles ABP and PA B ¢ ¢ we have, AB A B 40 40 3 = ¢ ¢ ( / ) \ A B AB ¢ ¢ = = 3 2 3 cm Similary, we can solve other parts also. 6. Simply apply : 1 1 1 v u f = = and m I o v u = =  for lateral magnification. If magnitfication is positive, image will be virtual. If magnification is negative, image will be real. 3 40 cm 20 cm a a a 2q q q q q A' F C M B A 20 cm Page 4 26. Reflection of Light Introductory Exercise 26.2 1. Total deviation produced d q = °  + °  180 2 180 2 i d q = °  + 360 2( ) i From figure q = °  90 i Þ d = °  +  360 2 90 [ ] i i = ° 180 Hence rays 1 and 2 are parallel (anti parallel). 2. v 0 2 = m/s for plane mirror v i = 2 m/s. Velocity of approach = + v v i 0 = 4 m/s. 3. 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 Introductory Exercise 26.1 1. Since c = 1 0 0 m e where c is the speed of light in vacuum hence unit of 1 0 0 m e is m/s. 2. Hence B x t y = ´ + ´  2 10 500 10 7 11 T 1.5 sin [ ] Comparing this equation with the standard wave eqution B B kx t y = + 0 sin [ ] w k =  500 1 m Þ k = 2p l Þ l p = 2 k m Þ l p = 2 500 m = p 250 metre w = ´ 1.5 10 11 rad/s Þ 2 10 11 pn = ´ 1.5 Þ n = ´ 1.5 2 10 11 p Hz Speed of the wave v k = = ´ w 1.5 10 500 11 = ´ 3 10 8 m/s Let E 0 be the amplitude of electric field. Then E cB 0 0 8 7 3 10 2 10 = = ´ ´ ´  = 60 V/m Since wave is propagating along xaxis and B along yaxis, hence E must be along zaxis Þ E = 60 V/m sin [ ] 500 10 11 x t + ´ 1.5 90° 90° 90° i q q i 180°–2i N 1 1 2 180°–2q N 2 = + = + EK KN KN 6 Now in DNKB, NK KB = f tan Þ NK KB = tan f = 4 tan f In DBC C ¢ we get, tan f = ¢ ¢ = = BC CC 2 2 1 f = ° 45 So, NK = ´ ° = 4 45 4 tan m Hence in minimum height = + = 6 4 10 m m m In DAC C ¢ tan q = = 4 2 2 In DL LA ¢ we get, LL LA ¢ = tan q Þ LL¢ = 4 2 Þ LL¢ = 8 m Maximum height = + ¢ = + = CA LL 8 8 16 m Introductory Exercise 26.3 1. Here f =  10 cm (concave mirror) (a) u =  25 cm Using mirror formula, 1 1 1 v u f + = Þ 1 1 1 v f u =  =  + 1 10 1 25 Þ 1 5 2 50 v =  + Þ v =  =  50 3 167 . 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 1 1 10 1 5 v f u =  =  + Þ 1 1 2 10 v =  + Þ v = 10 cm Hence, image is virtual, erect and two time of the object. 2. Here u =  3 m, f =  1 2 m, we have, (a) 1 1 1 v f u =  Þ 1 2 1 3 v =  + Þ 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 =  =  1 m 2 B A C K N L' L 8 m q q f f f C' C D 2 m pole = 4 m M N 1 2 m 6 m 2 m E Using h ut gt = + 1 2 2 Þ t h g = = ´ 2 2 2 9.8 = 0.639 s Similarly again images match at t = 0.78 s. 3. Since image is magnified, hence the mirror is concave. Here, m v u =  Þ  = v u 5 Þ v u =  5 …(i) Let distance between mirror and object is x. Since image is formed at a distance 5 m from mirror v x =  + ( ) 5 …(ii) From Eqs.(i) and (ii), we get  + =  ( ) 5 5 x x Þ 4 5 x = Þ x = 1.25 Hence mirror is placed at 1.25 m on right side of the object by mirror formula 1 1 1 v u f + = , we have 1 1 1 f =   6.25 1.25 Þ f =  6.25 6 , Hence R f = 2 Þ R =  =  6.25 2.08 3 m Thus mirror is concave mirror of radius of curvature 2.08 m. 4. Since the incident rays and reflected rays are parallel to each other therefore mirror is plane mirror. 5. Let us solve the first case : By applying the geometry we can prove that, PA v ¢ = = 40 3 cm Further, in triangles ABP and PA B ¢ ¢ we have, AB A B 40 40 3 = ¢ ¢ ( / ) \ A B AB ¢ ¢ = = 3 2 3 cm Similary, we can solve other parts also. 6. Simply apply : 1 1 1 v u f = = and m I o v u = =  for lateral magnification. If magnitfication is positive, image will be virtual. If magnification is negative, image will be real. 3 40 cm 20 cm a a a 2q q q q q A' F C M B A 20 cm AIEEE Corner ¢ Subjective Questions (Level 1) 1. Here v = 39.2 cm, hence v =  39.2 cm and magnification m = 1 Þ h h i o = = 4.85 Hence image is formed at 39.2 cm behind the mirror and height of image is = 4.85 cm. 2. From figure, angle of incident = ° 15 Let reflected ray makes an angle q with the horizontal, then q + ° + ° = ° 15 15 90 Þ q = ° 60 3. 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. 4. The distance of the object from images are 2 4 6 b b b , , ..... etc. Hence the images distance are 2 nb, where n = 1 2 , ,K . Ans. 5. Suppose mirror is rotated at angle q about its axis perpendicular to both the incident ray and normal as shown in figure In figure (b) I remain unchanged N and R shift to N¢ and R¢. From figure (a) angle of rotation = i, From figure (b) it is i  2q Thus, reflected ray has been rotated by angle 2q. 6. I is incident ray Ð = ° = Ð i r 30 From D PA A ¢ , we get 4 i i I N R y x (a) IV R' y x (b) I q q i–q i–q i–2q O'' b O''' 4b O' O''' b b 1 B D A C 4b 2 b O'' 30 cm 30 cm O''' 50 cm 1o cm 1o cm O' O''' 70 cm 40 cm B D A C 50 cm 150° q 150° 90° N Reflacted ray Horizontal 15° 15° Incident ray Mirror 30°30° 20 cm B B' x R I A 1.6m P A' Page 5 26. Reflection of Light Introductory Exercise 26.2 1. Total deviation produced d q = °  + °  180 2 180 2 i d q = °  + 360 2( ) i From figure q = °  90 i Þ d = °  +  360 2 90 [ ] i i = ° 180 Hence rays 1 and 2 are parallel (anti parallel). 2. v 0 2 = m/s for plane mirror v i = 2 m/s. Velocity of approach = + v v i 0 = 4 m/s. 3. 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 Introductory Exercise 26.1 1. Since c = 1 0 0 m e where c is the speed of light in vacuum hence unit of 1 0 0 m e is m/s. 2. Hence B x t y = ´ + ´  2 10 500 10 7 11 T 1.5 sin [ ] Comparing this equation with the standard wave eqution B B kx t y = + 0 sin [ ] w k =  500 1 m Þ k = 2p l Þ l p = 2 k m Þ l p = 2 500 m = p 250 metre w = ´ 1.5 10 11 rad/s Þ 2 10 11 pn = ´ 1.5 Þ n = ´ 1.5 2 10 11 p Hz Speed of the wave v k = = ´ w 1.5 10 500 11 = ´ 3 10 8 m/s Let E 0 be the amplitude of electric field. Then E cB 0 0 8 7 3 10 2 10 = = ´ ´ ´  = 60 V/m Since wave is propagating along xaxis and B along yaxis, hence E must be along zaxis Þ E = 60 V/m sin [ ] 500 10 11 x t + ´ 1.5 90° 90° 90° i q q i 180°–2i N 1 1 2 180°–2q N 2 = + = + EK KN KN 6 Now in DNKB, NK KB = f tan Þ NK KB = tan f = 4 tan f In DBC C ¢ we get, tan f = ¢ ¢ = = BC CC 2 2 1 f = ° 45 So, NK = ´ ° = 4 45 4 tan m Hence in minimum height = + = 6 4 10 m m m In DAC C ¢ tan q = = 4 2 2 In DL LA ¢ we get, LL LA ¢ = tan q Þ LL¢ = 4 2 Þ LL¢ = 8 m Maximum height = + ¢ = + = CA LL 8 8 16 m Introductory Exercise 26.3 1. Here f =  10 cm (concave mirror) (a) u =  25 cm Using mirror formula, 1 1 1 v u f + = Þ 1 1 1 v f u =  =  + 1 10 1 25 Þ 1 5 2 50 v =  + Þ v =  =  50 3 167 . 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 1 1 10 1 5 v f u =  =  + Þ 1 1 2 10 v =  + Þ v = 10 cm Hence, image is virtual, erect and two time of the object. 2. Here u =  3 m, f =  1 2 m, we have, (a) 1 1 1 v f u =  Þ 1 2 1 3 v =  + Þ 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 =  =  1 m 2 B A C K N L' L 8 m q q f f f C' C D 2 m pole = 4 m M N 1 2 m 6 m 2 m E Using h ut gt = + 1 2 2 Þ t h g = = ´ 2 2 2 9.8 = 0.639 s Similarly again images match at t = 0.78 s. 3. Since image is magnified, hence the mirror is concave. Here, m v u =  Þ  = v u 5 Þ v u =  5 …(i) Let distance between mirror and object is x. Since image is formed at a distance 5 m from mirror v x =  + ( ) 5 …(ii) From Eqs.(i) and (ii), we get  + =  ( ) 5 5 x x Þ 4 5 x = Þ x = 1.25 Hence mirror is placed at 1.25 m on right side of the object by mirror formula 1 1 1 v u f + = , we have 1 1 1 f =   6.25 1.25 Þ f =  6.25 6 , Hence R f = 2 Þ R =  =  6.25 2.08 3 m Thus mirror is concave mirror of radius of curvature 2.08 m. 4. Since the incident rays and reflected rays are parallel to each other therefore mirror is plane mirror. 5. Let us solve the first case : By applying the geometry we can prove that, PA v ¢ = = 40 3 cm Further, in triangles ABP and PA B ¢ ¢ we have, AB A B 40 40 3 = ¢ ¢ ( / ) \ A B AB ¢ ¢ = = 3 2 3 cm Similary, we can solve other parts also. 6. Simply apply : 1 1 1 v u f = = and m I o v u = =  for lateral magnification. If magnitfication is positive, image will be virtual. If magnification is negative, image will be real. 3 40 cm 20 cm a a a 2q q q q q A' F C M B A 20 cm AIEEE Corner ¢ Subjective Questions (Level 1) 1. Here v = 39.2 cm, hence v =  39.2 cm and magnification m = 1 Þ h h i o = = 4.85 Hence image is formed at 39.2 cm behind the mirror and height of image is = 4.85 cm. 2. From figure, angle of incident = ° 15 Let reflected ray makes an angle q with the horizontal, then q + ° + ° = ° 15 15 90 Þ q = ° 60 3. 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. 4. The distance of the object from images are 2 4 6 b b b , , ..... etc. Hence the images distance are 2 nb, where n = 1 2 , ,K . Ans. 5. Suppose mirror is rotated at angle q about its axis perpendicular to both the incident ray and normal as shown in figure In figure (b) I remain unchanged N and R shift to N¢ and R¢. From figure (a) angle of rotation = i, From figure (b) it is i  2q Thus, reflected ray has been rotated by angle 2q. 6. I is incident ray Ð = ° = Ð i r 30 From D PA A ¢ , we get 4 i i I N R y x (a) IV R' y x (b) I q q i–q i–q i–2q O'' b O''' 4b O' O''' b b 1 B D A C 4b 2 b O'' 30 cm 30 cm O''' 50 cm 1o cm 1o cm O' O''' 70 cm 40 cm B D A C 50 cm 150° q 150° 90° N Reflacted ray Horizontal 15° 15° Incident ray Mirror 30°30° 20 cm B B' x R I A 1.6m P A' x 20 30 = ° tan Þ x = ° 20 30 tan No. of reflection = = ´ ° AB x 160 20 30 cm cm tan = » 8 3 14 Hence the reflected ray reach other end after 14 reflections. 7. The deviation produced by mirror M 1 is = °  180 2 a and the deviation produced by mirror M 2 is =  180 2 Hence total deviation =  +  f 180 2 180 2 a =  + f 360 2 ( ) a In D ABC we get, 90 90 180  + +  f = a q Þ a q + f = Hence deviation produces =  180 2q. 8. Here f R =  =  =  2 22 2 11 cm Object height h 0 6 = mm u =  16.5 cm (a) The ray diagram is shown in figure Using mirror formula, 1 1 1 v u f + = Þ 1 1 1 v f u =  Þ 1 1 11 1 11 11 v =  + =  + ´ 16.5 165 16.5 Þ v =  ´ =  16.5 5.5 11 33 cm 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   m v u = ½ ½ ½ ½ = = 33 2 16.5 Hence size of image is h h i = ´ 2 0 = ´ = 2 6 12 mm. 9. Here u =  12 cm, f R = + 2 = + 10 cm Using mirror formula 1 1 1 v u f + = we get 1 1 1 1 10 1 12 v f u =  = + = + 6 5 60 Þ v = 60 11 cm = 5.46 cm The image is formed on right side of the vertex at a distance 60 11 cm. the image is virtual and erect the absolute magnification is given by   m v u = ½ ½ ½ ½ Þ   ( ) m = ´  ½ ½ ½ ½ ½ ½ = 60 11 12 5 11 Q m < 1 Hence image is demagnified. Height of image h m h i = ´   0 5 A A' B' u = 16.5 cm f B a 90°–a 180°–2a A Z' I 1 M 1 R 2 180°–2q C R 1 90°–f q a f fRead More
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