Page 1 J E E - P h y s i c s 56 E NODE6\E:\Data\2014\Kota\JEE-Advanced\SMP\Phy\Unit No-11\Ray-Optics\Eng\02. Ray theory-Part2.p65 Solution Ans. 1.5 For case (a) 1 1 v ? ? ? = 1 R ? ? , 2 1 1 v v R ? ? ? = 1 ? ? ? and v 2 = 2R m ? m=2 ? ? 1 ? ? µ For case (b) 2 1 v ? ? ? = 1 R ? ? ? and v 2 = R m 1 ? ? µ = m Therefore ? ? 2 1 ? ? ? ? µ ? 1 1 2 ? ? ? ? µ = 1.5 Example#27 In figure, L is half part of an equiconvex glass lens ( ? = 1.5) whose surfaces have radius of curvature R = 40 cm and its right surface is silvered. Normal to its principal axis a plane mirror M is placed on right of the lens. Distance between lens L and mirror M is b. A small object O is placed on left of the lens such that there is no parallax between final images formed by the lens and mirror. If transverse length of final image formed by lens is twice that of image formed by the mirror, calculate distance 'a' in cm between lens and object. O a b M \ \ \ \ \ \ \ \ \ \ \ \ \ \ Solution Ans. 5 Distance of image of object O from plane mirror = a + b. Since, there is no parallax between the images formed by the silvered lens L and plane mirror M, therefore, two images are formed at the same point. Distance of image = (a + 2b) behind lens. Since, length of image formed by L is twice the length of image formed by the mirror M and length of image formed by a plane mirror is always equal to length of the object, therefore, transverse magnification produced by the lens L is equal to 2.Since, distance of object from L is a, therefore, distance of image from L must be equal to 2a. ? (a + 2b) = 2a ? b = a 2 The silvered lens L may be assumed as a combination of an equi–convex lens and a concave mirror placed in contact with each other co–axially as shown in figure. \ \ \ \ \ \ \ \ \ \ \ \ \ \ Focal length of convex lens f 1 is given by 1 1 f = ( ?–1) 1 2 1 1 R R ? ? ? ? ? ? ? ? f 1 = 40 cm For concave mirror focal length, f m = R 2 = – 20 cm The combination L behaves like a mirror whose equivalent focal length F is given by m 1 1 1 2 F f f ? ? ? F = –10 cm, Hence, for the combination u = –a, v = +2a, F = –10 cm Using mirror formula, 1 1 1 v u F ? ? ? a = 5 cm JEEMAIN.GURU Page 2 J E E - P h y s i c s 56 E NODE6\E:\Data\2014\Kota\JEE-Advanced\SMP\Phy\Unit No-11\Ray-Optics\Eng\02. Ray theory-Part2.p65 Solution Ans. 1.5 For case (a) 1 1 v ? ? ? = 1 R ? ? , 2 1 1 v v R ? ? ? = 1 ? ? ? and v 2 = 2R m ? m=2 ? ? 1 ? ? µ For case (b) 2 1 v ? ? ? = 1 R ? ? ? and v 2 = R m 1 ? ? µ = m Therefore ? ? 2 1 ? ? ? ? µ ? 1 1 2 ? ? ? ? µ = 1.5 Example#27 In figure, L is half part of an equiconvex glass lens ( ? = 1.5) whose surfaces have radius of curvature R = 40 cm and its right surface is silvered. Normal to its principal axis a plane mirror M is placed on right of the lens. Distance between lens L and mirror M is b. A small object O is placed on left of the lens such that there is no parallax between final images formed by the lens and mirror. If transverse length of final image formed by lens is twice that of image formed by the mirror, calculate distance 'a' in cm between lens and object. O a b M \ \ \ \ \ \ \ \ \ \ \ \ \ \ Solution Ans. 5 Distance of image of object O from plane mirror = a + b. Since, there is no parallax between the images formed by the silvered lens L and plane mirror M, therefore, two images are formed at the same point. Distance of image = (a + 2b) behind lens. Since, length of image formed by L is twice the length of image formed by the mirror M and length of image formed by a plane mirror is always equal to length of the object, therefore, transverse magnification produced by the lens L is equal to 2.Since, distance of object from L is a, therefore, distance of image from L must be equal to 2a. ? (a + 2b) = 2a ? b = a 2 The silvered lens L may be assumed as a combination of an equi–convex lens and a concave mirror placed in contact with each other co–axially as shown in figure. \ \ \ \ \ \ \ \ \ \ \ \ \ \ Focal length of convex lens f 1 is given by 1 1 f = ( ?–1) 1 2 1 1 R R ? ? ? ? ? ? ? ? f 1 = 40 cm For concave mirror focal length, f m = R 2 = – 20 cm The combination L behaves like a mirror whose equivalent focal length F is given by m 1 1 1 2 F f f ? ? ? F = –10 cm, Hence, for the combination u = –a, v = +2a, F = –10 cm Using mirror formula, 1 1 1 v u F ? ? ? a = 5 cm JEEMAIN.GURU J E E - P h y s i c s E 55 NODE6\E:\Data\2014\Kota\JEE-Advanced\SMP\Phy\Unit No-11\Ray-Optics\Eng\02. Ray theory-Part2.p65 Distance of Bird as seen by fish x Bf = d + ?h h d By differentiating fB d(x ) dh 1 d(d) dt dt dt ? ? ? v FB ? 3 + 4 ? = 6 cm/s dh d(d) 4 5 2 3 cm / s, 2 ( 2) 4 cm / s, dt dt 3 ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? v BF = d(d) dh dt dt ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? 4 + 4 3 ? ? ? ? ? ? (3) ? 8 cm/s d(d) dt (for fish image after reflection = 0 ) ? 3 + 1 ? (0) = 3 cm/s Similarly speed of image of bird ? 4 cm/s Example#25 In the shown figure the focal length of equivalent system in the form of 50x 13 ? ? ? ? ? ? . Find the value of x. Ans. 2 1 1 3 1 1 1 1 f 2 10 10 10 ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ; 2 1 6 1 1 3 1 f 5 10 20 100 ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? and 3 1 8 1 1 3 1 f 5 20 20 50 ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? 1 2 3 1 1 1 1 1 3 3 f f f f 10 100 50 ? ? ? ? ? ? ? = 100 13 = 50x 13 ? x=2 Example#26 Quarter part of a transparent cylinder ABC of radius R is kept on a horizontal floor and a horizontal beam of light falls on the cylinder in the two different arrangement of cylinder as shown in the figure (a) & (b). In arrangement (a) light converges at point D, which is at a distance 2R/m from B. And in arrangement (b) light converges at point E, which is at a distance R/(m – 1) from C. Find out the refractive index of the material. C B A D 2R/m B C E R m -1 (a) (b) JEEMAIN.GURU Page 3 J E E - P h y s i c s 56 E NODE6\E:\Data\2014\Kota\JEE-Advanced\SMP\Phy\Unit No-11\Ray-Optics\Eng\02. Ray theory-Part2.p65 Solution Ans. 1.5 For case (a) 1 1 v ? ? ? = 1 R ? ? , 2 1 1 v v R ? ? ? = 1 ? ? ? and v 2 = 2R m ? m=2 ? ? 1 ? ? µ For case (b) 2 1 v ? ? ? = 1 R ? ? ? and v 2 = R m 1 ? ? µ = m Therefore ? ? 2 1 ? ? ? ? µ ? 1 1 2 ? ? ? ? µ = 1.5 Example#27 In figure, L is half part of an equiconvex glass lens ( ? = 1.5) whose surfaces have radius of curvature R = 40 cm and its right surface is silvered. Normal to its principal axis a plane mirror M is placed on right of the lens. Distance between lens L and mirror M is b. A small object O is placed on left of the lens such that there is no parallax between final images formed by the lens and mirror. If transverse length of final image formed by lens is twice that of image formed by the mirror, calculate distance 'a' in cm between lens and object. O a b M \ \ \ \ \ \ \ \ \ \ \ \ \ \ Solution Ans. 5 Distance of image of object O from plane mirror = a + b. Since, there is no parallax between the images formed by the silvered lens L and plane mirror M, therefore, two images are formed at the same point. Distance of image = (a + 2b) behind lens. Since, length of image formed by L is twice the length of image formed by the mirror M and length of image formed by a plane mirror is always equal to length of the object, therefore, transverse magnification produced by the lens L is equal to 2.Since, distance of object from L is a, therefore, distance of image from L must be equal to 2a. ? (a + 2b) = 2a ? b = a 2 The silvered lens L may be assumed as a combination of an equi–convex lens and a concave mirror placed in contact with each other co–axially as shown in figure. \ \ \ \ \ \ \ \ \ \ \ \ \ \ Focal length of convex lens f 1 is given by 1 1 f = ( ?–1) 1 2 1 1 R R ? ? ? ? ? ? ? ? f 1 = 40 cm For concave mirror focal length, f m = R 2 = – 20 cm The combination L behaves like a mirror whose equivalent focal length F is given by m 1 1 1 2 F f f ? ? ? F = –10 cm, Hence, for the combination u = –a, v = +2a, F = –10 cm Using mirror formula, 1 1 1 v u F ? ? ? a = 5 cm JEEMAIN.GURU J E E - P h y s i c s E 55 NODE6\E:\Data\2014\Kota\JEE-Advanced\SMP\Phy\Unit No-11\Ray-Optics\Eng\02. Ray theory-Part2.p65 Distance of Bird as seen by fish x Bf = d + ?h h d By differentiating fB d(x ) dh 1 d(d) dt dt dt ? ? ? v FB ? 3 + 4 ? = 6 cm/s dh d(d) 4 5 2 3 cm / s, 2 ( 2) 4 cm / s, dt dt 3 ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? v BF = d(d) dh dt dt ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? 4 + 4 3 ? ? ? ? ? ? (3) ? 8 cm/s d(d) dt (for fish image after reflection = 0 ) ? 3 + 1 ? (0) = 3 cm/s Similarly speed of image of bird ? 4 cm/s Example#25 In the shown figure the focal length of equivalent system in the form of 50x 13 ? ? ? ? ? ? . Find the value of x. Ans. 2 1 1 3 1 1 1 1 f 2 10 10 10 ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ; 2 1 6 1 1 3 1 f 5 10 20 100 ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? and 3 1 8 1 1 3 1 f 5 20 20 50 ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? 1 2 3 1 1 1 1 1 3 3 f f f f 10 100 50 ? ? ? ? ? ? ? = 100 13 = 50x 13 ? x=2 Example#26 Quarter part of a transparent cylinder ABC of radius R is kept on a horizontal floor and a horizontal beam of light falls on the cylinder in the two different arrangement of cylinder as shown in the figure (a) & (b). In arrangement (a) light converges at point D, which is at a distance 2R/m from B. And in arrangement (b) light converges at point E, which is at a distance R/(m – 1) from C. Find out the refractive index of the material. C B A D 2R/m B C E R m -1 (a) (b) JEEMAIN.GURU J E E - P h y s i c s 54 E NODE6\E:\Data\2014\Kota\JEE-Advanced\SMP\Phy\Unit No-11\Ray-Optics\Eng\02. Ray theory-Part2.p65 Example#23 Column-I contains a list of mirrors and position of object. Match this with Column-II describing the nature of image. Column I Column II (A) F C (P) real, inverted, enlarged (B) (Q) virtual, erect, enlarged (C) (R) virtual, erect, diminished (D) F (S) virtual, erect Solution Ans. (A) P (b)RS (C) S (D) QS Example#24 A bird in air is diving vertically over a tank with speed 5 cm/s, base of tank is silvered. A fish in the tank is rising upward along the same line with speed 2 cm/s. Water level is falling at rate of 2 cm/s. [Take : ? water = 4/3] Column I (cm/s) Column II (A) Speed of the image of fish as seen by the bird directly (P) 8 (B) Speed of the image of fish formed after reflection in (Q) 6 the mirror as seen by the bird (C) Speed of image of bird relative to the fish looking upwards (R) 3 (D) Speed of image of bird relative to the fish looking (S) 4 downwards in the mirror Solution Ans. (A) – (Q) ; (B) – (R) ; (C) – (P) ; (D) – (S) Distance of fish asseen by bird x fB = h + d ? JEEMAIN.GURU Page 4 J E E - P h y s i c s 56 E NODE6\E:\Data\2014\Kota\JEE-Advanced\SMP\Phy\Unit No-11\Ray-Optics\Eng\02. Ray theory-Part2.p65 Solution Ans. 1.5 For case (a) 1 1 v ? ? ? = 1 R ? ? , 2 1 1 v v R ? ? ? = 1 ? ? ? and v 2 = 2R m ? m=2 ? ? 1 ? ? µ For case (b) 2 1 v ? ? ? = 1 R ? ? ? and v 2 = R m 1 ? ? µ = m Therefore ? ? 2 1 ? ? ? ? µ ? 1 1 2 ? ? ? ? µ = 1.5 Example#27 In figure, L is half part of an equiconvex glass lens ( ? = 1.5) whose surfaces have radius of curvature R = 40 cm and its right surface is silvered. Normal to its principal axis a plane mirror M is placed on right of the lens. Distance between lens L and mirror M is b. A small object O is placed on left of the lens such that there is no parallax between final images formed by the lens and mirror. If transverse length of final image formed by lens is twice that of image formed by the mirror, calculate distance 'a' in cm between lens and object. O a b M \ \ \ \ \ \ \ \ \ \ \ \ \ \ Solution Ans. 5 Distance of image of object O from plane mirror = a + b. Since, there is no parallax between the images formed by the silvered lens L and plane mirror M, therefore, two images are formed at the same point. Distance of image = (a + 2b) behind lens. Since, length of image formed by L is twice the length of image formed by the mirror M and length of image formed by a plane mirror is always equal to length of the object, therefore, transverse magnification produced by the lens L is equal to 2.Since, distance of object from L is a, therefore, distance of image from L must be equal to 2a. ? (a + 2b) = 2a ? b = a 2 The silvered lens L may be assumed as a combination of an equi–convex lens and a concave mirror placed in contact with each other co–axially as shown in figure. \ \ \ \ \ \ \ \ \ \ \ \ \ \ Focal length of convex lens f 1 is given by 1 1 f = ( ?–1) 1 2 1 1 R R ? ? ? ? ? ? ? ? f 1 = 40 cm For concave mirror focal length, f m = R 2 = – 20 cm The combination L behaves like a mirror whose equivalent focal length F is given by m 1 1 1 2 F f f ? ? ? F = –10 cm, Hence, for the combination u = –a, v = +2a, F = –10 cm Using mirror formula, 1 1 1 v u F ? ? ? a = 5 cm JEEMAIN.GURU J E E - P h y s i c s E 55 NODE6\E:\Data\2014\Kota\JEE-Advanced\SMP\Phy\Unit No-11\Ray-Optics\Eng\02. Ray theory-Part2.p65 Distance of Bird as seen by fish x Bf = d + ?h h d By differentiating fB d(x ) dh 1 d(d) dt dt dt ? ? ? v FB ? 3 + 4 ? = 6 cm/s dh d(d) 4 5 2 3 cm / s, 2 ( 2) 4 cm / s, dt dt 3 ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? v BF = d(d) dh dt dt ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? 4 + 4 3 ? ? ? ? ? ? (3) ? 8 cm/s d(d) dt (for fish image after reflection = 0 ) ? 3 + 1 ? (0) = 3 cm/s Similarly speed of image of bird ? 4 cm/s Example#25 In the shown figure the focal length of equivalent system in the form of 50x 13 ? ? ? ? ? ? . Find the value of x. Ans. 2 1 1 3 1 1 1 1 f 2 10 10 10 ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ; 2 1 6 1 1 3 1 f 5 10 20 100 ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? and 3 1 8 1 1 3 1 f 5 20 20 50 ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? 1 2 3 1 1 1 1 1 3 3 f f f f 10 100 50 ? ? ? ? ? ? ? = 100 13 = 50x 13 ? x=2 Example#26 Quarter part of a transparent cylinder ABC of radius R is kept on a horizontal floor and a horizontal beam of light falls on the cylinder in the two different arrangement of cylinder as shown in the figure (a) & (b). In arrangement (a) light converges at point D, which is at a distance 2R/m from B. And in arrangement (b) light converges at point E, which is at a distance R/(m – 1) from C. Find out the refractive index of the material. C B A D 2R/m B C E R m -1 (a) (b) JEEMAIN.GURU J E E - P h y s i c s 54 E NODE6\E:\Data\2014\Kota\JEE-Advanced\SMP\Phy\Unit No-11\Ray-Optics\Eng\02. Ray theory-Part2.p65 Example#23 Column-I contains a list of mirrors and position of object. Match this with Column-II describing the nature of image. Column I Column II (A) F C (P) real, inverted, enlarged (B) (Q) virtual, erect, enlarged (C) (R) virtual, erect, diminished (D) F (S) virtual, erect Solution Ans. (A) P (b)RS (C) S (D) QS Example#24 A bird in air is diving vertically over a tank with speed 5 cm/s, base of tank is silvered. A fish in the tank is rising upward along the same line with speed 2 cm/s. Water level is falling at rate of 2 cm/s. [Take : ? water = 4/3] Column I (cm/s) Column II (A) Speed of the image of fish as seen by the bird directly (P) 8 (B) Speed of the image of fish formed after reflection in (Q) 6 the mirror as seen by the bird (C) Speed of image of bird relative to the fish looking upwards (R) 3 (D) Speed of image of bird relative to the fish looking (S) 4 downwards in the mirror Solution Ans. (A) – (Q) ; (B) – (R) ; (C) – (P) ; (D) – (S) Distance of fish asseen by bird x fB = h + d ? JEEMAIN.GURU J E E - P h y s i c s E 53 NODE6\E:\Data\2014\Kota\JEE-Advanced\SMP\Phy\Unit No-11\Ray-Optics\Eng\02. Ray theory-Part2.p65 Example#19 to 21 There is a spherical glass ball of refractive index ? 1 and another glass ball of refractive index ? 2 inside it as shown in figure. The radius of the outer ball is R 1 and that of inner ball is R 2 . A ray is incident on the outer surface of the ball at an angle i 1 . ? 1 ? 2 R 1 r 1 i 2 B C O R 2 A i 1 D r 2 1 9 . Find the value of r 1 (A) 1 1 1 sin i sin ? ? ? ? ? ? ? ? (B) ? ? 1 1 1 sin sin i ? ? (C) 1 1 1 sin sin i ? ? ? ? ? ? ? ? (D) 1 1 1 1 sin sin i ? ? ? ? ? ? ? ? 2 0 . Find the value of i 2 (A) 1 2 1 1 1 R sin i sin R ? ? ? ? ? ? ? ? (B) ] 1 1 1 2 2 R sin i sin R ? ? ? ? ? ? ? ? (C) 1 1 1 2 1 R sin i sin R ? ? ? ? ? ? ? ? (D) 1 2 1 1 2 R sin i sin R ? ? ? ? ? ? ? ? 2 1 . Find the value of r 2 (A) 1 1 1 2 2 R sin sin i R ? ? ? ? ? ? ? ? (B) 1 2 1 2 1 R sin sin i R ? ? ? ? ? ? ? ? (C) 1 1 1 2 1 R 1 sin R sin i ? ? ? ? ? ? ? ? (D) 1 2 1 1 1 R sin sin i R ? ? ? ? ? ? ? ? Solution 1 9 . Ans. (A) ? 1 sin r 1 = sin i 1 ? r 1 = 1 1 1 sin i sin ? ? ? ? ? ? ? ? 2 0 . Ans. (C) Using sine rule ? ? 1 2 2 1 sin r sin(180 i ) R R ? ??sin i 2 = 1 2 R R sin r 1 = 1 1 2 1 R sin i R ? ? ? ? ? ? ? 1 1 1 2 2 1 R sin i i sin R ? ? ? ? ? ? ? ? ? ? 2 1 . Ans. (A) 1 2 ? ? sin i 2 = sin r 2 ; 1 1 1 2 2 1 R sin i R ? ? ? = sin r 2 ; r 2 = 1 1 1 2 2 R sin sin i R ? ? ? ? ? ? ? ? Example#22 Consider a an equilateral prism ABC of glass 3 2 ? ? ? ? ? ? ? ? placed in water 4 3 ? ? ? ? ? ? ? ? A water =4/3 ? glass =3/2 ? Column-I Column-II (A) FG is parallel to BC (P) Maximum deviation (B) i 1 = 90 0 (Q) Minimum deviation (C) i 1 = i 2 = sin -1 9 16 ? ? ? ? ? ? (R) TIR will take place at surface AC (D) EF is perpendicular to AB (S) No TIR will take place at surface BC Solution Ans. (A) QS, (B) PS, (C) QS, (D) S At minimum deviation i 1 = i 2 , EF BC; At maximum deviation i 1 = 90 0 or i 2 =90 0 For i 1 =0, TIR will not take place at AC JEEMAIN.GURU Page 5 J E E - P h y s i c s 56 E NODE6\E:\Data\2014\Kota\JEE-Advanced\SMP\Phy\Unit No-11\Ray-Optics\Eng\02. Ray theory-Part2.p65 Solution Ans. 1.5 For case (a) 1 1 v ? ? ? = 1 R ? ? , 2 1 1 v v R ? ? ? = 1 ? ? ? and v 2 = 2R m ? m=2 ? ? 1 ? ? µ For case (b) 2 1 v ? ? ? = 1 R ? ? ? and v 2 = R m 1 ? ? µ = m Therefore ? ? 2 1 ? ? ? ? µ ? 1 1 2 ? ? ? ? µ = 1.5 Example#27 In figure, L is half part of an equiconvex glass lens ( ? = 1.5) whose surfaces have radius of curvature R = 40 cm and its right surface is silvered. Normal to its principal axis a plane mirror M is placed on right of the lens. Distance between lens L and mirror M is b. A small object O is placed on left of the lens such that there is no parallax between final images formed by the lens and mirror. If transverse length of final image formed by lens is twice that of image formed by the mirror, calculate distance 'a' in cm between lens and object. O a b M \ \ \ \ \ \ \ \ \ \ \ \ \ \ Solution Ans. 5 Distance of image of object O from plane mirror = a + b. Since, there is no parallax between the images formed by the silvered lens L and plane mirror M, therefore, two images are formed at the same point. Distance of image = (a + 2b) behind lens. Since, length of image formed by L is twice the length of image formed by the mirror M and length of image formed by a plane mirror is always equal to length of the object, therefore, transverse magnification produced by the lens L is equal to 2.Since, distance of object from L is a, therefore, distance of image from L must be equal to 2a. ? (a + 2b) = 2a ? b = a 2 The silvered lens L may be assumed as a combination of an equi–convex lens and a concave mirror placed in contact with each other co–axially as shown in figure. \ \ \ \ \ \ \ \ \ \ \ \ \ \ Focal length of convex lens f 1 is given by 1 1 f = ( ?–1) 1 2 1 1 R R ? ? ? ? ? ? ? ? f 1 = 40 cm For concave mirror focal length, f m = R 2 = – 20 cm The combination L behaves like a mirror whose equivalent focal length F is given by m 1 1 1 2 F f f ? ? ? F = –10 cm, Hence, for the combination u = –a, v = +2a, F = –10 cm Using mirror formula, 1 1 1 v u F ? ? ? a = 5 cm JEEMAIN.GURU J E E - P h y s i c s E 55 NODE6\E:\Data\2014\Kota\JEE-Advanced\SMP\Phy\Unit No-11\Ray-Optics\Eng\02. Ray theory-Part2.p65 Distance of Bird as seen by fish x Bf = d + ?h h d By differentiating fB d(x ) dh 1 d(d) dt dt dt ? ? ? v FB ? 3 + 4 ? = 6 cm/s dh d(d) 4 5 2 3 cm / s, 2 ( 2) 4 cm / s, dt dt 3 ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? v BF = d(d) dh dt dt ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? 4 + 4 3 ? ? ? ? ? ? (3) ? 8 cm/s d(d) dt (for fish image after reflection = 0 ) ? 3 + 1 ? (0) = 3 cm/s Similarly speed of image of bird ? 4 cm/s Example#25 In the shown figure the focal length of equivalent system in the form of 50x 13 ? ? ? ? ? ? . Find the value of x. Ans. 2 1 1 3 1 1 1 1 f 2 10 10 10 ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ; 2 1 6 1 1 3 1 f 5 10 20 100 ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? and 3 1 8 1 1 3 1 f 5 20 20 50 ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? 1 2 3 1 1 1 1 1 3 3 f f f f 10 100 50 ? ? ? ? ? ? ? = 100 13 = 50x 13 ? x=2 Example#26 Quarter part of a transparent cylinder ABC of radius R is kept on a horizontal floor and a horizontal beam of light falls on the cylinder in the two different arrangement of cylinder as shown in the figure (a) & (b). In arrangement (a) light converges at point D, which is at a distance 2R/m from B. And in arrangement (b) light converges at point E, which is at a distance R/(m – 1) from C. Find out the refractive index of the material. C B A D 2R/m B C E R m -1 (a) (b) JEEMAIN.GURU J E E - P h y s i c s 54 E NODE6\E:\Data\2014\Kota\JEE-Advanced\SMP\Phy\Unit No-11\Ray-Optics\Eng\02. Ray theory-Part2.p65 Example#23 Column-I contains a list of mirrors and position of object. Match this with Column-II describing the nature of image. Column I Column II (A) F C (P) real, inverted, enlarged (B) (Q) virtual, erect, enlarged (C) (R) virtual, erect, diminished (D) F (S) virtual, erect Solution Ans. (A) P (b)RS (C) S (D) QS Example#24 A bird in air is diving vertically over a tank with speed 5 cm/s, base of tank is silvered. A fish in the tank is rising upward along the same line with speed 2 cm/s. Water level is falling at rate of 2 cm/s. [Take : ? water = 4/3] Column I (cm/s) Column II (A) Speed of the image of fish as seen by the bird directly (P) 8 (B) Speed of the image of fish formed after reflection in (Q) 6 the mirror as seen by the bird (C) Speed of image of bird relative to the fish looking upwards (R) 3 (D) Speed of image of bird relative to the fish looking (S) 4 downwards in the mirror Solution Ans. (A) – (Q) ; (B) – (R) ; (C) – (P) ; (D) – (S) Distance of fish asseen by bird x fB = h + d ? JEEMAIN.GURU J E E - P h y s i c s E 53 NODE6\E:\Data\2014\Kota\JEE-Advanced\SMP\Phy\Unit No-11\Ray-Optics\Eng\02. Ray theory-Part2.p65 Example#19 to 21 There is a spherical glass ball of refractive index ? 1 and another glass ball of refractive index ? 2 inside it as shown in figure. The radius of the outer ball is R 1 and that of inner ball is R 2 . A ray is incident on the outer surface of the ball at an angle i 1 . ? 1 ? 2 R 1 r 1 i 2 B C O R 2 A i 1 D r 2 1 9 . Find the value of r 1 (A) 1 1 1 sin i sin ? ? ? ? ? ? ? ? (B) ? ? 1 1 1 sin sin i ? ? (C) 1 1 1 sin sin i ? ? ? ? ? ? ? ? (D) 1 1 1 1 sin sin i ? ? ? ? ? ? ? ? 2 0 . Find the value of i 2 (A) 1 2 1 1 1 R sin i sin R ? ? ? ? ? ? ? ? (B) ] 1 1 1 2 2 R sin i sin R ? ? ? ? ? ? ? ? (C) 1 1 1 2 1 R sin i sin R ? ? ? ? ? ? ? ? (D) 1 2 1 1 2 R sin i sin R ? ? ? ? ? ? ? ? 2 1 . Find the value of r 2 (A) 1 1 1 2 2 R sin sin i R ? ? ? ? ? ? ? ? (B) 1 2 1 2 1 R sin sin i R ? ? ? ? ? ? ? ? (C) 1 1 1 2 1 R 1 sin R sin i ? ? ? ? ? ? ? ? (D) 1 2 1 1 1 R sin sin i R ? ? ? ? ? ? ? ? Solution 1 9 . Ans. (A) ? 1 sin r 1 = sin i 1 ? r 1 = 1 1 1 sin i sin ? ? ? ? ? ? ? ? 2 0 . Ans. (C) Using sine rule ? ? 1 2 2 1 sin r sin(180 i ) R R ? ??sin i 2 = 1 2 R R sin r 1 = 1 1 2 1 R sin i R ? ? ? ? ? ? ? 1 1 1 2 2 1 R sin i i sin R ? ? ? ? ? ? ? ? ? ? 2 1 . Ans. (A) 1 2 ? ? sin i 2 = sin r 2 ; 1 1 1 2 2 1 R sin i R ? ? ? = sin r 2 ; r 2 = 1 1 1 2 2 R sin sin i R ? ? ? ? ? ? ? ? Example#22 Consider a an equilateral prism ABC of glass 3 2 ? ? ? ? ? ? ? ? placed in water 4 3 ? ? ? ? ? ? ? ? A water =4/3 ? glass =3/2 ? Column-I Column-II (A) FG is parallel to BC (P) Maximum deviation (B) i 1 = 90 0 (Q) Minimum deviation (C) i 1 = i 2 = sin -1 9 16 ? ? ? ? ? ? (R) TIR will take place at surface AC (D) EF is perpendicular to AB (S) No TIR will take place at surface BC Solution Ans. (A) QS, (B) PS, (C) QS, (D) S At minimum deviation i 1 = i 2 , EF BC; At maximum deviation i 1 = 90 0 or i 2 =90 0 For i 1 =0, TIR will not take place at AC JEEMAIN.GURU J E E - P h y s i c s 52 E NODE6\E:\Data\2014\Kota\JEE-Advanced\SMP\Phy\Unit No-11\Ray-Optics\Eng\02. Ray theory-Part2.p65 Solution 13. Ans. (A) Time t = distance speed = ? ? 2 2 2 2 1 2 Y x x Y c ? ? ? ? ? 1 4 . Ans. (A) For least time dt dx =0 ? 2 2 1 2x Y x ? – ? ? ? ? 2 2 2 2 x x Y ? ? ? ? ? = 0 ? 1 2 sin sin ? ? ? ? 1 2 ? ? ? 1 5 . Ans. (A) Example#16 to 18 One hard and stormy night you find yourself lost in the forest when you come upon a small hut. Entering it you see a crooked old woman in the corner hunched over a crystal ball. You are about to make a hasty exit when you hear the howl of wolves outside. Taking another look at the gypsy you decide to take your chances with the wolves, but the door is jammed shut. Resigned to a bad situation you approach her slowly, wondering just what is the focal length of that nifty crystal ball. 1 6 . If the crystal ball is 20 cm in diameter with R.I. = 1.5, the gypsy lady is 1.2 m from the ball, where is the image of the gypsy in focus as you walk towards her? (A) 6.9 cm from the crystal ball (B) 7.9 cm from the crystal ball (C) 8.9 cm from the crystal ball (D) None of these 1 7 . The image of old lady is (A) real, inverted and enlarged (B) erect, virtual and small (C) erect, virtual and magnified (D) real, inverted and small 1 8 . The old lady moves the crystal ball closer to her wrinkled old face. At some point you can no longer get an image of her. At what object distance will there be no change of the gypsy formed? (A) 10cm (B) 5 cm (C) 15 cm (D) None of these Solution 1 6 . Ans. (A) For refraction at 1 st surface 1 1 1.5 1 1.5 1 v 36cm v 120 10 ? ? ? ? ? ? ? Lady 120cm R=10cm (1) (2) for refraction at 2 nd surface 1 1.5 1 1.5 80 v 6.9cm v (36 20) 10 11.5 ? ? ? ? ? ? ? ? 1 7 . Ans. (D) Total magnification = m 1 m 2 = 1 1 2 2 1 1 v v v u v u ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? = negative e 1 8 . Ans. (B) At this point image will formed at infinity For refraction at second surface 1 1 1 1.5 1 1.5 v 30cm v 10 ? ? ? ? ? ? ? ? x 20cm (1) (2) For refraction at first surface 1.5 1 1.5 1 x 5cm 10 x 10 ? ? ? ? ? ? ? ? JEEMAIN.GURURead More

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