Page 1 19.1 SOLUTIONS TO CONCEPTS CHAPTER 19 1. The visual angles made by the tree with the eyes can be calculated be below. ? = A Height of the tree AB 2 0.04 Distance from the eye OB 50 ? ? ? ? ? similarly, ? B = 2.5 / 80 = 0.03125 ? C = 1.8 / 70 = 0.02571 ? D = 2.8 / 100 = 0.028 Since, ? A > ?? B > ?? D > ?? C , the arrangement in decreasing order is given by A, B, D and C. ? 2. For the given simple microscope, f = 12 cm and D = 25 cm For maximum angular magnification, the image should be produced at least distance of clear vision. So, v = – D = –25 cm Now, 1 1 1 v u f ? ? ? 1 1 1 1 1 37 u v f 25 12 300 ? ? ? ? ? ? ? ? u = –8.1 cm So, the object should be placed 8.1 cm away from the lens. 3. The simple microscope has, m = 3, when image is formed at D = 25 cm a) m = D 1 f ? ? 25 3 1 f ? ? ? f = 25/2 = 12.5 cm b) When the image is formed at infinity (normal adjustment) Magnifying power = D 25 f 12.5 ? = 2.0 4. The child has D = 10 cm and f = 10 cm The maximum angular magnification is obtained when the image is formed at near point. m = D 10 1 1 f 10 ? ? ? = 1 + 1 = 2 5. The simple microscope has magnification of 5 for normal relaxed eye (D = 25 cm). Because, the eye is relaxed the image is formed at infinity (normal adjustment) So, m = 5 = D 25 f f ? ? f = 5 cm For the relaxed farsighted eye, D = 40 cm So, m = D 40 f 5 ? = 8 So, its magnifying power is 8X. A ? A +ve B B ? D=25cm (Simple Microscope) Distance A height B ? ? Page 2 19.1 SOLUTIONS TO CONCEPTS CHAPTER 19 1. The visual angles made by the tree with the eyes can be calculated be below. ? = A Height of the tree AB 2 0.04 Distance from the eye OB 50 ? ? ? ? ? similarly, ? B = 2.5 / 80 = 0.03125 ? C = 1.8 / 70 = 0.02571 ? D = 2.8 / 100 = 0.028 Since, ? A > ?? B > ?? D > ?? C , the arrangement in decreasing order is given by A, B, D and C. ? 2. For the given simple microscope, f = 12 cm and D = 25 cm For maximum angular magnification, the image should be produced at least distance of clear vision. So, v = – D = –25 cm Now, 1 1 1 v u f ? ? ? 1 1 1 1 1 37 u v f 25 12 300 ? ? ? ? ? ? ? ? u = –8.1 cm So, the object should be placed 8.1 cm away from the lens. 3. The simple microscope has, m = 3, when image is formed at D = 25 cm a) m = D 1 f ? ? 25 3 1 f ? ? ? f = 25/2 = 12.5 cm b) When the image is formed at infinity (normal adjustment) Magnifying power = D 25 f 12.5 ? = 2.0 4. The child has D = 10 cm and f = 10 cm The maximum angular magnification is obtained when the image is formed at near point. m = D 10 1 1 f 10 ? ? ? = 1 + 1 = 2 5. The simple microscope has magnification of 5 for normal relaxed eye (D = 25 cm). Because, the eye is relaxed the image is formed at infinity (normal adjustment) So, m = 5 = D 25 f f ? ? f = 5 cm For the relaxed farsighted eye, D = 40 cm So, m = D 40 f 5 ? = 8 So, its magnifying power is 8X. A ? A +ve B B ? D=25cm (Simple Microscope) Distance A height B ? ? Chapter 19 19.2 6. For the given compound microscope f 0 = 1 25 diopter = 0.04 m = 4 cm, f e = 1 5 diopter = 0.2 m = 20 cm D = 25 cm, separation between objective and eyepiece = 30 cm The magnifying power is maximum when the image is formed by the eye piece at least distance of clear vision i.e. D = 25 cm for the eye piece, v e = –25 cm, f e = 20 cm For lens formula, e e e 1 1 1 v u f ? ? ? e e e 1 1 1 1 1 u v f 25 20 ? ? ? ? ? ? u e = 11.11 cm So, for the objective lens, the image distance should be v 0 = 30 – (11.11) = 18.89 cm Now, for the objective lens, v 0 = +18.89 cm (because real image is produced) f 0 = 4 cm So, o o o 1 1 1 1 1 u v f 18.89 4 ? ? ? ? = 0.053 – 0.25 = –0.197 ? u o = –5.07 cm So, the maximum magnificent power is given by m = o o e v D 18.89 25 1 1 u f 5.07 20 ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? = 3.7225 ? 2.25 = 8.376 7. For the given compound microscope f o = 1 cm, f e = 6 cm, D = 24 cm For the eye piece, v e = –24 cm, f e = 6 cm Now, e e e 1 1 1 v u f ? ? ? e e e 1 1 1 1 1 5 u v f 24 6 24 ? ? ? ? ? ? ? ? ? ? ? ? ? ? u e = –4.8 cm a) When the separation between objective and eye piece is 9.8 cm, the image distance for the objective lens must be (9.8) – (4.8) = 5.0 cm Now, 0 0 u 1 v 1 ? = 0 f 1 ? 0 u 1 = 0 0 f 1 v 1 ? = 1 1 5 1 ? = – 5 4 ? u 0 = – 4 5 = – 1.25 cm So, the magnifying power is given by, m = ? ? ? ? ? ? ? f D 1 u v o 0 = ? ? ? ? ? ? ? ? ? 6 24 1 25 . 1 5 = 4 × 5 = 20 (b) When the separation is 11.8 cm, v 0 = 11.8 – 4.8 = 7.0 cm, f 0 = 1 cm ? 0 u 1 = 0 0 f 1 v 1 ? = 1 1 7 1 ? = 7 6 ? f o = 0.04m objective A ? B A B ? B ? ? 11.11 cm 25cm 30cm A ?? f e = 0.2m objective Fig-A 5 cm ? 9.8cm ? 4.8cm ? 24 cm ? Fig-B 7 cm ? 11.8cm ? 4.8cm ? 24 cm ? Page 3 19.1 SOLUTIONS TO CONCEPTS CHAPTER 19 1. The visual angles made by the tree with the eyes can be calculated be below. ? = A Height of the tree AB 2 0.04 Distance from the eye OB 50 ? ? ? ? ? similarly, ? B = 2.5 / 80 = 0.03125 ? C = 1.8 / 70 = 0.02571 ? D = 2.8 / 100 = 0.028 Since, ? A > ?? B > ?? D > ?? C , the arrangement in decreasing order is given by A, B, D and C. ? 2. For the given simple microscope, f = 12 cm and D = 25 cm For maximum angular magnification, the image should be produced at least distance of clear vision. So, v = – D = –25 cm Now, 1 1 1 v u f ? ? ? 1 1 1 1 1 37 u v f 25 12 300 ? ? ? ? ? ? ? ? u = –8.1 cm So, the object should be placed 8.1 cm away from the lens. 3. The simple microscope has, m = 3, when image is formed at D = 25 cm a) m = D 1 f ? ? 25 3 1 f ? ? ? f = 25/2 = 12.5 cm b) When the image is formed at infinity (normal adjustment) Magnifying power = D 25 f 12.5 ? = 2.0 4. The child has D = 10 cm and f = 10 cm The maximum angular magnification is obtained when the image is formed at near point. m = D 10 1 1 f 10 ? ? ? = 1 + 1 = 2 5. The simple microscope has magnification of 5 for normal relaxed eye (D = 25 cm). Because, the eye is relaxed the image is formed at infinity (normal adjustment) So, m = 5 = D 25 f f ? ? f = 5 cm For the relaxed farsighted eye, D = 40 cm So, m = D 40 f 5 ? = 8 So, its magnifying power is 8X. A ? A +ve B B ? D=25cm (Simple Microscope) Distance A height B ? ? Chapter 19 19.2 6. For the given compound microscope f 0 = 1 25 diopter = 0.04 m = 4 cm, f e = 1 5 diopter = 0.2 m = 20 cm D = 25 cm, separation between objective and eyepiece = 30 cm The magnifying power is maximum when the image is formed by the eye piece at least distance of clear vision i.e. D = 25 cm for the eye piece, v e = –25 cm, f e = 20 cm For lens formula, e e e 1 1 1 v u f ? ? ? e e e 1 1 1 1 1 u v f 25 20 ? ? ? ? ? ? u e = 11.11 cm So, for the objective lens, the image distance should be v 0 = 30 – (11.11) = 18.89 cm Now, for the objective lens, v 0 = +18.89 cm (because real image is produced) f 0 = 4 cm So, o o o 1 1 1 1 1 u v f 18.89 4 ? ? ? ? = 0.053 – 0.25 = –0.197 ? u o = –5.07 cm So, the maximum magnificent power is given by m = o o e v D 18.89 25 1 1 u f 5.07 20 ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? = 3.7225 ? 2.25 = 8.376 7. For the given compound microscope f o = 1 cm, f e = 6 cm, D = 24 cm For the eye piece, v e = –24 cm, f e = 6 cm Now, e e e 1 1 1 v u f ? ? ? e e e 1 1 1 1 1 5 u v f 24 6 24 ? ? ? ? ? ? ? ? ? ? ? ? ? ? u e = –4.8 cm a) When the separation between objective and eye piece is 9.8 cm, the image distance for the objective lens must be (9.8) – (4.8) = 5.0 cm Now, 0 0 u 1 v 1 ? = 0 f 1 ? 0 u 1 = 0 0 f 1 v 1 ? = 1 1 5 1 ? = – 5 4 ? u 0 = – 4 5 = – 1.25 cm So, the magnifying power is given by, m = ? ? ? ? ? ? ? f D 1 u v o 0 = ? ? ? ? ? ? ? ? ? 6 24 1 25 . 1 5 = 4 × 5 = 20 (b) When the separation is 11.8 cm, v 0 = 11.8 – 4.8 = 7.0 cm, f 0 = 1 cm ? 0 u 1 = 0 0 f 1 v 1 ? = 1 1 7 1 ? = 7 6 ? f o = 0.04m objective A ? B A B ? B ? ? 11.11 cm 25cm 30cm A ?? f e = 0.2m objective Fig-A 5 cm ? 9.8cm ? 4.8cm ? 24 cm ? Fig-B 7 cm ? 11.8cm ? 4.8cm ? 24 cm ? Chapter 19 19.3 So, m = – ? ? ? ? ? ? ? f D 1 u v o 0 = ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? 6 24 1 6 7 7 = 6 × 5 = 30 So, the range of magnifying power will be 20 to 30. 8. For the given compound microscope. f 0 = D 20 1 = 0.05 m = 5 cm, f e = D 10 1 = 0.1 m = 10 cm. D = 25 cm, separation between objective & eyepiece= 20 cm For the minimum separation between two points which can be distinguished by eye using the microscope, the magnifying power should be maximum. For the eyepiece, v 0 = –25 cm, f e = 10 cm So, e u 1 = e e f 1 v 1 ? = 10 1 25 1 ? ? = – ? ? ? ? ? ? ? 50 5 2 ? u e = – 7 50 cm So, the image distance for the objective lens should be, V 0 = 20 – 7 50 = 7 90 cm Now, for the objective lens, 0 0 0 f 1 v 1 u 1 ? ? = 5 1 90 7 ? = – 90 11 ? u 0 = – 11 90 cm So, the maximum magnifying power is given by, m = ? ? ? ? ? ? ? ? e 0 0 f D 1 u v = ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? 10 25 1 11 90 7 90 = 5 . 3 7 11 ? = 5.5 Thus, minimum separation eye can distinguish = 5 . 5 22 . 0 mm = 0.04 mm 9. For the give compound microscope, f 0 = 0.5cm, tube length = 6.5cm magnifying power = 100 (normal adjustment) Since, the image is formed at infinity, the real image produced by the objective lens should lie on the focus of the eye piece. So, v 0 + f e = 6.5 cm …(1) Again, magnifying power= e 0 0 f D u v ? [for normal adjustment] ? m = – e 0 0 f D f v 1 ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? 0 0 0 0 f v 1 u v ? ? 100 = – e 0 f 25 5 . 0 v 1 ? ? ? ? ? ? ? ? [Taking D = 25 cm] ? 100 f e = –(1 – 2v 0 ) × 25 ? 2v 0 – 4f e = 1 …(2) ? ? v 0 ? Eye piece ? f e ? F e ? Objective ? Page 4 19.1 SOLUTIONS TO CONCEPTS CHAPTER 19 1. The visual angles made by the tree with the eyes can be calculated be below. ? = A Height of the tree AB 2 0.04 Distance from the eye OB 50 ? ? ? ? ? similarly, ? B = 2.5 / 80 = 0.03125 ? C = 1.8 / 70 = 0.02571 ? D = 2.8 / 100 = 0.028 Since, ? A > ?? B > ?? D > ?? C , the arrangement in decreasing order is given by A, B, D and C. ? 2. For the given simple microscope, f = 12 cm and D = 25 cm For maximum angular magnification, the image should be produced at least distance of clear vision. So, v = – D = –25 cm Now, 1 1 1 v u f ? ? ? 1 1 1 1 1 37 u v f 25 12 300 ? ? ? ? ? ? ? ? u = –8.1 cm So, the object should be placed 8.1 cm away from the lens. 3. The simple microscope has, m = 3, when image is formed at D = 25 cm a) m = D 1 f ? ? 25 3 1 f ? ? ? f = 25/2 = 12.5 cm b) When the image is formed at infinity (normal adjustment) Magnifying power = D 25 f 12.5 ? = 2.0 4. The child has D = 10 cm and f = 10 cm The maximum angular magnification is obtained when the image is formed at near point. m = D 10 1 1 f 10 ? ? ? = 1 + 1 = 2 5. The simple microscope has magnification of 5 for normal relaxed eye (D = 25 cm). Because, the eye is relaxed the image is formed at infinity (normal adjustment) So, m = 5 = D 25 f f ? ? f = 5 cm For the relaxed farsighted eye, D = 40 cm So, m = D 40 f 5 ? = 8 So, its magnifying power is 8X. A ? A +ve B B ? D=25cm (Simple Microscope) Distance A height B ? ? Chapter 19 19.2 6. For the given compound microscope f 0 = 1 25 diopter = 0.04 m = 4 cm, f e = 1 5 diopter = 0.2 m = 20 cm D = 25 cm, separation between objective and eyepiece = 30 cm The magnifying power is maximum when the image is formed by the eye piece at least distance of clear vision i.e. D = 25 cm for the eye piece, v e = –25 cm, f e = 20 cm For lens formula, e e e 1 1 1 v u f ? ? ? e e e 1 1 1 1 1 u v f 25 20 ? ? ? ? ? ? u e = 11.11 cm So, for the objective lens, the image distance should be v 0 = 30 – (11.11) = 18.89 cm Now, for the objective lens, v 0 = +18.89 cm (because real image is produced) f 0 = 4 cm So, o o o 1 1 1 1 1 u v f 18.89 4 ? ? ? ? = 0.053 – 0.25 = –0.197 ? u o = –5.07 cm So, the maximum magnificent power is given by m = o o e v D 18.89 25 1 1 u f 5.07 20 ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? = 3.7225 ? 2.25 = 8.376 7. For the given compound microscope f o = 1 cm, f e = 6 cm, D = 24 cm For the eye piece, v e = –24 cm, f e = 6 cm Now, e e e 1 1 1 v u f ? ? ? e e e 1 1 1 1 1 5 u v f 24 6 24 ? ? ? ? ? ? ? ? ? ? ? ? ? ? u e = –4.8 cm a) When the separation between objective and eye piece is 9.8 cm, the image distance for the objective lens must be (9.8) – (4.8) = 5.0 cm Now, 0 0 u 1 v 1 ? = 0 f 1 ? 0 u 1 = 0 0 f 1 v 1 ? = 1 1 5 1 ? = – 5 4 ? u 0 = – 4 5 = – 1.25 cm So, the magnifying power is given by, m = ? ? ? ? ? ? ? f D 1 u v o 0 = ? ? ? ? ? ? ? ? ? 6 24 1 25 . 1 5 = 4 × 5 = 20 (b) When the separation is 11.8 cm, v 0 = 11.8 – 4.8 = 7.0 cm, f 0 = 1 cm ? 0 u 1 = 0 0 f 1 v 1 ? = 1 1 7 1 ? = 7 6 ? f o = 0.04m objective A ? B A B ? B ? ? 11.11 cm 25cm 30cm A ?? f e = 0.2m objective Fig-A 5 cm ? 9.8cm ? 4.8cm ? 24 cm ? Fig-B 7 cm ? 11.8cm ? 4.8cm ? 24 cm ? Chapter 19 19.3 So, m = – ? ? ? ? ? ? ? f D 1 u v o 0 = ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? 6 24 1 6 7 7 = 6 × 5 = 30 So, the range of magnifying power will be 20 to 30. 8. For the given compound microscope. f 0 = D 20 1 = 0.05 m = 5 cm, f e = D 10 1 = 0.1 m = 10 cm. D = 25 cm, separation between objective & eyepiece= 20 cm For the minimum separation between two points which can be distinguished by eye using the microscope, the magnifying power should be maximum. For the eyepiece, v 0 = –25 cm, f e = 10 cm So, e u 1 = e e f 1 v 1 ? = 10 1 25 1 ? ? = – ? ? ? ? ? ? ? 50 5 2 ? u e = – 7 50 cm So, the image distance for the objective lens should be, V 0 = 20 – 7 50 = 7 90 cm Now, for the objective lens, 0 0 0 f 1 v 1 u 1 ? ? = 5 1 90 7 ? = – 90 11 ? u 0 = – 11 90 cm So, the maximum magnifying power is given by, m = ? ? ? ? ? ? ? ? e 0 0 f D 1 u v = ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? 10 25 1 11 90 7 90 = 5 . 3 7 11 ? = 5.5 Thus, minimum separation eye can distinguish = 5 . 5 22 . 0 mm = 0.04 mm 9. For the give compound microscope, f 0 = 0.5cm, tube length = 6.5cm magnifying power = 100 (normal adjustment) Since, the image is formed at infinity, the real image produced by the objective lens should lie on the focus of the eye piece. So, v 0 + f e = 6.5 cm …(1) Again, magnifying power= e 0 0 f D u v ? [for normal adjustment] ? m = – e 0 0 f D f v 1 ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? 0 0 0 0 f v 1 u v ? ? 100 = – e 0 f 25 5 . 0 v 1 ? ? ? ? ? ? ? ? [Taking D = 25 cm] ? 100 f e = –(1 – 2v 0 ) × 25 ? 2v 0 – 4f e = 1 …(2) ? ? v 0 ? Eye piece ? f e ? F e ? Objective ? Chapter 19 19.4 Solving equation (1) and (2) we can get, V 0 = 4.5 cm and f e = 2 cm So, the focal length of the eye piece is 2cm. 10. Given that, f o = = 1 cm, f e = 5 cm, u 0 = 0.5 cm, v e = 30 cm For the objective lens, u 0 = – 0.5 cm, f 0 = 1 cm. From lens formula, 0 0 0 f 1 u 1 v 1 ? ? ? 0 0 0 f 1 u 1 v 1 ? ? = 1 1 5 . 0 1 ? ? = – 1 ? v 0 = – 1 cm So, a virtual image is formed by the objective on the same side as that of the object at a distance of 1 cm from the objective lens. This image acts as a virtual object for the eyepiece. For the eyepiece, 0 0 0 f 1 u 1 v 1 ? ? ? 0 0 0 f 1 v 1 u 1 ? ? = 5 1 30 1 ? = 30 5 ? = 6 1 ? ? u 0 = – 6 cm So, as shown in figure, Separation between the lenses = u 0 – v 0 = 6 – 1 = 5 cm 11. The optical instrument has f 0 = D 25 1 = 0.04 m = 4 cm f e = D 20 1 = 0.05 m = 5 cm tube length = 25 cm (normal adjustment) (a) The instrument must be a microscope as f 0 < f e (b) Since the final image is formed at infinity, the image produced by the objective should lie on the focal plane of the eye piece. So, image distance for objective = v 0 = 25 – 5 = 20 cm Now, using lens formula. 0 0 0 f 1 u 1 v 1 ? ? ? 0 0 0 f 1 v 1 u 1 ? ? = 4 1 20 1 ? = 20 4 ? = 5 1 ? ? u 0 = – 5 cm So, angular magnification = m = – e 0 0 f D u v ? [Taking D = 25 cm] = – 5 25 5 20 ? ? = 20 12. For the astronomical telescope in normal adjustment. Magnifying power = m = 50, length of the tube = L = 102 cm Let f 0 and f e be the focal length of objective and eye piece respectively. m = e 0 f f = 50 ? f 0 = 50 f e …(1) and, L = f 0 + f e = 102 cm …(2) Putting the value of f 0 from equation (1) in (2), we get, f 0 + f e = 102 ? 51f e = 102 ? f e = 2 cm = 0.02 m So, f 0 = 100 cm = 1 m ? Power of the objective lens = 0 f 1 = 1D And Power of the eye piece lens = e f 1 = 02 . 0 1 = 50D B ? A ? ? ? 20cm ? 5cm ? F e ? ? ? f 0 ? Eye piece ? f e ? F e ? Objective ? A ? A ? ? 0.5cm ? Eye piece ? 1cm ? Objective ? 30cm ? 60cm ? B ? ? A ? ? B ? ? B ? Page 5 19.1 SOLUTIONS TO CONCEPTS CHAPTER 19 1. The visual angles made by the tree with the eyes can be calculated be below. ? = A Height of the tree AB 2 0.04 Distance from the eye OB 50 ? ? ? ? ? similarly, ? B = 2.5 / 80 = 0.03125 ? C = 1.8 / 70 = 0.02571 ? D = 2.8 / 100 = 0.028 Since, ? A > ?? B > ?? D > ?? C , the arrangement in decreasing order is given by A, B, D and C. ? 2. For the given simple microscope, f = 12 cm and D = 25 cm For maximum angular magnification, the image should be produced at least distance of clear vision. So, v = – D = –25 cm Now, 1 1 1 v u f ? ? ? 1 1 1 1 1 37 u v f 25 12 300 ? ? ? ? ? ? ? ? u = –8.1 cm So, the object should be placed 8.1 cm away from the lens. 3. The simple microscope has, m = 3, when image is formed at D = 25 cm a) m = D 1 f ? ? 25 3 1 f ? ? ? f = 25/2 = 12.5 cm b) When the image is formed at infinity (normal adjustment) Magnifying power = D 25 f 12.5 ? = 2.0 4. The child has D = 10 cm and f = 10 cm The maximum angular magnification is obtained when the image is formed at near point. m = D 10 1 1 f 10 ? ? ? = 1 + 1 = 2 5. The simple microscope has magnification of 5 for normal relaxed eye (D = 25 cm). Because, the eye is relaxed the image is formed at infinity (normal adjustment) So, m = 5 = D 25 f f ? ? f = 5 cm For the relaxed farsighted eye, D = 40 cm So, m = D 40 f 5 ? = 8 So, its magnifying power is 8X. A ? A +ve B B ? D=25cm (Simple Microscope) Distance A height B ? ? Chapter 19 19.2 6. For the given compound microscope f 0 = 1 25 diopter = 0.04 m = 4 cm, f e = 1 5 diopter = 0.2 m = 20 cm D = 25 cm, separation between objective and eyepiece = 30 cm The magnifying power is maximum when the image is formed by the eye piece at least distance of clear vision i.e. D = 25 cm for the eye piece, v e = –25 cm, f e = 20 cm For lens formula, e e e 1 1 1 v u f ? ? ? e e e 1 1 1 1 1 u v f 25 20 ? ? ? ? ? ? u e = 11.11 cm So, for the objective lens, the image distance should be v 0 = 30 – (11.11) = 18.89 cm Now, for the objective lens, v 0 = +18.89 cm (because real image is produced) f 0 = 4 cm So, o o o 1 1 1 1 1 u v f 18.89 4 ? ? ? ? = 0.053 – 0.25 = –0.197 ? u o = –5.07 cm So, the maximum magnificent power is given by m = o o e v D 18.89 25 1 1 u f 5.07 20 ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? = 3.7225 ? 2.25 = 8.376 7. For the given compound microscope f o = 1 cm, f e = 6 cm, D = 24 cm For the eye piece, v e = –24 cm, f e = 6 cm Now, e e e 1 1 1 v u f ? ? ? e e e 1 1 1 1 1 5 u v f 24 6 24 ? ? ? ? ? ? ? ? ? ? ? ? ? ? u e = –4.8 cm a) When the separation between objective and eye piece is 9.8 cm, the image distance for the objective lens must be (9.8) – (4.8) = 5.0 cm Now, 0 0 u 1 v 1 ? = 0 f 1 ? 0 u 1 = 0 0 f 1 v 1 ? = 1 1 5 1 ? = – 5 4 ? u 0 = – 4 5 = – 1.25 cm So, the magnifying power is given by, m = ? ? ? ? ? ? ? f D 1 u v o 0 = ? ? ? ? ? ? ? ? ? 6 24 1 25 . 1 5 = 4 × 5 = 20 (b) When the separation is 11.8 cm, v 0 = 11.8 – 4.8 = 7.0 cm, f 0 = 1 cm ? 0 u 1 = 0 0 f 1 v 1 ? = 1 1 7 1 ? = 7 6 ? f o = 0.04m objective A ? B A B ? B ? ? 11.11 cm 25cm 30cm A ?? f e = 0.2m objective Fig-A 5 cm ? 9.8cm ? 4.8cm ? 24 cm ? Fig-B 7 cm ? 11.8cm ? 4.8cm ? 24 cm ? Chapter 19 19.3 So, m = – ? ? ? ? ? ? ? f D 1 u v o 0 = ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? 6 24 1 6 7 7 = 6 × 5 = 30 So, the range of magnifying power will be 20 to 30. 8. For the given compound microscope. f 0 = D 20 1 = 0.05 m = 5 cm, f e = D 10 1 = 0.1 m = 10 cm. D = 25 cm, separation between objective & eyepiece= 20 cm For the minimum separation between two points which can be distinguished by eye using the microscope, the magnifying power should be maximum. For the eyepiece, v 0 = –25 cm, f e = 10 cm So, e u 1 = e e f 1 v 1 ? = 10 1 25 1 ? ? = – ? ? ? ? ? ? ? 50 5 2 ? u e = – 7 50 cm So, the image distance for the objective lens should be, V 0 = 20 – 7 50 = 7 90 cm Now, for the objective lens, 0 0 0 f 1 v 1 u 1 ? ? = 5 1 90 7 ? = – 90 11 ? u 0 = – 11 90 cm So, the maximum magnifying power is given by, m = ? ? ? ? ? ? ? ? e 0 0 f D 1 u v = ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? 10 25 1 11 90 7 90 = 5 . 3 7 11 ? = 5.5 Thus, minimum separation eye can distinguish = 5 . 5 22 . 0 mm = 0.04 mm 9. For the give compound microscope, f 0 = 0.5cm, tube length = 6.5cm magnifying power = 100 (normal adjustment) Since, the image is formed at infinity, the real image produced by the objective lens should lie on the focus of the eye piece. So, v 0 + f e = 6.5 cm …(1) Again, magnifying power= e 0 0 f D u v ? [for normal adjustment] ? m = – e 0 0 f D f v 1 ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? 0 0 0 0 f v 1 u v ? ? 100 = – e 0 f 25 5 . 0 v 1 ? ? ? ? ? ? ? ? [Taking D = 25 cm] ? 100 f e = –(1 – 2v 0 ) × 25 ? 2v 0 – 4f e = 1 …(2) ? ? v 0 ? Eye piece ? f e ? F e ? Objective ? Chapter 19 19.4 Solving equation (1) and (2) we can get, V 0 = 4.5 cm and f e = 2 cm So, the focal length of the eye piece is 2cm. 10. Given that, f o = = 1 cm, f e = 5 cm, u 0 = 0.5 cm, v e = 30 cm For the objective lens, u 0 = – 0.5 cm, f 0 = 1 cm. From lens formula, 0 0 0 f 1 u 1 v 1 ? ? ? 0 0 0 f 1 u 1 v 1 ? ? = 1 1 5 . 0 1 ? ? = – 1 ? v 0 = – 1 cm So, a virtual image is formed by the objective on the same side as that of the object at a distance of 1 cm from the objective lens. This image acts as a virtual object for the eyepiece. For the eyepiece, 0 0 0 f 1 u 1 v 1 ? ? ? 0 0 0 f 1 v 1 u 1 ? ? = 5 1 30 1 ? = 30 5 ? = 6 1 ? ? u 0 = – 6 cm So, as shown in figure, Separation between the lenses = u 0 – v 0 = 6 – 1 = 5 cm 11. The optical instrument has f 0 = D 25 1 = 0.04 m = 4 cm f e = D 20 1 = 0.05 m = 5 cm tube length = 25 cm (normal adjustment) (a) The instrument must be a microscope as f 0 < f e (b) Since the final image is formed at infinity, the image produced by the objective should lie on the focal plane of the eye piece. So, image distance for objective = v 0 = 25 – 5 = 20 cm Now, using lens formula. 0 0 0 f 1 u 1 v 1 ? ? ? 0 0 0 f 1 v 1 u 1 ? ? = 4 1 20 1 ? = 20 4 ? = 5 1 ? ? u 0 = – 5 cm So, angular magnification = m = – e 0 0 f D u v ? [Taking D = 25 cm] = – 5 25 5 20 ? ? = 20 12. For the astronomical telescope in normal adjustment. Magnifying power = m = 50, length of the tube = L = 102 cm Let f 0 and f e be the focal length of objective and eye piece respectively. m = e 0 f f = 50 ? f 0 = 50 f e …(1) and, L = f 0 + f e = 102 cm …(2) Putting the value of f 0 from equation (1) in (2), we get, f 0 + f e = 102 ? 51f e = 102 ? f e = 2 cm = 0.02 m So, f 0 = 100 cm = 1 m ? Power of the objective lens = 0 f 1 = 1D And Power of the eye piece lens = e f 1 = 02 . 0 1 = 50D B ? A ? ? ? 20cm ? 5cm ? F e ? ? ? f 0 ? Eye piece ? f e ? F e ? Objective ? A ? A ? ? 0.5cm ? Eye piece ? 1cm ? Objective ? 30cm ? 60cm ? B ? ? A ? ? B ? ? B ? Chapter 19 19.5 13. For the given astronomical telescope in normal adjustment, F e = 10 cm, L = 1 m = 100cm S0, f 0 = L – f e = 100 – 10 = 90 cm and, magnifying power = e 0 f f = 10 90 = 9 14. For the given Galilean telescope, (When the image is formed at infinity) f 0 = 30 cm, L = 27 cm Since L = f 0 – e f [Since, concave eyepiece lens is used in Galilean Telescope] ? f e = f 0 – L = 30 – 27 = 3 cm 15. For the far sighted person, u = – 20 cm, v = – 50 cm from lens formula f 1 u 1 v 1 ? ? f 1 = 20 1 50 1 ? ? ? = 50 1 20 1 ? = 100 3 ? f = 3 100 cm = 3 1 m So, power of the lens = f 1 = 3 Diopter 16. For the near sighted person, u = ? and v = – 200 cm = – 2m So, u 1 v 1 f 1 ? ? = ? ? ? 1 2 1 = – 2 1 = – 0.5 So, power of the lens is –0.5D 17. The person wears glasses of power –2.5D So, the person must be near sighted. u = ?, v = far point, f= 5 . 2 1 ? = – 0.4m = – 40 cm Now, f 1 u 1 v 1 ? ? ? f 1 u 1 v 1 ? ? = 40 1 0 ? ? ? v = – 40 cm So, the far point of the person is 40 cm 18. On the 50 th birthday, he reads the card at a distance 25cm using a glass of +2.5D. Ten years later, his near point must have changed. So after ten years, u = – 50 cm, f = D 5 . 2 1 = 0.4m = 40 cm v = near point Now, f 1 u 1 v 1 ? ? ? f 1 u 1 v 1 ? ? = 40 1 50 1 ? ? = 200 1 So, near point = v = 200cm To read the farewell letter at a distance of 25 cm, U = – 25 cm For lens formula, f 1 u 1 v 1 ? ? ? f 1 = 25 200 1 ? ? ? = 25 1 200 1 ? = 200 9 ? f = 9 200 cm = 9 2 m ? Power of the lens = f 1 = 2 9 = 4.5D ?He has to use a lens of power +4.5D.Read More

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