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Page 1 20.1 SOLUTIONS TO CONCEPTS CHAPTER – 20 1. Given that, Refractive index of flint glass = ? f = 1.620 Refractive index of crown glass = ? c = 1.518 Refracting angle of flint prism = A f = 6.0° For zero net deviation of mean ray ( ? f – 1)A f = ( ? c – 1) A c ? A c = f c f A 1 1 ? ? ? ? = ? ? ? ) 0 . 6 ( 1 518 . 1 1 620 . 1 = 7.2° 2. Given that ? r = 1.56, ? y = 1.60, and ? v = 1.68 (a) Dispersive power = ? = 1 y r v ? ? ? ? ? = 1 60 . 1 56 . 1 68 . 1 ? ? = 0.2 (b) Angular dispersion = ( ? v - ? r )A = 0.12 × 6° = 7.2° ? 3. The focal length of a lens is given by f 1 = ( ? – 1) ? ? ? ? ? ? ? ? ? 2 1 R 1 R 1 ? ( ? – 1) = ? ? ? ? ? ? ? ? ? ? 2 1 R 1 R 1 1 f 1 = f K …(1) So, ? r – 1 = 100 K …(2) ? y – 1 = 98 K …(3) And ? v – 1 = 96 K (4) So, Dispersive power = ??= 1 y r v ? ? ? ? ? = ) 1 ( ) 1 ( ) 1 ( y r v ? ? ? ? ? ? ? = 98 K 100 K 96 K ? = 9600 4 98 ? ?= 0.0408 ? 4. Given that, ? v – ? r = 0.014 Again, ? y = depth Apparent depth al Re = 30 . 1 00 . 2 = 1.515 So, dispersive power = 1 y r v ? ? ? ? ? = 1 515 . 1 014 . 0 ? = 0.027 5. Given that, ? r = 1.61, ? v = 1.65, ? = 0.07 and ? y = 4° Now, ? = 1 y r v ? ? ? ? ? ? ? 0.07 = 1 61 . 1 65 . 1 y ? ? ? ? ? ? y – 1 = 07 . 0 04 . 0 = 7 4 Again, ? = ( ? – 1) A ? A = 1 y y ? ? ? = ) 7 / 4 ( 4 = 7° 2cm Image Object 1.32cm Page 2 20.1 SOLUTIONS TO CONCEPTS CHAPTER – 20 1. Given that, Refractive index of flint glass = ? f = 1.620 Refractive index of crown glass = ? c = 1.518 Refracting angle of flint prism = A f = 6.0° For zero net deviation of mean ray ( ? f – 1)A f = ( ? c – 1) A c ? A c = f c f A 1 1 ? ? ? ? = ? ? ? ) 0 . 6 ( 1 518 . 1 1 620 . 1 = 7.2° 2. Given that ? r = 1.56, ? y = 1.60, and ? v = 1.68 (a) Dispersive power = ? = 1 y r v ? ? ? ? ? = 1 60 . 1 56 . 1 68 . 1 ? ? = 0.2 (b) Angular dispersion = ( ? v - ? r )A = 0.12 × 6° = 7.2° ? 3. The focal length of a lens is given by f 1 = ( ? – 1) ? ? ? ? ? ? ? ? ? 2 1 R 1 R 1 ? ( ? – 1) = ? ? ? ? ? ? ? ? ? ? 2 1 R 1 R 1 1 f 1 = f K …(1) So, ? r – 1 = 100 K …(2) ? y – 1 = 98 K …(3) And ? v – 1 = 96 K (4) So, Dispersive power = ??= 1 y r v ? ? ? ? ? = ) 1 ( ) 1 ( ) 1 ( y r v ? ? ? ? ? ? ? = 98 K 100 K 96 K ? = 9600 4 98 ? ?= 0.0408 ? 4. Given that, ? v – ? r = 0.014 Again, ? y = depth Apparent depth al Re = 30 . 1 00 . 2 = 1.515 So, dispersive power = 1 y r v ? ? ? ? ? = 1 515 . 1 014 . 0 ? = 0.027 5. Given that, ? r = 1.61, ? v = 1.65, ? = 0.07 and ? y = 4° Now, ? = 1 y r v ? ? ? ? ? ? ? 0.07 = 1 61 . 1 65 . 1 y ? ? ? ? ? ? y – 1 = 07 . 0 04 . 0 = 7 4 Again, ? = ( ? – 1) A ? A = 1 y y ? ? ? = ) 7 / 4 ( 4 = 7° 2cm Image Object 1.32cm Chapter 20 20.2 6. Given that, ? r = 38.4°, ? y = 38.7° and ? v = 39.2° Dispersive power = 1 y r v ? ? ? ? ? = ) 1 ( ) 1 ( ) 1 ( y r v ? ? ? ? ? ? ? = ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? A A A v r v [ ? ? = ( ? – 1) A] = y r v ? ? ? ? = 7 . 38 4 . 38 2 . 39 ? = 0.0204 7. Two prisms of identical geometrical shape are combined. Let A = Angle of the prisms ?? v = 1.52 and ? v = 1.62, ? v = 1° ? v = ( ? v – 1)A – ( ?? v – 1) A [since A = A ?] ? ? v = ( ? v – ?? v )A ? A = v v v ? ? ? ? ? = 52 . 1 62 . 1 1 ? = 10° 8. Total deviation for yellow ray produced by the prism combination is ? y = ? cy – ? fy + ? cy = 2 ?? cy – ? fy = 2( ? cy – 1)A – ( ? cy – 1)A ? Similarly the angular dispersion produced by the combination is ? v – ? r = [( ? vc – 1)A – ( ? vf – 1)A ? + ( ? vc – 1)A] – [( ? rc – 1) A – ( ? rf – 1)A ? + ( ? r – 1) A)] = 2( ? vc – 1)A – ( ? vf – 1)A ? (a) For net angular dispersion to be zero, ? v – ? r = 0 ? 2( ? vc – 1)A = ( ? vf – 1)A ?? ? A A ? = ) ( ) ( 2 rf vf rc cv ? ? ? ? ? ? = ) ( ) ( 2 r v r v ? ? ? ? ? ? ? ? (b) For net deviation in the yellow ray to be zero, ? y = 0 ? 2( ? cy – 1)A = ( ? fy – 1)A ? ? A A ? = ) 1 ( ) 1 ( 2 fy cy ? ? ? ? = ) 1 ( ) 1 ( 2 y y ? ? ? ? ? ? 9. Given that, ? cr = 1.515, ? cv = 1.525 and ? fr = 1.612, ? fv = 1.632 and A = 5° Since, they are similarly directed, the total deviation produced is given by, ? = ? c + ? r = ( ? c – 1)A + ( ? r – 1) A = ( ? c + ? r – 2)A So, angular dispersion of the combination is given by, ? v – ? y = ( ? cv + ? fv – 2)A – ( ? cr + ? fr – 2)A = ( ? cv + ? fv – ? cr – ? fr )A = (1.525 + 1.632 – 1.515 – 1.612) 5 = 0.15° ? 10. Given that, A ? = 6°, ?? = 0.07, ?? y = 1.50 A = ? ? = 0.08, ? y = 1.60 The combination produces no deviation in the mean ray. (a) ? y = ( ? y – 1)A – ( ?? y – 1)A ? = 0 [Prism must be oppositely directed] ? (1.60 – 1)A = ((1.50 – 1)A ? ? A = 60 . 0 6 50 . 0 ? ? = 5° ? (b) When a beam of white light passes through it, Net angular dispersion = ( ? y – 1) ?A – ( ?? y – 1) ??A ? ? (1.60 – 1)(0.08)(5°) – (1.50 – 1)(0.07)(6°) ? 0.24° – 0.21° = 0.03° (c) If the prisms are similarly directed, ? y = ( ? y – 1)A + ( ?? y – 1)A = (1.60 – 1)5° + (1.50 – 1)6° = 3° + 3° = 6° ? (d) Similarly, if the prisms are similarly directed, the net angular dispersion is given by, ? v – ? r = ( ? y – 1) ?A – ( ?? y – 1) ??A ? = 0.24° + 0.21° = 0.45° ? 5° 6° 5° 6° 5° 5° Prism2 Prism1 Crown Crown A A A ? Flint Page 3 20.1 SOLUTIONS TO CONCEPTS CHAPTER – 20 1. Given that, Refractive index of flint glass = ? f = 1.620 Refractive index of crown glass = ? c = 1.518 Refracting angle of flint prism = A f = 6.0° For zero net deviation of mean ray ( ? f – 1)A f = ( ? c – 1) A c ? A c = f c f A 1 1 ? ? ? ? = ? ? ? ) 0 . 6 ( 1 518 . 1 1 620 . 1 = 7.2° 2. Given that ? r = 1.56, ? y = 1.60, and ? v = 1.68 (a) Dispersive power = ? = 1 y r v ? ? ? ? ? = 1 60 . 1 56 . 1 68 . 1 ? ? = 0.2 (b) Angular dispersion = ( ? v - ? r )A = 0.12 × 6° = 7.2° ? 3. The focal length of a lens is given by f 1 = ( ? – 1) ? ? ? ? ? ? ? ? ? 2 1 R 1 R 1 ? ( ? – 1) = ? ? ? ? ? ? ? ? ? ? 2 1 R 1 R 1 1 f 1 = f K …(1) So, ? r – 1 = 100 K …(2) ? y – 1 = 98 K …(3) And ? v – 1 = 96 K (4) So, Dispersive power = ??= 1 y r v ? ? ? ? ? = ) 1 ( ) 1 ( ) 1 ( y r v ? ? ? ? ? ? ? = 98 K 100 K 96 K ? = 9600 4 98 ? ?= 0.0408 ? 4. Given that, ? v – ? r = 0.014 Again, ? y = depth Apparent depth al Re = 30 . 1 00 . 2 = 1.515 So, dispersive power = 1 y r v ? ? ? ? ? = 1 515 . 1 014 . 0 ? = 0.027 5. Given that, ? r = 1.61, ? v = 1.65, ? = 0.07 and ? y = 4° Now, ? = 1 y r v ? ? ? ? ? ? ? 0.07 = 1 61 . 1 65 . 1 y ? ? ? ? ? ? y – 1 = 07 . 0 04 . 0 = 7 4 Again, ? = ( ? – 1) A ? A = 1 y y ? ? ? = ) 7 / 4 ( 4 = 7° 2cm Image Object 1.32cm Chapter 20 20.2 6. Given that, ? r = 38.4°, ? y = 38.7° and ? v = 39.2° Dispersive power = 1 y r v ? ? ? ? ? = ) 1 ( ) 1 ( ) 1 ( y r v ? ? ? ? ? ? ? = ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? A A A v r v [ ? ? = ( ? – 1) A] = y r v ? ? ? ? = 7 . 38 4 . 38 2 . 39 ? = 0.0204 7. Two prisms of identical geometrical shape are combined. Let A = Angle of the prisms ?? v = 1.52 and ? v = 1.62, ? v = 1° ? v = ( ? v – 1)A – ( ?? v – 1) A [since A = A ?] ? ? v = ( ? v – ?? v )A ? A = v v v ? ? ? ? ? = 52 . 1 62 . 1 1 ? = 10° 8. Total deviation for yellow ray produced by the prism combination is ? y = ? cy – ? fy + ? cy = 2 ?? cy – ? fy = 2( ? cy – 1)A – ( ? cy – 1)A ? Similarly the angular dispersion produced by the combination is ? v – ? r = [( ? vc – 1)A – ( ? vf – 1)A ? + ( ? vc – 1)A] – [( ? rc – 1) A – ( ? rf – 1)A ? + ( ? r – 1) A)] = 2( ? vc – 1)A – ( ? vf – 1)A ? (a) For net angular dispersion to be zero, ? v – ? r = 0 ? 2( ? vc – 1)A = ( ? vf – 1)A ?? ? A A ? = ) ( ) ( 2 rf vf rc cv ? ? ? ? ? ? = ) ( ) ( 2 r v r v ? ? ? ? ? ? ? ? (b) For net deviation in the yellow ray to be zero, ? y = 0 ? 2( ? cy – 1)A = ( ? fy – 1)A ? ? A A ? = ) 1 ( ) 1 ( 2 fy cy ? ? ? ? = ) 1 ( ) 1 ( 2 y y ? ? ? ? ? ? 9. Given that, ? cr = 1.515, ? cv = 1.525 and ? fr = 1.612, ? fv = 1.632 and A = 5° Since, they are similarly directed, the total deviation produced is given by, ? = ? c + ? r = ( ? c – 1)A + ( ? r – 1) A = ( ? c + ? r – 2)A So, angular dispersion of the combination is given by, ? v – ? y = ( ? cv + ? fv – 2)A – ( ? cr + ? fr – 2)A = ( ? cv + ? fv – ? cr – ? fr )A = (1.525 + 1.632 – 1.515 – 1.612) 5 = 0.15° ? 10. Given that, A ? = 6°, ?? = 0.07, ?? y = 1.50 A = ? ? = 0.08, ? y = 1.60 The combination produces no deviation in the mean ray. (a) ? y = ( ? y – 1)A – ( ?? y – 1)A ? = 0 [Prism must be oppositely directed] ? (1.60 – 1)A = ((1.50 – 1)A ? ? A = 60 . 0 6 50 . 0 ? ? = 5° ? (b) When a beam of white light passes through it, Net angular dispersion = ( ? y – 1) ?A – ( ?? y – 1) ??A ? ? (1.60 – 1)(0.08)(5°) – (1.50 – 1)(0.07)(6°) ? 0.24° – 0.21° = 0.03° (c) If the prisms are similarly directed, ? y = ( ? y – 1)A + ( ?? y – 1)A = (1.60 – 1)5° + (1.50 – 1)6° = 3° + 3° = 6° ? (d) Similarly, if the prisms are similarly directed, the net angular dispersion is given by, ? v – ? r = ( ? y – 1) ?A – ( ?? y – 1) ??A ? = 0.24° + 0.21° = 0.45° ? 5° 6° 5° 6° 5° 5° Prism2 Prism1 Crown Crown A A A ? Flint Chapter 20 20.3 11. Given that, ?? v – ?? r = 0.014 and ? v – ? r = 0.024 A ? = 5.3° and A = 3.7° (a) When the prisms are oppositely directed, angular dispersion = ( ? v – ? r )A – ( ?? v – ?? r )A ? = 0.024 × 3.7° – 0.014 × 5.3° = 0.0146° (b) When they are similarly directed, angular dispersion = ( ? v – ? r )A + ( ?? v – ?? r )A ? = 0.024 × 3.7° + 0.014 × 5.3° = 0.163° ? ? ? ? ? 3.7° 5.3° 5.3° 3.7°Read More
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FAQs on HC Verma Solutions: Chapter 20 - Dispersion & Spectra - Physics Class 11 - NEET
| 1. What is dispersion in the context of optics? |  |
Ans. Dispersion refers to the phenomenon where different colors of light separate and spread out when passing through a medium. This occurs because the refractive index of the medium varies with the wavelength of light, causing different colors to bend at different angles.
| 2. How does a prism disperse white light? | |
Ans. A prism disperses white light by refracting different colors of light at different angles. When white light enters a prism, it undergoes multiple refractions and reflections within the prism. The different colors of light are refracted at different angles, causing them to separate and form a spectrum.
| 3. What is the relationship between the angle of deviation and the refractive index of a medium? | |
Ans. The angle of deviation is directly proportional to the refractive index of a medium. This means that as the refractive index increases, the angle of deviation also increases. The relationship between the two can be expressed mathematically as: Angle of Deviation = (Refractive Index - 1) x Angle of Prism.
| 4. How does dispersion affect the formation of rainbows? | |
Ans. Dispersion plays a crucial role in the formation of rainbows. When sunlight passes through water droplets in the atmosphere, it undergoes dispersion, causing the different colors of light to separate. This separation of colors creates the beautiful arc of a rainbow, with red on the outer edge and violet on the inner edge.
| 5. What are the applications of dispersion in everyday life? | |
Ans. Dispersion has several practical applications in everyday life. Some examples include:
- The functioning of prisms in optical devices such as cameras, binoculars, and spectrometers.
- The creation of colorful displays in devices like televisions and computer screens.
- The design of optical fibers used in telecommunications to transmit different wavelengths of light simultaneously.
- The production of colorful pigments in paints, dyes, and printing inks.
- The analysis of elements and compounds in spectroscopy by studying the unique spectral lines produced by their dispersed light.
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