NEET Exam > NEET Notes > Physics Class 11 > HC Verma Solutions: Chapter 20 - Dispersion & Spectra
HC Verma Solutions: Chapter 20 - Dispersion & Spectra | Physics Class 11 - NEET PDF Download
| Download, print and study this document offline |
Please wait while the PDF view is loading
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
|
97 videos|378 docs|103 tests
|
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.
About this Document
|
4.87/5 Rating |
|
Dec 18, 2024 Last updated |
Document Description: HC Verma Solutions: Chapter 20 - Dispersion & Spectra for NEET 2024 is part of Physics Class 11 preparation.
The notes and questions for HC Verma Solutions: Chapter 20 - Dispersion & Spectra have been prepared according to the NEET exam syllabus. Information about HC Verma Solutions: Chapter 20 - Dispersion & Spectra covers topics
like and HC Verma Solutions: Chapter 20 - Dispersion & Spectra Example, for NEET 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises and tests below for HC Verma Solutions: Chapter 20 - Dispersion & Spectra.
Introduction of HC Verma Solutions: Chapter 20 - Dispersion & Spectra in English is available as part of our Physics Class 11
for NEET & HC Verma Solutions: Chapter 20 - Dispersion & Spectra in Hindi for Physics Class 11 course.
Download more important topics related with notes, lectures and mock test series for NEET
Exam by signing up for free. NEET: HC Verma Solutions: Chapter 20 - Dispersion & Spectra | Physics Class 11 - NEET
Description
Full syllabus notes, lecture & questions for HC Verma Solutions: Chapter 20 - Dispersion & Spectra | Physics Class 11 - NEET - NEET | Plus excerises question with solution to help you revise complete syllabus for Physics Class 11 | Best notes, free PDF download
Information about HC Verma Solutions: Chapter 20 - Dispersion & Spectra
In this doc you can find the meaning of HC Verma Solutions: Chapter 20 - Dispersion & Spectra defined & explained in the simplest way possible. Besides explaining types of
HC Verma Solutions: Chapter 20 - Dispersion & Spectra theory, EduRev gives you an ample number of questions to practice HC Verma Solutions: Chapter 20 - Dispersion & Spectra tests, examples and also practice NEET
tests
|
Explore Courses for NEET exam
|
|
Signup for Free!
Signup to see your scores go up within 7 days! Learn & Practice with 1000+ FREE Notes, Videos & Tests.
Related Searches
Free
,ppt
,Semester Notes
,MCQs
,Objective type Questions
,mock tests for examination
,Previous Year Questions with Solutions
,past year papers
,Viva Questions
,Exam
,video lectures
,Extra Questions
,shortcuts and tricks
,Sample Paper
,HC Verma Solutions: Chapter 20 - Dispersion & Spectra | Physics Class 11 - NEET
,Important questions
,practice quizzes
,study material
,HC Verma Solutions: Chapter 20 - Dispersion & Spectra | Physics Class 11 - NEET
,HC Verma Solutions: Chapter 20 - Dispersion & Spectra | Physics Class 11 - NEET
,Summary
;
Additional Information about HC Verma Solutions: Chapter 20 - Dispersion & Spectra for NEET Preparation
HC Verma Solutions: Chapter 20 - Dispersion & Spectra Free PDF Download
The HC Verma Solutions: Chapter 20 - Dispersion & Spectra is an invaluable resource that delves deep into the core of the NEET exam.
These study notes are curated by experts and cover all the essential topics and concepts, making your preparation more efficient and effective.
With the help of these notes, you can grasp complex subjects quickly, revise important points easily,
and reinforce your understanding of key concepts. The study notes are presented in a concise and easy-to-understand manner,
allowing you to optimize your learning process. Whether you're looking for best-recommended books, sample papers, study material,
or toppers' notes, this PDF has got you covered. Download the HC Verma Solutions: Chapter 20 - Dispersion & Spectra now and kickstart your journey towards success in the NEET exam.
Importance of HC Verma Solutions: Chapter 20 - Dispersion & Spectra
The importance of HC Verma Solutions: Chapter 20 - Dispersion & Spectra cannot be overstated, especially for NEET aspirants.
This document holds the key to success in the NEET exam.
It offers a detailed understanding of the concept, providing invaluable insights into the topic.
By knowing the concepts well in advance, students can plan their preparation effectively.
Utilize this indispensable guide for a well-rounded preparation and achieve your desired results.
HC Verma Solutions: Chapter 20 - Dispersion & Spectra Notes
HC Verma Solutions: Chapter 20 - Dispersion & Spectra Notes offer in-depth insights into the specific topic to help you master it with ease.
This comprehensive document covers all aspects related to HC Verma Solutions: Chapter 20 - Dispersion & Spectra.
It includes detailed information about the exam syllabus, recommended books, and study materials for a well-rounded preparation.
Practice papers and question papers enable you to assess your progress effectively.
Additionally, the paper analysis provides valuable tips for tackling the exam strategically.
Access to Toppers' notes gives you an edge in understanding complex concepts.
Whether you're a beginner or aiming for advanced proficiency, HC Verma Solutions: Chapter 20 - Dispersion & Spectra Notes on EduRev are your ultimate resource for success.
HC Verma Solutions: Chapter 20 - Dispersion & Spectra NEET Questions
The "HC Verma Solutions: Chapter 20 - Dispersion & Spectra NEET Questions" guide is a valuable resource for all aspiring students preparing for the
NEET exam. It focuses on providing a wide range of practice questions to help students gauge
their understanding of the exam topics. These questions cover the entire syllabus, ensuring comprehensive preparation.
The guide includes previous years' question papers for students to familiarize themselves with the exam's format and difficulty level.
Additionally, it offers subject-specific question banks, allowing students to focus on weak areas and improve their performance.
Study HC Verma Solutions: Chapter 20 - Dispersion & Spectra on the App
Students of NEET can study HC Verma Solutions: Chapter 20 - Dispersion & Spectra alongwith tests & analysis from the EduRev app,
which will help them while preparing for their exam. Apart from the HC Verma Solutions: Chapter 20 - Dispersion & Spectra,
students can also utilize the EduRev App for other study materials such as previous year question papers, syllabus, important questions, etc.
The EduRev App will make your learning easier as you can access it from anywhere you want.
The content of HC Verma Solutions: Chapter 20 - Dispersion & Spectra is prepared as per the latest NEET syllabus.
|
© EduRev
|
Education Revolution
|
Follow Us
|
Signup to see your scores
go up within 7 days!
Access 1000+ FREE Docs, Videos and Tests
Takes less than 10 seconds to signup