Courses
Courses for Kids
Free study material
Offline Centres
More
Store Icon
Store

NCERT Solutions For Class 10 Science Chapter 11 Human Eye and Colourful World in Hindi - 2025-26

ffImage
banner

Solved NCERT Questions For Class 10 Science Chapter 11 In Hindi - Free PDF

Download the Class 10 Science NCERT Solutions in Hindi medium and English medium as well offered by the leading e-learning platform Vedantu. If you are a student of Class 10, you have reached the right platform. The NCERT Solutions for Class 10 Science in Hindi provided by us are designed in a simple, straightforward language, which are easy to memorise. 

toc-symbolTable of Content
toggle-arrow


You will also be able to download the PDF file for NCERT Solutions for Class 10 Science in English and Hindi from our website at absolutely free of cost. Download NCERT Solutions for Class 10 Maths from Vedantu, which are curated by master teachers.


Class:

NCERT Solutions for Class 10

Subject:

Class 10 Science

Chapter Name:

Chapter 11 - Human Eye And Colourful World

Content-Type:

Text, Videos, Images and PDF Format

Academic Year:

2024-25

Medium:

English and Hindi

Available Materials:

  • Chapter Wise

  • Exercise Wise

Other Materials

  • Important Questions

  • Revision Notes



NCERT, which stands for The National Council of Educational Research and Training, is responsible for designing and publishing textbooks for all the classes and subjects. NCERT textbooks covered all the topics and are applicable to the Central Board of Secondary Education (CBSE) and various state boards. 


We, at Vedantu, offer free NCERT Solutions in English medium and Hindi medium for all the classes as well. Created by subject matter experts, these NCERT Solutions in Hindi are very helpful to the students of all classes.

Watch videos on

NCERT Solutions For Class 10 Science Chapter 11 Human Eye and Colourful World in Hindi - 2025-26
Previous
Next
Vedantu 9&10
Subscribe
Download Notes
iconShare
The Human Eye and The Colourful World L-6 [🧐Score Booster MCQ Quiz] CBSE Class 10 Science Chapter 11
5.3K likes
65.4K Views
3 years ago
Vedantu 9&10
Subscribe
Download Notes
iconShare
The Human Eye and The Colourful World in One Shot (Full Chapter) CBSE 10 Science Chapter 11(Term 1)
14.2K likes
313.5K Views
3 years ago

Access NCERT Solutions for Class 10 Science Chapter 11- मानव नेत्र तथा रंगबिरंगा संसार

अभ्यास

1. मानव नेत्र अभिनेत्र लेंस की फोकस दूरी को समायोजित करके विभिन्न दूरियों पर रखी वस्तुओं को फोकसित कर सकता है ऐसा हो पाने का कारण है-

(a) जरा-दूरदृष्टिता

(b) समंजन

(c) निकट-दृष्टि

(d) दीर्घ-दृष्टि

उत्तर: (b) समंजन


2. मानव नेत्र जिस भाग पर किसी वस्तु का प्रतिबिंब बनाते हैं वह है-

(a) कॉर्निया

(b) परितारिका

(c) पुतली

(d) दृष्टिपटल

उत्तर: (d) दृष्टिपटल


3. सामान्य दृष्टि के वयस्क के लिए सुस्पष्ट दर्शन की अल्पतम दूरी होती है, लगभग-

(a) 25 m

(b) 2.5 cm

(c) 25 cm

(d) 2.5 m

उत्तर: (c) 25 cm


4. अभिनेत्र लेंस की फोकस दूरी में परिवर्तन किया जाता है-

(a) पुतली द्वारा

(b) दृष्टिपटल द्वारा

(c) पक्ष्माभी द्वारा

(d) परितारिका द्वारा

उत्तर: (c) पक्ष्माभी द्वारा


5. किसी व्यक्ति को अपनी दूर की दृष्टि को संशोधित करने के लिए -5.5 डाइऑप्टर क्षमता के लेंस की आवश्यकता है। अपनी निकट की दृष्टि को संशोधित करने के लिए उसे +1.5 डाइऑप्टर क्षमता के लेंस की आवश्यकता है। संशोधित करने के लिए आवश्यक लेंस की फोकस दूरी क्या होगी?

(i) दूर की दृष्टि के लिए 

उत्तर: दूर की दृष्टि को संशोधित करने वाले लेंस की क्षमता P1 = – 5.5D

इसलिए, फोकस दूरी

f1 = 1/P1 = 1/(-5.5)m

अतः

f1 = -0.18m


(ii) निकट की दृष्टि के लिए

उत्तर: निकट दृष्टि को संशोधित करने वाले लेंस की क्षमता

P2 = +1.5D

इसलिए,

f2 = 1/P2 = 1/(1.5)m

f2 = 10/15 = 0.67m

अतः

f2  = +0.67m


6. किसी निकट-दृष्टि दोष से पीड़ित व्यक्ति का दूर बिंदु नेत्र के सामने 80 cm दूरी पर है। इस दोष को संशोधित करने के लिए आवश्यक लेंस की प्रकृति तथा क्षमता क्या होगी?

उत्तर: निकट-दृष्टि दोष से पीड़ित व्यक्ति के लिए :

u = -∞

v = -80cm 

( ताकि अनंत दूरी पर स्थिति वस्तु का प्रतिबिंब  v = -80cm पर बन जाए |)

संशोधित लेंस की फोकस दूरी, f = ?

1/v - 1/u = 1/f ( लेंस सूत्र द्वारा )

1/(-80) - 1/(-∞) = 1/f

1/(-80) = 1/f

f = -80cm = -0.80m ( अवतल लेंस )

∴ संशोधित लेंस की क्षमता 

P = 1/f = 1/(-0.80m) = -1.25D

अतः इस दोष को संशोधित करने के लिए -1.25 D क्षमता वाले एक अवतल लेंस का प्रयोग करना चाहिए।


7. चित्र बनाकर दर्शाइए कि दीर्घ-दृष्टि दोष कैसे संशोधित किया जाता है। एक दीर्घ-दृष्टि दोषयुक्त नेत्र का निकट बिंदु 1 m है। इस दोष को संशोधित करने के लिए आवश्यक लेंस की क्षमता क्या होगी? यह मान लीजिए कि सामान्य नेत्र का निकट बिंदु 25 cm है।

उत्तर: हम जानते हैं कि दीर्घ-दृष्टि दोष युक्त नेत्र दूर की वस्तुओं को तो स्पष्ट देख लेता है, लेकिन नजदीक की वस्तुओं को स्पष्ट नहीं देख पाता है, इसके निवारण के लिए उचित क्षमता का उत्तल लेंस प्रयुक्त करते हैं ताकि पास से आने वाली प्रकाश किरणें रेटिना पर फोकसित हो जाए।

यह उत्तल लेंस 25 cm पर रखी वस्तु N’ का आभासी प्रतिबिंब N बना देता है। अब पीड़ित आँख N बिंदु से आने वाली प्रकाश किरणों को रेटिना पर फोकसित कर देती है।


(i) Eye picture with hyperopia. (ii) Correction of Hyper Sightedness


u = – 25 cm, v = – 1 m = – 100 cm

∴ 1/v - 1/u = 1/f ( लेंस सूत्र द्वारा )

1/(-100cm) - 1/(-25cm) = 1/f

1/f = -1/100 + 1/25 

= (-1 + 4)/100

= 3/100 cm

f = 100/3 = (1/3)m

अतः संशोधित लेंस की क्षमता

P = 1/f = 1/(1/3) = +3D ( उत्तल लेंस )


8. सामान्य नेत्र 25 cm से निकट रखी वस्तुओं को सुस्पष्ट क्यों नहीं देख पाते हैं?

उत्तर: पक्ष्माभी पेशियाँ अभिनेत्र लेंस की फोकस दूरी को एक निश्चित न्यूनतम सीमा से कम नहीं कर पातीं। इसलिए सामान्य नेत्र भी स्पष्ट दर्शन की न्यूनतम दूरी 25 cm से कम पर रखी वस्तुओं को सुस्पष्ट नहीं देख पाती हैं।


9. जब हम नेत्र से किसी वस्तु की दूरी को बढ़ा देते हैं तो नेत्र में प्रतिबिंब-दूरी का क्या होता है?

उत्तर: नेत्र के सामने किसी वस्तु को 25 cm तथा अनंत के बीच कहीं भी रखें, प्रतिबिंब सदैव रेटिना पर ही बनेगा। अत: नेत्र से वस्तु की दूरी बढ़ाने पर भी प्रतिबिंब-दूरी अपरिवर्तित रहती है।


10. तारे क्यों टिमटिमाते हैं?

उत्तर: तारों से आने वाले प्रकाश के वायुमंडलीय अपवर्तन के कारण तारे टिमटिमाते हुए प्रतीत होते हैं। पृथ्वी के वायुमंडल में प्रवेश करने के बाद तारे के प्रकाश को विभिन्न अपवर्तनांक वाले वायुमंडल से गुजरना होता है, इसलिए प्रकाश का लगातार अपवर्तन होते रहने के कारण प्रकाश की दिशा बदलती रहती है, जिससे तारे टिमटिमाते हुए प्रतीत होते हैं।


11. व्याख्या कीजिए कि ग्रह क्यों नहीं टिमटिमाते?

उत्तर: हम जानते हैं कि ग्रह तारों की अपेक्षा पृथ्वी के बहुत पास हैं और ये प्रकाश के विस्तृत स्रोत की भाँति माने जाते हैं। यदि हम ग्रह को बिंदु आकार के अनेक प्रकाश स्रोतों का संग्रह मान लें तो सभी बिंदु आकार के प्रकाश स्रोतों से हमारी आँखों में आने वाले प्रकाश की मात्रा में कुल परिवर्तन का औसत मान शून्य होगा, जिसके कारण ग्रहों के टिमटिमाने का प्रभाव लगभग शून्य हो जाता है।


12. सूर्योदय के समय सूर्य रक्ताभ क्यों प्रतीत होता है?

उत्तर: सूर्योदय के समय सूर्य क्षितिज के पास होता है, जहाँ से आने वाले प्रकाश को वायुमंडल की मोटी परतों से होकर गुजरना पड़ता है तथा अधिक दूरी तय करनी पड़ती है। नीले तथा कम तरंगदैर्ध्य के प्रकाश का अधिकांश भाग कणों द्वारा प्रकीर्णित हो जाता है और सिर्फ अधिक तरंगदैर्ध्य वाले प्रकाश जैसे लाल रंग ही हम तक पहुँचते है। अतः सूर्य रक्ताभ प्रतीत होता है।


13. किसी अंतरिक्ष यात्री को आकाश नीले की अपेक्षा काला क्यों प्रतीत होता है?

उत्तर: आकाश का नीला रंग पृथ्वी पर स्थित वायुमंडल के सूक्ष्म कणों द्वारा प्रकाश के प्रकीर्णन के कारण होता है। अंतरिक्ष यात्री को आकाश नीले की अपेक्षा काला इसलिए दिखाई देता है, क्योंकि वे अत्यधिक ऊँचाई पर पृथ्वी की परिक्रमा करते हैं, जहाँ वायुमंडल नहीं होता। परिणामस्वरूप प्रकाश का प्रकीर्णन नहीं हो पाता है और आकाश काला प्रतीत होता है।


NCERT Solutions for Class 10 Science Chapter 11 Human Eye and Colourful World in Hindi

Chapter-wise NCERT Solutions are provided everywhere on the internet with an aim to help the students to gain a comprehensive understanding. Class 10 Science Chapter 11 solution Hindi mediums are created by our in-house experts keeping the understanding ability of all types of candidates in mind. NCERT textbooks and solutions are built to give a strong foundation to every concept. These NCERT Solutions for Class 10 Science Chapter 11 in Hindi ensure a smooth understanding of all the concepts including the advanced concepts covered in the textbook.

NCERT Solutions for Class 10 Science Chapter 11 in Hindi medium PDF download are easily available on our official website (vedantu.com). Upon visiting the website, you have to register on the website with your phone number and email address. Then you will be able to download all the study materials of your preference in a click. You can also download the Class 10 Science Human Eye and Colourful World solution Hindi medium from Vedantu app as well by following the similar procedures, but you have to download the app from Google play store before doing that. 

NCERT Solutions in Hindi medium have been created keeping those students in mind who are studying in a Hindi medium school. These NCERT Solutions for Class 10 Science Human Eye and Colourful World in Hindi medium pdf download have innumerable benefits as these are created in simple and easy-to-understand language. The best feature of these solutions is a free download option. Students of Class 10 can download these solutions at any time as per their convenience for self-study purpose. 

These solutions are nothing but a compilation of all the answers to the questions of the textbook exercises. The answers/ solutions are given in a stepwise format and very well researched by the subject matter experts who have relevant experience in this field. Relevant diagrams, graphs, illustrations are provided along with the answers wherever required. In nutshell, NCERT Solutions for Class 10 Science in Hindi come really handy in exam preparation and quick revision as well prior to the final examinations. 

WhatsApp Banner
Best Seller - Grade 10
View More>
Previous
Next

FAQs on NCERT Solutions For Class 10 Science Chapter 11 Human Eye and Colourful World in Hindi - 2025-26

1. Where can I find reliable, step-by-step NCERT Solutions for Class 10 Science Chapter 11 for the CBSE 2025-26 session?

You can find comprehensive and accurate NCERT Solutions for Class 10 Science Chapter 11, 'The Human Eye and the Colourful World,' on Vedantu. These solutions are crafted by subject matter experts and provide a detailed, step-by-step methodology for solving every in-text and exercise question as per the latest CBSE 2025-26 guidelines. They focus on helping you understand the correct approach to framing answers and solving numericals.

2. What is the correct method to solve numerical problems on myopia and hypermetropia from the Chapter 11 NCERT exercises?

To correctly solve numericals on vision defects from the NCERT textbook, follow these steps:

  • First, identify the defect. For myopia (nearsightedness), the person's far point is less than infinity. For hypermetropia (farsightedness), the near point is more than 25 cm.
  • Use the lens formula: 1/f = 1/v - 1/u.
  • For myopia correction, the object (u) is at infinity (∞), and the image (v) must form at the person's far point. The required lens will be a concave lens, so the focal length (f) and power (P) will be negative.
  • For hypermetropia correction, the object (u) is at the normal near point (25 cm), and the image (v) must form at the person's actual near point. The required lens is a convex lens, so the focal length (f) and power (P) will be positive.
  • Always use the correct sign convention (u is always negative). Finally, calculate power using P = 1/f (where f is in metres).

3. How should I structure my answer for the NCERT question, 'Why do stars twinkle but planets do not?' to score full marks?

For the NCERT solution to this question, a well-structured answer should include these key points:

  • Source of Light: Stars are distant, point-sized sources of light, while planets are closer and appear as extended sources (a collection of many point sources).
  • Atmospheric Refraction: Explain that the twinkling of stars is due to the continuous atmospheric refraction of starlight as it passes through Earth's atmosphere, which has varying optical densities.
  • Apparent Position: Mention that this refraction causes the apparent position of the star to fluctuate slightly, and the path of the light rays changes continuously. This varying amount of light entering the eye causes the twinkling effect.
  • Effect on Planets: Conclude by explaining that planets, being extended sources, have light rays from multiple points. The dimming effect from some points is nullified by the brightening effect from others, averaging out to a constant brightness. Thus, the twinkling effect is cancelled, and planets do not appear to twinkle.

4. While solving an NCERT question on correcting vision, what are the essential elements to include in a ray diagram?

To draw a complete and correct ray diagram for an NCERT solution on vision defects, you must include:

  • The human eye lens (a convex lens).
  • The retina, where the image should ideally be formed.
  • For a defective eye, show where the image is incorrectly formed (in front of the retina for myopia, behind for hypermetropia).
  • The corrective lens (concave for myopia, convex for hypermetropia) placed in front of the eye.
  • Show at least two light rays coming from the object (e.g., from infinity for myopia correction) and trace their path through the corrective lens and then the eye lens, converging correctly on the retina.
  • Clearly label all parts and use arrows to indicate the direction of light.

5. How does the concept of 'power of accommodation' help in solving NCERT questions about the range of vision?

The 'power of accommodation' is a crucial concept for solving NCERT questions about vision range because it explains the eye's ability to adjust its focal length. For a normal eye, the range is from infinity (the far point) to 25 cm (the near point). The eye accommodates by making the lens thicker (more curved) to see nearby objects and thinner (less curved) for distant objects. When a problem states a person's near or far point has shifted (e.g., near point is 1m), it implies their power of accommodation is weakened or faulty, which is the root cause of the refractive defect that needs correction.

6. For NCERT solutions explaining atmospheric phenomena, what is the key difference between using 'scattering' and 'refraction' in an answer?

When solving NCERT questions, it's vital to distinguish between these two phenomena. Atmospheric refraction is the bending of light as it passes through atmospheric layers of different densities. Use this concept to explain why stars twinkle or why sunrise appears advanced. In contrast, scattering of light is the process where light is absorbed and re-emitted in different directions by particles in the atmosphere. Use scattering (specifically Rayleigh scattering) to explain why the sky is blue (blue light is scattered most) and why the sun appears red at sunrise/sunset (most blue light is scattered away, leaving red light to reach the observer).

7. For an NCERT problem where a person cannot see objects beyond 1.2 m, how do I determine the defect and the correct sign convention for the lens power calculation?

First, identify the defect. A normal eye can see up to infinity. Since this person's far point is only 1.2 m, they are suffering from myopia (nearsightedness). The goal is to use a lens that makes an object at infinity (u = -∞) appear at the person's far point (v = -1.2 m). Using the lens formula, 1/f = 1/v - 1/u, we get 1/f = 1/(-1.2) - 1/(-∞). Since 1/∞ is 0, the focal length f = -1.2 m. The negative sign confirms a concave lens is needed. The power P = 1/f = 1/(-1.2) = -0.83 D. The sign convention is critical: both 'v' and 'f' are negative for myopic correction.

8. What is the step-by-step solution for the NCERT question explaining why danger signals are red?

To answer this question as per the NCERT solutions, follow these steps:

  • Start by stating the core principle: the phenomenon of light scattering.
  • Explain that according to Rayleigh's law of scattering, the intensity of scattered light is inversely proportional to the fourth power of its wavelength (I ∝ 1/λ⁴).
  • Mention that red light has the longest wavelength among all colours in the visible spectrum, while blue/violet has the shortest.
  • Therefore, red light is scattered the least by particles of dust, fog, or smoke in the atmosphere.
  • Conclude that because it is scattered the least, red light can travel the farthest distance through the atmosphere without losing its intensity, making it visible from a great distance. This is why it is used for danger signals.