Light – Reflection and Refraction (Class 10): Notes और MCQ Practice

Class 10 Science • Class 10

आसान भाषा में:

Light एक energy है जो हमें दुनिया देखने में मदद करती है। जब light किसी shiny surface (जैसे mirror) से टकराकर वापस लौटती है, तो उसे Reflection कहते हैं। और जब light एक transparent medium से दूसरे में जाते हुए अपनी speed बदलने के कारण मुड़ जाती है (bend होती है), तो उसे Refraction कहते हैं।

Important Formulas & Definitions:

  • Laws of Reflection: Angle of incidence (i) = Angle of reflection (r).
  • Mirror Formula: 1/v + 1/u = 1/f
  • Lens Formula: 1/v − 1/u = 1/f
  • Magnification (m):
  • Mirrors: m = h' ÷ h = −v ÷ u
  • Lenses: m = h' ÷ h = v ÷ u

(m negative मतलब Real & Inverted; m positive मतलब Virtual & Erect)

  • Snell's Law: sin i ÷ sin r = Constant (Refractive Index)
  • Absolute Refractive Index (n): n = c ÷ v (Speed of light in vacuum ÷ Speed in medium)
  • Power of Lens (P): P = 1 ÷ f (f focal length हमेशा meters में). Unit: Dioptre (D).
  • Convex lens: P is positive (+)
  • Concave lens: P is negative (−)

2 Memory Tricks:

  1. Formula Sign Trick: "Mirror जोड़ता है (+), Lens घटाता है (−)" — Mirror formula में बीच में '+' आता है (1/v + 1/u), Lens formula में बीच में '−' आता है (1/v − 1/u). Magnification में इसके ठीक उल्टे signs होते हैं!
  2. Focal Length Rule: "Con-CAVE = CAVE (गुफा = Negative/अंधेरा)" — Concave lens/mirror की focal length (f) हमेशा negative (−) होती है। Convex की हमेशा positive (+) होती है।

2 Common Board-Exam Mistakes:

  1. Power Calculation में Unit की गलती: Students focal length (f) को cm में रखकर ही P = 1 ÷ f निकाल देते हैं। 'f' को पहले meter में convert करें, तभी unit Dioptre (D) बनेगी।
  2. Ray Diagrams में Arrows छोड़ना: Light rays की direction दिखाने वाले arrows ना बनाने पर सीधे marks कटते हैं। हमेशा incident और refracted/reflected ray पर arrow बनाएं।

Light – Reflection and Refraction: Class 10 Notes

1. Chapter in Short

Light is a form of energy that enables us to see objects around us. An object reflects light that falls on it, and this reflected light, when received by our eyes, enables us to see.

This chapter covers two fundamental phenomena of light:

  1. Reflection of Light: The bouncing back of light rays into the same medium when they strike a smooth surface (like plane mirrors and curved/spherical mirrors).
  2. Refraction of Light: The bending of light when it passes obliquely from one transparent medium to another due to a change in its speed (like through glass slabs, water, and spherical lenses).

You will also learn the Sign Convention, ray tracing, mirror and lens formulas, magnification, and the power of a lens.

2. Key Concepts Explained Simply

A. Reflection of Light & Spherical Mirrors

  • Spherical Mirrors: Mirrors whose reflecting surfaces are curved spherical parts of a hollow glass sphere.
  • Concave Mirror: Reflecting surface curves inwards (facing towards the centre of the sphere). It converges parallel rays of light.
  • Everyday uses: Shaving/makeup mirrors (gives an enlarged view when held close), dentist mirrors, car headlights, and solar furnaces.
  • Convex Mirror: Reflecting surface curves outwards. It diverges parallel rays of light.
  • Everyday uses: Rear-view mirrors in cars and security mirrors in shops (gives an erect, diminished image and provides a very wide field of view).

B. Sign Convention (New Cartesian Sign Convention)

To solve numerical problems correctly, remember these rules:

  • The Pole (P) of the mirror or the Optical Centre (O) of the lens is taken as the origin (0, 0).
  • The principal axis is taken along the x-axis.
  • Light always travels from left to right.
  • All distances measured to the left of the origin are taken as negative (-). Thus, object distance (u) is almost always negative.
  • All distances measured to the right of the origin are taken as positive (+).
  • Heights measured upwards (perpendicular to principal axis) are positive (+).
  • Heights measured downwards are negative (-).

C. Refraction of Light

  • When light travels from one medium to another, its speed changes. This change in speed causes a change in direction (bending).
  • Optically Rarer to Denser (e.g., Air to Glass): Light slows down and bends towards the normal.
  • Optically Denser to Rarer (e.g., Glass to Air): Light speeds up and bends away from the normal.
  • Everyday Examples: A pencil partially dipped in water appears bent at the surface; a coin placed at the bottom of a water cup appears raised.

D. Refractive Index

  • It measures how much the speed of light changes in a given medium compared to another.
  • Absolute Refractive Index (n): Ratio of speed of light in air/vacuum to speed of light in the medium.

E. Spherical Lenses

  • Convex Lens (Converging): Thicker at the middle, thinner at the edges. It converges light rays.
  • Concave Lens (Diverging): Thinner at the middle, thicker at the edges. It diverges light rays.

F. Power of a Lens

  • The degree of convergence or divergence of light rays achieved by a lens is expressed in terms of its power.
  • It is the reciprocal of its focal length in metres.

3. Important Definitions, Formulas, and Facts

Laws of Reflection:

  1. The angle of incidence equals the angle of reflection: ∠i = ∠r.
  2. The incident ray, the reflected ray, and the normal to the reflecting surface at the point of incidence, all lie in the same plane.

Laws of Refraction:

  1. The incident ray, the refracted ray, and the normal to the interface of two transparent media at the point of incidence, all lie in the same plane.
  2. Snell's Law: The ratio of sine of angle of incidence to the sine of angle of refraction is a constant for a given pair of media and color of light:
  • (sin i) ÷ (sin r) = Constant = n₂₁ (refractive index of medium 2 with respect to medium 1)

Key Formulas:

  • Radius of Curvature and Focal Length:
  • R = 2 × f (or f = R ÷ 2)
  • Mirror Formula:
  • 1 ÷ v + 1 ÷ u = 1 ÷ f
  • Magnification by Mirror (m):
  • m = Height of image (h') ÷ Height of object (h) = -v ÷ u
  • Refractive Index (n₂₁):
  • n₂₁ = (Speed of light in medium 1, v₁) ÷ (Speed of light in medium 2, v₂)
  • Absolute refractive index: n = c ÷ v (where c = speed of light in air/vacuum = 3 × 10⁸ m/s)
  • Lens Formula:
  • 1 ÷ v - 1 ÷ u = 1 ÷ f
  • Magnification by Lens (m):
  • m = Height of image (h') ÷ Height of object (h) = v ÷ u
  • Power of a Lens (P):
  • P = 1 ÷ f (in metres)
  • SI Unit of power: Dioptre (D), where 1 D = 1 m⁻¹ (1 dioptre is the power of a lens of focal length 1 metre).

Important Sign Rules:

  • Focal length of Concave mirror / Concave lens = Negative (-)
  • Focal length of Convex mirror / Convex lens = Positive (+)
  • Magnification (m): If negative, image is real and inverted. If positive, image is virtual and erect.

4. Solved Examples (Board-Exam Style)

Example 1: Convex Mirror (Vehicle Rear-view)

Question: A convex mirror used for a rear-view on an automobile has a radius of curvature of 3.00 m. If a bus is located at 5.00 m from this mirror, find the position, nature, and size of the image.

Solution:

  • Radius of curvature, R = +3.00 m
  • Focal length, f = R ÷ 2 = +1.50 m
  • Object distance, u = -5.00 m
  • Mirror formula: 1 ÷ v + 1 ÷ u = 1 ÷ f
  • 1 ÷ v = 1 ÷ f - 1 ÷ u
  • 1 ÷ v = 1 ÷ 1.50 - 1 ÷ (-5.00) = 1 ÷ 1.50 + 1 ÷ 5.00
  • 1 ÷ v = (5.00 + 1.50) ÷ 7.50 = 6.50 ÷ 7.50
  • v = 7.50 ÷ 6.50 = +1.15 m
  • Magnification, m = -v ÷ u = -(1.15) ÷ (-5.00) = +0.23
  • Answer: The image is formed at 1.15 m behind the mirror. The image is virtual, erect, and diminished to 0.23 times the size of the bus.

Example 2: Concave Lens

Question: A concave lens has a focal length of 15 cm. At what distance should the object from the lens be placed so that it forms an image at 10 cm from the lens? Also, find the magnification.

Solution:

  • Concave lens forms a virtual image on the same side as the object, so v = -10 cm.
  • Focal length of concave lens, f = -15 cm.
  • Lens formula: 1 ÷ v - 1 ÷ u = 1 ÷ f
  • -1 ÷ u = 1 ÷ f - 1 ÷ v
  • 1 ÷ u = 1 ÷ v - 1 ÷ f
  • 1 ÷ u = 1 ÷ (-10) - 1 ÷ (-15) = -1 ÷ 10 + 1 ÷ 15
  • 1 ÷ u = (-3 + 2) ÷ 30 = -1 ÷ 30
  • u = -30 cm
  • Magnification, m = v ÷ u = (-10) ÷ (-30) = +1 ÷ 3 = +0.33
  • Answer: The object should be placed at a distance of 30 cm in front of the lens. The positive sign of m shows the image is virtual and erect, reduced to one-third of the object's size.

Example 3: Power of a Lens

Question: A doctor prescribes a corrective lens of power +2.0 D. Find the focal length of the lens. Is the prescribed lens diverging or converging?

Solution:

  • Power, P = +2.0 D
  • Formula: P = 1 ÷ f (in metres)
  • f = 1 ÷ P = 1 ÷ (+2.0) = +0.5 m = +50 cm
  • Since the power and focal length are positive, it is a converging lens (convex lens).
  • Answer: Focal length is +0.5 m (or +50 cm); the lens is converging.

Example 4: Conceptual Question

Question: Why do we prefer a convex mirror as a rear-view mirror in vehicles?

Answer:

  1. A convex mirror always gives an erect though diminished image of objects behind.
  2. Because it is curved outwards, it offers a much wider field of view than a plane mirror, allowing the driver to see a larger area of traffic behind.

5. Common Mistakes and Quick Revision Points

  • Sign of Object Distance (u): u is always taken as negative (-) for both mirrors and lenses because objects are placed to the left.
  • Formula Confusion:
  • Mirror formula has a plus sign: 1/v + 1/u = 1/f
  • Lens formula has a minus sign: 1/v - 1/u = 1/f
  • Magnification Sign Confusion:
  • Mirror: m = -v / u (notice the minus sign!)
  • Lens: m = +v / u (no minus sign!)
  • Power of Lens Unit Error: Always convert focal length f into metres (m) before calculating P = 1 ÷ f. If f is in cm, use P = 100 ÷ f (in cm).
  • Image Nature Clues:
  • Real images are always inverted (below the axis, m is negative).
  • Virtual images are always erect (above the axis, m is positive).

Practice MCQ (15)

  1. एक गोलीय दर्पण की वक्रता त्रिज्या 32 cm है। इसकी फोकस दूरी क्या है?

    • (A) 16 cm
    • (B) 32 cm
    • (C) 64 cm
    • (D) 8 cm

    उत्तर: (A)

  2. वाहनों में पश्च-दृश्य (रियर-व्यू) दर्पण के रूप में किस प्रकार के दर्पण को प्राथमिकता दी जाती है क्योंकि यह हमेशा सीधा प्रतिबिंब देता है और दृष्टि-क्षेत्र विस्तृत प्रदान करता है?

    • (A) Concave mirror
    • (B) Plane mirror
    • (C) Convex mirror
    • (D) Cylindrical mirror

    उत्तर: (C)

  3. लेंस की क्षमता (पावर) का SI मात्रक क्या है?

    • (A) Metre
    • (B) Watt
    • (C) Dioptre
    • (D) Joule

    उत्तर: (C)

  4. जल का निरपेक्ष अपवर्तनांक 4 ÷ 3 है। यदि निर्वात में प्रकाश की चाल 3 × 10⁸ m/s है, तो जल में प्रकाश की चाल क्या होगी?

    • (A) 2.25 × 10⁸ m/s
    • (B) 4.0 × 10⁸ m/s
    • (C) 1.5 × 10⁸ m/s
    • (D) 2.0 × 10⁸ m/s

    उत्तर: (A)

  5. अवतल दर्पण के वक्रता केंद्र से होकर गुजरने वाली प्रकाश की किरण परावर्तन के बाद अपने ही मार्ग पर वापस लौट जाती है क्योंकि:

    • (A) The ray falls normally on the reflecting surface, so the angle of incidence is 0°
    • (B) The ray is parallel to the principal axis
    • (C) The mirror absorbs light passing through its centre
    • (D) The angle of reflection is 90°

    उत्तर: (A)

  6. एक गोलीय दर्पण द्वारा उत्पन्न रेखीय आवर्धन −2 है। प्रतिबिंब के बारे में क्या निष्कर्ष निकाला जा सकता है?

    • (A) The image is real, inverted, and diminished
    • (B) The image is real, inverted, and magnified
    • (C) The image is virtual, erect, and magnified
    • (D) The image is virtual, erect, and diminished

    उत्तर: (B)

  7. एक वस्तु 15 cm फोकस दूरी वाले अवतल दर्पण के सामने 20 cm की दूरी पर रखी गई है। दर्पण से कितनी दूरी पर प्रतिबिंब बनेगा?

    • (A) −60 cm
    • (B) +60 cm
    • (C) −30 cm
    • (D) +30 cm

    उत्तर: (A)

  8. एक उत्तल लेंस किसी सुई का वास्तविक तथा उल्टा प्रतिबिंब उससे 50 cm की दूरी पर बनाता है। यदि प्रतिबिंब का आकार सुई के आकार के बराबर है, तो सुई कहाँ रखी गई है और लेंस की फोकस दूरी क्या है?

    • (A) Object distance = −50 cm, focal length = +25 cm
    • (B) Object distance = −25 cm, focal length = +50 cm
    • (C) Object distance = −50 cm, focal length = +50 cm
    • (D) Object distance = −100 cm, focal length = +25 cm

    उत्तर: (A)

  9. वायु के सापेक्ष जल और कांच के अपवर्तनांक क्रमशः 4 ÷ 3 और 3 ÷ 2 हैं। जल के सापेक्ष कांच का अपवर्तनांक क्या है?

    • (A) 8 ÷ 9
    • (B) 9 ÷ 8
    • (C) 1 ÷ 2
    • (D) 2

    उत्तर: (B)

  10. एक नेत्र रोग विशेषज्ञ संपर्क में रखे दो पतले लेंसों के संयोजन का सुझाव देता है जिनकी क्षमताएं +2.5 D और −1.5 D हैं। इस संयोजन की फोकस दूरी और प्रकाशिक व्यवहार क्या है?

    • (A) +100 cm, converging
    • (B) −100 cm, diverging
    • (C) +50 cm, converging
    • (D) −50 cm, diverging

    उत्तर: (A)

  11. जब कोई प्रकाश किरण वायु से एक आयताकार कांच की सिल्ली (स्लैब) में तिरछी प्रवेश करती है और पुनः वायु में निर्गत होती है, तो निर्गत किरण:

    • (A) Bent towards the normal and parallel to the incident ray
    • (B) Parallel to the incident ray but laterally displaced
    • (C) Perpendicular to the incident ray
    • (D) Along the same straight line without any shift

    उत्तर: (B)

  12. एक प्रकाशिक युक्ति 'A' अपने सामने रखी वस्तु की सभी वास्तविक स्थितियों के लिए हमेशा एक आभासी, सीधा और छोटा प्रतिबिंब बनाती है। युक्ति 'A' क्या हो सकती है?

    • (A) Only a convex mirror
    • (B) Only a concave lens
    • (C) Either a convex mirror or a concave lens
    • (D) Either a concave mirror or a convex lens

    उत्तर: (C)

  13. एक अवतल लेंस की फोकस दूरी 15 cm है। किसी वस्तु को लेंस से कितनी दूरी पर रखा जाना चाहिए ताकि उसका प्रतिबिंब लेंस से 10 cm की दूरी पर बने, और उत्पन्न आवर्धन क्या होगा?

    • (A) u = −30 cm, m = +1 ÷ 3
    • (B) u = −30 cm, m = −1 ÷ 3
    • (C) u = −20 cm, m = +1 ÷ 2
    • (D) u = −60 cm, m = +1 ÷ 6

    उत्तर: (A)

  14. प्रकाश तीन पारदर्शी माध्यमों A, B और C से क्रमशः 1.8 × 10⁸ m/s, 2.4 × 10⁸ m/s और 2.0 × 10⁸ m/s की चाल से गुजरता है। यदि प्रकाश की एक किरण माध्यम B से माध्यम A में प्रवेश करती है, और एक अन्य समान किरण समान आपतन कोण पर माध्यम B से माध्यम C में प्रवेश करती है:

    • (A) The beam bends more towards the normal in medium A than in medium C
    • (B) The beam bends away from the normal in medium A and towards the normal in medium C
    • (C) The beam bends more towards the normal in medium C than in medium A
    • (D) The beam bends away from the normal in both media

    उत्तर: (A)

  15. 4.0 cm ऊँचाई की एक वस्तु 30 cm वक्रता त्रिज्या वाले एक उत्तल दर्पण के सामने 20 cm की दूरी पर रखी है। प्रतिबिंब की स्थिति और ऊँचाई क्या है?

    • (A) v = +8.57 cm, height = +1.71 cm
    • (B) v = −8.57 cm, height = −1.71 cm
    • (C) v = +12.0 cm, height = +2.40 cm
    • (D) v = −12.0 cm, height = +1.71 cm

    उत्तर: (A)