Light Reflection and Refraction for CTET Paper II (Class 6-8 Science)

intermediate 22 min read

Concept

Light behaves like a very disciplined traveler — it always takes the fastest path, and when it hits a boundary between two materials, it makes a decision: bounce back (reflection) or cross over and bend (refraction).

Think of reflection like a ball bouncing off a wall at the same angle it came in. The wall doesn't absorb it, doesn't bend it — just redirects it. Refraction is more like a soldier marching in formation across a boundary where the terrain changes: the soldier on the slow side lags behind, and the whole column pivots. That pivoting is bending of light.

In a classroom (Class 6-8), you demonstrate reflection with a plane mirror showing a student their image, and you show refraction with a pencil appearing broken in a glass of water. Both are everyday encounters with the same physics.

Reflection follows one clean rule: the angle of incidence (measured from the normal, not the surface) equals the angle of reflection. The incident ray, reflected ray, and normal all lie in the same plane. Always. No exceptions.

Refraction follows Snell's Law: n₁ sin θ₁ = n₂ sin θ₂. When light moves from a rarer medium (air, n ≈ 1) to a denser medium (glass, n ≈ 1.5), it slows down and bends toward the normal. When it exits back into air, it speeds up and bends away from the normal. This is why a straw in water looks bent — you're seeing the air-water refraction in action.

The refractive index n of a medium is the ratio of the speed of light in vacuum to its speed in that medium: n = c/v. It's always ≥ 1 for any real medium. Glass (n ≈ 1.5), water (n ≈ 1.33), diamond (n ≈ 2.42) — higher n means the medium bends light more sharply.

For CTET, you're expected to know the concepts well enough to teach them (so the underlying logic matters), and also to calculate image positions, heights, and refractive indices — because numerical questions appear in the paper.


Deep Dive

Laws of Reflection — The Foundation

Both the laws of reflection are testable at the conceptual level:

  1. Angle of incidence = Angle of reflection (∠i = ∠r), both measured from the normal.
  2. The incident ray, the reflected ray, and the normal at the point of incidence — all lie in the same plane.

For a plane mirror: the image is virtual, erect, laterally inverted, and as far behind the mirror as the object is in front. No formula needed — just the geometry.

Minimum mirror length to see your full body: exactly H/2, where H is your height. The proof uses the midpoint principle — rays from the top of your head and your feet both hit the mirror at the midpoint between those extremes and your eyes. This is a classic classroom question.

Mirror Formula and Sign Convention

Use the New Cartesian Sign Convention: object is always to the left, distances measured from the pole. Distances in the direction of incident light are positive; against it are negative. For a concave mirror, f is negative; for a convex mirror, f is positive.

Mirror formula: 1/v + 1/u = 1/f

Focal length of a mirror: f = R/2 (where R is radius of curvature).

Magnification: m = -v/u

Concave mirror (converging): Used in torches, headlights, shaving mirrors. Can form real and virtual images depending on object position.

Convex mirror (diverging): Always forms virtual, erect, diminished image. This is why it's used as a rear-view mirror — gives a wider field of view, but image is always smaller than actual.

Refraction and Snell's Law

n₁ sin θ₁ = n₂ sin θ₂

Or equivalently, for light going from air (n=1) into a medium: n = sin i / sin r.

The refractive index of medium 2 with respect to medium 1: ₁n₂ = n₂/n₁

This gives you the relation: n_glass(wrt water) = n_glass(wrt air) / n_water(wrt air).

So if you know any two of the three, you can find the third. This appears in PYQs directly.

Lens Formula and Magnification

For lenses, the sign convention is similar but with a key difference — distances are measured from the optical centre.

Lens formula: 1/v - 1/u = 1/f

Note the subtraction (not addition like in mirrors).

Magnification for lens: m = v/u (no negative sign here)

Lens power: P = 1/f (in metres), unit is dioptre (D). Convex lenses have positive power, concave lenses have negative power. Combined power of lenses in contact: P = P₁ + P₂.

Convex lens image positions — know these for the exam:

| Object position | Image position | Image nature | |---|---|---| | Beyond 2F | Between F and 2F | Real, inverted, diminished | | At 2F | At 2F | Real, inverted, same size | | Between F and 2F | Beyond 2F | Real, inverted, magnified | | At F | At infinity | — | | Between F and lens | Same side as object | Virtual, erect, magnified |

Concave lens: Always forms virtual, erect, diminished image — regardless of object position. Same property as convex mirror.

Total Internal Reflection (TIR)

When light travels from a denser to a rarer medium (e.g., glass to air), at angles beyond a threshold called the critical angle (θ_c), no refraction occurs — all light is reflected back into the denser medium.

Condition for TIR:

  1. Light must go from denser → rarer medium.
  2. Angle of incidence must exceed the critical angle.

sin θ_c = n_rarer / n_denser = 1/n (for glass-air interface)

For glass (n = 1.5): sin θ_c = 1/1.5 = 2/3, so θ_c ≈ 42°.

Applications: optical fibers (internet cables, endoscopes), diamonds cut to maximize TIR for sparkle, mirage formation.


Memory Tricks & Shortcuts

patternMirror vs Lens Formula — The Subtraction Signal

Mirrors use addition: 1/v + 1/u = 1/f. Lenses use subtraction: 1/v - 1/u = 1/f. Remember: Mirror = Meet (plus), Lens = Leave (minus). The "L" in Lens matches "Less" (subtract). Standard method: students confuse the two and waste 60–90 seconds rechecking. This mnemonic locks the correct form in under 5 seconds, zero rechecking needed.

substitutionRefractive Index Chain — Divide Through the Common Medium

When you see "n of A with respect to B = X" and "n of A with respect to C = Y", find "n of B with respect to C" using: n_B(wrt C) = n_A(wrt C) / n_A(wrt B). Just cancel A from numerator-denominator style. Example from PYQ: n_water(wrt air) = n_glass(wrt air) / n_glass(wrt water) = 1.5 ÷ (9/8) = 1.5 × 8/9 = 4/3. Standard method (going through definitions): 3–4 steps. This direct ratio: 2 steps, ~20 seconds saved.

patternf = R/2 for Mirrors — Only Mirrors, Never Lenses

Students sometimes apply f = R/2 to lenses. Lenses don't have a single center of curvature like mirrors — they have two curved surfaces. The f = R/2 rule is exclusively for mirrors. Stamp this: Mirror + R/2 = focal length. Lens + focal length = given directly. Catching this distinction eliminates one of the top-3 wrong-answer traps in optics questions. Saves re-reading time of ~30 seconds per question.

patternConvex Mirror and Concave Lens — The Twin Divergers

Both convex mirrors and concave lenses always produce virtual, erect, diminished images — for any object position. No calculation needed for these image-nature questions. If an MCQ asks for the nature of image in a convex mirror or concave lens, mark "virtual, erect, diminished" immediately. Standard method: students trace rays or apply formula. This pattern recognition: under 5 seconds versus 60+ seconds.

eliminationCritical Angle Check — One Inequality, One Decision

For Total Internal Reflection questions, you only need two checks: (1) Is light going from denser to rarer? (2) Is angle of incidence > critical angle? If both YES → TIR. If either NO → refraction occurs (partial or full). Don't engage with Snell's law calculations. Example: glass-air, critical angle 42°, incidence 50° — both conditions met, TIR. Decision in under 10 seconds versus setting up and solving Snell's equation (~45 seconds).


Fast-Solving Framework

Here's how to approach any optics question in the exam hall:

Step 1 — Identify the device. Mirror or lens? Concave or convex? This takes 3 seconds and sets every formula and sign.

Step 2 — Is it a "nature of image" question? If convex mirror or concave lens → mark "virtual, erect, diminished" and move on. If concave mirror or convex lens → you need object position relative to F or 2F to read off the table.

Step 3 — Is it a numerical? Assign signs first (Cartesian convention), then plug into the correct formula. Mirror: 1/v + 1/u = 1/f. Lens: 1/v - 1/u = 1/f. Don't calculate until signs are assigned.

Step 4 — Refractive index chain? Write out the ratio equation directly: n_A(wrt C) = n_A(wrt B) × n_B(wrt C). No re-deriving Snell's law from scratch.

Step 5 — TIR question? Two conditions only (dense → rare; angle > critical). Done.

If the question involves f = R/2, remember that's mirrors only. If it gives focal length in cm and asks for power, convert to metres first.


Solved PYQs

Why this question: Tests the refractive index chain relationship — a direct formula-application question that trips students who re-derive instead of using the ratio shortcut.

Previous Year Questionपिछले वर्ष का प्रश्न
The refractive index of glass with respect to water is 9/8. If the refractive index of glass with respect to air is 1.5, what is the refractive index of water with respect to air?
जल के सापेक्ष कांच का अपवर्तनांक 9/8 है। यदि वायु के सापेक्ष कांच का अपवर्तनांक 1.5 है, तो वायु के सापेक्ष जल का अपवर्तनांक क्या होगा?
  1. 1.125
  2. 1.333
  3. 0.750
  4. 1.688
  1. 1.125
  2. 1.333
  3. 0.750
  4. 1.688
Solutionसमाधान
n_glass/n_water = 9/8, so n_water = n_glass × 8/9 = 1.5 × 8/9 = 12/9 = 4/3 ≈ 1.333. The refractive index of water with respect to air is 4/3.
n_glass/n_water = 9/8 ⟹ n_water = 1.5 × 8/9 = 12/9 = 4/3 ≈ 1.333. वायु के सापेक्ष जल का अपवर्तनांक 4/3 होता है।

Solving path: You know n_glass(wrt water) = 9/8 and n_glass(wrt air) = 1.5. You want n_water(wrt air). Use the chain: n_water(wrt air) = n_glass(wrt air) / n_glass(wrt water) = 1.5 ÷ (9/8) = 1.5 × 8/9 = 12/9 = 4/3 ≈ 1.333. Answer: (B).


Why this question: Tests lens formula + magnification together — a two-step numerical that rewards students who know both formulas cold. The sign of magnification (negative → inverted) is an additional trap.

Previous Year Questionपिछले वर्ष का प्रश्न
An object 4 cm tall is placed at 30 cm from a convex lens of focal length 20 cm. What is the height of the image formed?
4 cm ऊँची एक वस्तु को 20 cm फोकस दूरी वाले उत्तल लेंस से 30 cm दूर रखा गया है। बनने वाले प्रतिबिम्ब की ऊँचाई क्या होगी?
  1. 4 cm
  2. 8 cm
  3. 12 cm
  4. 6 cm
  1. 4 cm
  2. 8 cm
  3. 12 cm
  4. 6 cm
Solutionसमाधान
Using lens formula: 1/v − 1/u = 1/f. u = −30 cm, f = +20 cm. 1/v = 1/20 + 1/(−30) = 1/20 − 1/30 = 1/60. So v = 60 cm. Magnification m = v/u = 60/(−30) = −2. Height of image = |m| × object height = 2 × 4 = 8 cm (real and inverted).
लेंस सूत्र: 1/v − 1/u = 1/f. u = −30 cm, f = +20 cm. 1/v = 1/20 − 1/30 = 1/60, अतः v = 60 cm. आवर्धन m = v/u = 60/(−30) = −2. प्रतिबिम्ब की ऊँचाई = 2 × 4 = 8 cm (वास्तविक और उल्टा)।

Solving path: Object at u = -30 cm, f = +20 cm (convex lens). Lens formula: 1/v = 1/f + 1/u = 1/20 + 1/(-30) = 3/60 - 2/60 = 1/60. So v = 60 cm. Magnification m = v/u = 60/(-30) = -2. Image height = |m| × 4 = 8 cm. Answer: (B).


Why this question: A pure concept question — but only if you've locked in the convex mirror rule. Students who second-guess themselves and try to remember ray diagrams waste 90 seconds here.

Previous Year Questionपिछले वर्ष का प्रश्न
Which of the following correctly describes the image formed by a convex mirror irrespective of the position of the object?
निम्नलिखित में से कौन सा वाक्य उत्तल दर्पण द्वारा बनाए गए प्रतिबिम्ब का सही वर्णन करता है, चाहे वस्तु कहीं भी रखी हो?
  1. Real, inverted and diminished
  2. Real, erect and magnified
  3. Virtual, inverted and diminished
  4. Virtual, erect and diminished
  1. वास्तविक, उल्टा और छोटा
  2. वास्तविक, सीधा और आवर्धित
  3. आभासी, उल्टा और छोटा
  4. आभासी, सीधा और छोटा
Solutionसमाधान
A convex mirror always forms a virtual, erect and diminished image for any position of the object in front of it. This is because reflected rays always diverge and appear to meet behind the mirror. This property makes convex mirrors ideal for rear-view mirrors in vehicles.
उत्तल दर्पण हमेशा, वस्तु की किसी भी स्थिति के लिए, आभासी, सीधा और छोटा प्रतिबिम्ब बनाता है। परावर्तित किरणें हमेशा अपसारी होती हैं और दर्पण के पीछे मिलती प्रतीत होती हैं। इसीलिए वाहनों में पीछे देखने वाले दर्पण के रूप में उत्तल दर्पण का उपयोग होता है।

Solving path: No calculation needed. Convex mirror = always virtual, erect, diminished. Answer: (D). The reasoning is that reflected rays from a convex mirror always diverge — they never actually meet, so the image is always virtual, behind the mirror, and smaller than the object.


Why this question: Tests TIR conditions — angle of incidence versus critical angle. The distractor options are designed to lure students into thinking partial reflection occurs at 50°.

Previous Year Questionपिछले वर्ष का प्रश्न
A student claims that critical angle for glass-air interface is 42°. If a ray of light hits the interface from within glass at an angle of incidence of 50°, what will happen?
एक छात्र का दावा है कि कांच-वायु सीमा के लिए क्रांतिक कोण 42° है। यदि कांच के भीतर से कोई किरण 50° के आपतन कोण पर इस सीमा से टकराए, तो क्या होगा?
  1. Total internal reflection will occur
  2. Partial refraction and partial reflection will occur with the refracted ray bending toward the normal
  3. The ray will refract into air at an angle less than 90°
  4. The ray will pass straight through without bending
  1. पूर्ण आंतरिक परावर्तन होगा
  2. आंशिक अपवर्तन और आंशिक परावर्तन होगा तथा अपवर्तित किरण अभिलंब की ओर झुकेगी
  3. किरण 90° से कम कोण पर वायु में अपवर्तित हो जाएगी
  4. किरण बिना मुड़े सीधे निकल जाएगी
Solutionसमाधान
Total internal reflection occurs when light travels from a denser to a rarer medium and the angle of incidence exceeds the critical angle. Here, 50° > 42° (critical angle), so total internal reflection takes place. The ray does not enter air at all and is completely reflected back into glass.
पूर्ण आंतरिक परावर्तन तब होता है जब प्रकाश सघन से विरल माध्यम में जाता है और आपतन कोण क्रांतिक कोण से अधिक हो जाता है। यहाँ 50° > 42° (क्रांतिक कोण), इसलिए पूर्ण आंतरिक परावर्तन होगा। किरण वायु में बिल्कुल प्रवेश नहीं करती और पूरी तरह वापस कांच में परावर्तित हो जाती है।

Solving path: Light is going from glass (denser) to air (rarer) — condition 1 met. Angle of incidence 50° > critical angle 42° — condition 2 met. Therefore: Total Internal Reflection. No light enters air. Answer: (A). No Snell's law calculation required — just the two-condition check.


Why this question: A concave lens numerical — the sign convention trap is that both f and v are negative for concave lenses (virtual image, same side as object). Students who get v sign wrong get an absurd answer.

Previous Year Questionपिछले वर्ष का प्रश्न
A concave lens of focal length 25 cm forms an image at a distance of 20 cm from the lens. Where is the object placed?
25 cm फोकस दूरी वाले अवतल लेंस द्वारा लेंस से 20 cm दूर प्रतिबिम्ब बनता है। वस्तु कहाँ रखी है?
  1. 500 cm from the lens
  2. 45 cm from the lens
  3. 100 cm from the lens
  4. 50 cm from the lens
  1. लेंस से 500 cm दूर
  2. लेंस से 45 cm दूर
  3. लेंस से 100 cm दूर
  4. लेंस से 50 cm दूर
Solutionसमाधान
For a concave lens, f = −25 cm. The image is always virtual and on the same side as the object, so v = −20 cm. Using lens formula: 1/u = 1/v − 1/f = 1/(−20) − 1/(−25) = −1/20 + 1/25 = (−5 + 4)/100 = −1/100. So u = −100 cm, meaning the object is 100 cm in front of the lens.
अवतल लेंस के लिए f = −25 cm. प्रतिबिम्ब हमेशा आभासी और वस्तु की ही ओर बनता है, अतः v = −20 cm. लेंस सूत्र से: 1/u = 1/(−20) − 1/(−25) = −1/20 + 1/25 = −1/100. अतः u = −100 cm, यानी वस्तु लेंस से 100 cm सामने रखी है।

Solving path: Concave lens: f = -25 cm. Image is virtual, same side as object: v = -20 cm. Lens formula: 1/u = 1/v - 1/f = 1/(-20) - 1/(-25) = -1/20 + 1/25 = (-5 + 4)/100 = -1/100. So u = -100 cm. Object is 100 cm in front of the lens. Answer: (C).


Why this question: A straightforward focal length question, but students sometimes confuse f = R with f = R/2. The "distant object → image at focus" rule is also separately testable.

Previous Year Questionपिछले वर्ष का प्रश्न
Light from a distant object falls on a concave mirror of radius of curvature 40 cm. At what distance from the mirror will the image be formed?
एक दूरस्थ वस्तु से आने वाला प्रकाश 40 cm वक्रता त्रिज्या वाले अवतल दर्पण पर पड़ता है। दर्पण से कितनी दूरी पर प्रतिबिम्ब बनेगा?
  1. 40 cm in front of the mirror
  2. 20 cm in front of the mirror
  3. 10 cm in front of the mirror
  4. 20 cm behind the mirror
  1. दर्पण के सामने 40 cm
  2. दर्पण के सामने 20 cm
  3. दर्पण के सामने 10 cm
  4. दर्पण के पीछे 20 cm
Solutionसमाधान
The focal length of a concave mirror = R/2 = 40/2 = 20 cm. For a distant object, rays are parallel (object at infinity), and the image is formed at the focus. Thus the image forms 20 cm in front of the mirror at the focal point. The image is real, inverted and point-sized.
अवतल दर्पण की फोकस दूरी = R/2 = 40/2 = 20 cm. दूरस्थ वस्तु के लिए किरणें समांतर होती हैं (वस्तु अनंत पर), और प्रतिबिम्ब फोकस पर बनता है। अतः प्रतिबिम्ब दर्पण के सामने 20 cm पर बनता है। यह प्रतिबिम्ब वास्तविक, उल्टा और बिंदु आकार का होता है।

Solving path: f = R/2 = 40/2 = 20 cm. Distant object means parallel rays → image forms at focus. For a concave mirror, the focus is in front of the mirror. Image is 20 cm in front. Answer: (B).


Why this question: A direct Snell's law calculation — tests whether you know n = sin i / sin r for air-to-medium refraction, and whether you know sin 60° and sin 30° cold.

Previous Year Questionपिछले वर्ष का प्रश्न
A ray of light enters a rectangular glass slab at an angle of incidence of 60°. The refracted ray inside the slab makes an angle of 30° with the normal. What is the refractive index of the glass?
प्रकाश की एक किरण 60° के आपतन कोण पर आयताकार कांच की पट्टिका में प्रवेश करती है। पट्टिका के अंदर अपवर्तित किरण अभिलंब से 30° का कोण बनाती है। कांच का अपवर्तनांक क्या है?
  1. 2.0
  2. √2
  3. √3
  4. 1.5
  1. 2.0
  2. √2
  3. √3
  4. 1.5
Solutionसमाधान
By Snell's law: n = sin i / sin r = sin 60° / sin 30° = (√3/2) / (1/2) = √3 ≈ 1.732. The refractive index of the glass is √3.
स्नेल के नियम से: n = sin i / sin r = sin 60° / sin 30° = (√3/2) / (1/2) = √3 ≈ 1.732. कांच का अपवर्तनांक √3 है।

Solving path: n = sin i / sin r = sin 60° / sin 30° = (√3/2) / (1/2) = √3. Answer: (C). You need sin 60° = √3/2 and sin 30° = 1/2 as instant recall — not lookup values.


Why this question: A pedagogical question disguised as optics — tests whether you understand the geometric proof well enough to teach it. The answer H/2 surprises students who haven't thought through the midpoint geometry.

Previous Year Questionपिछले वर्ष का प्रश्न
A teacher asks students to find the minimum length of a plane mirror required for a person of height 'H' to see their full image. The correct answer is:
एक शिक्षक छात्रों से पूछते हैं कि 'H' ऊँचाई के व्यक्ति को अपना पूरा प्रतिबिम्ब देखने के लिए समतल दर्पण की न्यूनतम लंबाई कितनी होनी चाहिए? सही उत्तर है:
  1. H
  2. 2H
  3. H/2
  4. H/4
  1. H
  2. 2H
  3. H/2
  4. H/4
Solutionसमाधान
Due to the law of reflection, the mirror only needs to be half the height of the person. The top of the mirror should be at the midpoint between the person's eyes and the top of their head, and the bottom at the midpoint between the eyes and the feet. Rays from both extremes meet the mirror at these midpoints before reaching the eyes. So minimum mirror length = H/2.
परावर्तन के नियम के कारण, दर्पण की लंबाई व्यक्ति की ऊँचाई की केवल आधी होनी चाहिए। दर्पण का ऊपरी सिरा आँखों और सिर के शीर्ष के मध्य बिंदु पर, तथा निचला सिरा आँखों और पैरों के मध्य बिंदु पर होना चाहिए। अतः न्यूनतम दर्पण की लंबाई = H/2।

Solving path: The top of the mirror only needs to be at the height halfway between your eyes and the top of your head (the reflected ray from the top of your head travels half the vertical distance to reach your eyes). Similarly, the bottom of the mirror sits at the midpoint between your eyes and your feet. Total mirror height needed = half of total body height = H/2. Answer: (C).


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