Chapter 1 · Solved paper40 marks

Model Test – Solved ഉത്തരങ്ങൾ

Every question of the model paper, in the same order, with the full answer, the marking scheme, an animated ray diagram or working for every diagram and numerical question, and a link to the page where the idea is taught. Try each one again before you open the answer, and mark your own paper honestly.

📐 How the ray diagrams are drawn

Rays are red with arrows; the normal (ലംബം) is a dashed line drawn at the point of incidence; every angle is measured from the normal. Media are coloured as in the textbook: air grey, water blue, glass green. The angles in the animations are worked out with the real law of refraction, so they are drawn to scale.

SECTION AAnswer all · 8 × 1 = 8 marks
1.
The ratio of the speed of light in vacuum to the speed of light in a medium is called the of that medium.
(1 mark)

💡 n = c/v – what is n called?

✅ Answer: refractive index (അപവർത്തനാങ്കം)

Refractive index (അപവർത്തനാങ്കം) n = c/v, where c = 3 × 108 m/s is the speed of light in vacuum (ശൂന്യത) and v its speed in the medium.

Marking: 1 mark.

📖 Review this on Page 6 · Refractive index

2.
Which of these media has the highest optical density?
(a) air(b) water(c) glass(d) diamond
(1 mark)

💡 In which medium is light slowest?

✅ Answer: (d) diamond

Light is slowest in diamond (1.25 × 108 m/s in Table 1.1), so diamond is the optically densest (പ്രകാശികസാന്ദ്രത കൂടിയ മാധ്യമം). Order: air < water < glass < diamond.

Marking: 1 mark for (d).

📖 Review this on Page 5 · Table 1.1 and optical density

3.
The critical angle of glass is 42°. Inside a glass block, total internal reflection takes place when the angle of incidence is
(a) 30°(b) 40°(c) 42°(d) 50°
(1 mark)

💡 It must be greater than – not equal to – the critical angle.

✅ Answer: (d) 50°

Total internal reflection (പൂർണ്ണാന്തര പ്രതിപതനം) needs an angle of incidence greater than the critical angle. Only 50° is greater than 42°. At exactly 42° the ray grazes along the surface (r = 90°).

Marking: 1 mark for (d).

📖 Review this on Page 10 · Critical angle · Page 14 · Let us assess Q7

4.
A ray of light falls perpendicularly on the surface of a glass slab. Its angle of refraction is .
(1 mark)

💡 Angles are measured from the normal. Where is the ray?

✅ Answer: 0° (no change of direction)

A ray falling along the normal (ലംബമായി പതിക്കുന്ന) has i = 0°; it goes straight on, r = 0° – it is not deviated.

Marking: 1 mark.

📖 Review this on Page 7 · Rule (iv)

5.
Find the relation and fill in:   air → water : bends towards the normal  ::  water → air :
(1 mark)

💡 Is water → air rarer → denser or denser → rarer?

✅ Answer: bends away from the normal

Rarer → denser: towards the normal; denser → rarer: away from the normal (ലംബത്തിൽ നിന്ന് അകലുന്നു).

Marking: 1 mark.

📖 Review this on Page 7 · Table 1.3

6.
The twinkling of stars is due to .
(1 mark)

💡 Light from a star passes through many layers of air.

✅ Answer: atmospheric refraction (അന്തരീക്ഷത്തിലെ അപവർത്തനം)

The layers of the atmosphere keep changing in temperature and pressure, so their optical density keeps changing. Starlight is refracted irregularly, and the apparent position and brightness of the star keep changing – it twinkles.

Marking: 1 mark (“refraction” alone: ½ mark).

📖 Review this on Page 9 · Twinkling stars

7.
The unit of refractive index is
(a) m/s(b) s/m(c) metre(d) it has no unit
(1 mark)

💡 What is (m/s) ÷ (m/s)?

✅ Answer: (d) no unit

n = c/v is speed ÷ speed: the units m/s cancel. Refractive index is a pure number (glass 1.5).

Marking: 1 mark for (d).

📖 Review this on Page 6 · Key box 4 · Page 3 · Ratios have no unit

8.
The 2009 Nobel Prize in Physics for achievements in the transmission of light through optical fibres went to .
(1 mark)

💡 The textbook side box on p. 20.

✅ Answer: Charles K. Kao

Charles K. Kao showed that very pure glass fibres (ഒപ്റ്റിക്കൽ ഫൈബർ) could carry light signals over long distances – the basis of today’s internet cables.

Marking: 1 mark.

📖 Review this on Page 11 · Optical fibres

SECTION BAnswer all · 5 × 2 = 10 marks
9.
Light travels through glycerine at a speed of 2.04 × 108 m/s. Find the refractive index of glycerine. (Speed of light in vacuum c = 3 × 108 m/s)
(2 marks)

💡 Divide c by v; the powers of ten cancel.

🎬 Step by step
✅ Answer: 1.47

Given: v = 2.04 × 108 m/s; c = 3 × 108 m/s

Formula: n = c/v

Substitution: n = (3 × 108) ÷ (2.04 × 108) = 3 ÷ 2.04

Answer: n ≈ 1.47 (no unit)

The 108 cancels; the answer matches glycerine in Table 1.5.

Marking: ½ for the formula, ½ for substitution, 1 for the answer (writing a unit: lose ½).

📖 Review this on Page 6 · n = c/v · Page 3 · Dividing powers of ten

10.
A ray of light passes obliquely from air into water (figure). (a) Complete the diagram showing the refracted ray and mark the angle of incidence i and the angle of refraction r. (b) Which is greater, i or r?
(2 marks)

💡 Rarer → denser: towards or away from the normal?

🎬 Step by step
✅ Answer: (a) bends towards the normal   (b) i > r

(a) Draw the normal at the point of incidence. Air → water is rarer → denser, so the refracted ray in the water bends towards the normal. Mark i between the incident ray and the normal (in air) and r between the refracted ray and the normal (in water).

(b) i is greater than r (for i = 45°, r is about 32°).

Marking: 1 mark for the correct refracted ray with the normal, ½ for marking i and r from the normal, ½ for (b).

📖 Review this on Page 7 · Towards the normal · Page 2 · Measure from the normal

11.
Write the two conditions necessary for total internal reflection.
(2 marks)

💡 Which way must the light travel, and how big must the angle be?

🎬 Step by step
✅ Answer: denser → rarer; i > critical angle

(1) Light must travel from an optically denser medium to an optically rarer medium (for example glass → air).

(2) The angle of incidence must be greater than the critical angle (ക്രിട്ടിക്കൽ കോൺ) of the pair of media (42° for glass–air).

Marking: 1 mark for each condition.

📖 Review this on Page 10 · Key box 6, Fig 1.18

12.
A boy fishing with a bow and arrow in a paddy-field canal aims a little below the place where he sees the fish. Why? Explain with the help of the figure.
(2 marks)

💡 Where does the light from the fish bend, and which way?

🎬 Step by step
✅ Answer: the fish appears raised

Light from the fish passes from water (denser) into air (rarer) and bends away from the normal at the surface. The boy’s eye traces the ray back in a straight line, so he sees the fish at a raised position, higher than where it really is. So he must aim below the image he sees.

Marking: 1 mark for “bends away from the normal at the water surface”, 1 mark for “the fish appears higher than its real position”.

📖 Review this on Page 8 · Coin, pond and fishing

13.
Write two differences between reflection from a plane mirror and total internal reflection.
(2 marks)

💡 Think of Table 1.4 in the textbook.

✅ Answer: any two differences
Reflection from a plane mirror സമതല ദർപ്പണംTotal internal reflection പൂർണ്ണാന്തര പ്രതിപതനം
Light is not totally reflected – some is absorbedLight is totally reflected
Happens at any angle of incidenceHappens only when the angle of incidence is greater than the critical angle
Light may come from any medium onto the mirrorHappens only when light goes from a denser to a rarer medium

Marking: 1 mark for each correct difference (Table 1.4).

📖 Review this on Page 11 · Table 1.4

SECTION CAnswer all · 4 × 3 = 12 marks
14.
The speed of light in three media P, Q and R is given below.
MediumPQR
Speed of light (m/s)2.25 × 1081.25 × 1082 × 108
(a) Find the refractive index of Q (c = 3 × 108 m/s). (b) Which medium is optically the densest? (c) A ray passes obliquely from R into P. Does it bend towards or away from the normal? Why?
(3 marks)

💡 Lower speed = higher refractive index = optically denser.

🎬 Step by step
✅ Answer: (a) 2.4   (b) Q   (c) away from the normal

Given: v(Q) = 1.25 × 108 m/s; c = 3 × 108 m/s

Formula: n = c/v

Substitution: (a) n = 3 ÷ 1.25

Answer: (a) n(Q) = 2.4

(b) Light is slowest in Q, so Q is optically the densest (it is like diamond).

(c) n(R) = 3/2 = 1.5 and n(P) = 3/2.25 ≈ 1.33. Light is faster in P, so R → P is denser → rarer: the ray bends away from the normal.

Marking: 1 mark for each part; in (c) ½ for the answer, ½ for the reason.

📖 Review this on Page 6 · Table 1.2 · Page 5 · Optical density · Page 7 · The rules

15.
A ray of light inside a semicircular glass block is aimed at the centre of its flat face at three different angles of incidence: (i) 35° (ii) 42° (iii) 50°. In (i) the angle of refraction in air is 60°. (a) What happens to the ray in (ii)? What is this angle called? (b) What happens in (iii)? Name the phenomenon. (c) Why does the ray in (i) bend away from the normal?
(3 marks)

💡 Compare each angle with 42°.

🎬 Step by step
✅ Answer: (a) grazes along the surface, r = 90°; critical angle   (b) total internal reflection   (c) glass → air is denser → rarer

(a) At i = 42° the refracted ray goes along the surface – the angle of refraction is 90°. This angle of incidence is the critical angle of glass.

(b) At 50° (greater than 42°) the ray is not refracted; it is reflected completely back into the glass at 50° – total internal reflection.

(c) The light goes from glass (denser) to air (rarer), so it bends away from the normal: r (60°) is greater than i (35°).

Marking: 1 mark for each part.

📖 Review this on Page 10 · Fig 1.17 critical-angle lab

16.
A ray of light PQ falls on a rectangular glass slab ABCD at Q with an angle of incidence of 50° (figure). (a) At which points does refraction take place? (b) Draw the path of the ray through the slab and out of it. (c) How is the emergent ray related to the incident ray PQ?
(3 marks)

💡 Two surfaces – two refractions.

🎬 Step by step
✅ Answer: (a) at Q and at R   (b) towards the normal at Q, away at R   (c) parallel to PQ

(a) At Q (air → glass) and at R, where the ray leaves the slab (glass → air).

(b) At Q it bends towards the normal (about 31° inside the glass); at R it bends away from the normal and leaves along RS.

(c) The emergent ray RS is parallel to the incident ray PQ, but shifted sideways.

Marking: 1 mark for each part; in (b) normals must be drawn at both surfaces.

📖 Review this on Page 8 · Glass slab, Fig 1.9

17.
On a hot summer afternoon the road far ahead appears wet, but there is no water when we reach the spot. (a) Name this phenomenon. (b) Explain how it happens. (c) Why is the image seen upside down?
(3 marks)

💡 Which air is optically rarer – hot or cool?

🎬 Step by step
✅ Answer: (a) mirage (മരീചിക)   (b) refraction through hot air layers + total internal reflection   (c) the rays cross over

(a) A mirage (മരീചിക).

(b) The air near the hot road is hotter and optically rarer; the layers above are cooler and denser. Light from the sky or a distant tree travelling down passes from denser to rarer layers and bends away from the normal more and more. Near the road the angle of incidence becomes greater than the critical angle, and the light is totally internally reflected upwards into our eyes.

(c) The light from the top of the tree arrives at the eye from lower down than the light from the bottom, so the image appears inverted, below the road – just like a reflection in a pool of water.

Marking: ½ for (a), 2 marks for (b) (rarer hot air ½, bending away ½, total internal reflection 1), ½ for (c).

📖 Review this on Page 11 · Mirage · Page 15 · Mirage lab

SECTION DAnswer all · 2 × 5 = 10 marks
18.
Refractive indices of some media are given below.
Mediumwaterkerosenecrown glassdiamond
Refractive index1.331.441.522.42
(a) In which medium is the speed of light the highest? (b) Find the speed of light in diamond (c = 3 × 108 m/s). (c) A ray passes obliquely from kerosene into water. Does it bend towards or away from the normal? Why? (d) The critical angle of diamond is 24.4° and that of glass is 42°. Which of the two sparkles more? Explain.
(5 marks)

💡 Higher n → slower light → smaller critical angle.

🎬 Step by step
✅ Answer: (a) water   (b) 1.24 × 108 m/s   (c) away from the normal   (d) diamond

(a) Water – it has the lowest refractive index, and v = c/n.

Given: n = 2.42; c = 3 × 108 m/s

Formula: v = c/n

Substitution: v = (3 × 108) ÷ 2.42

Answer: (b) v ≈ 1.24 × 108 m/s

(c) Kerosene (1.44) is denser than water (1.33): denser → rarer, so it bends away from the normal. (Kerosene floats on water – it has a lower mass density – but it is optically denser.)

(d) Diamond. Its critical angle (24.4°) is much smaller, so light inside it meets the faces at angles greater than the critical angle far more often. It is totally internally reflected many times and comes out through the top towards our eyes.

Marking: (a) 1, (b) 1 (formula ½, answer with unit ½), (c) 1 (½ answer, ½ reason), (d) 2 (1 for diamond, 1 for the reason).

📖 Review this on Page 14 · Let us assess Q9 · Page 10 · Critical angle · Page 15 · Sparkle lab

19.
Right-angled isosceles glass prisms (critical angle 42°) are used in cycle reflectors and in periscopes. (a) A ray falls normally on the face PQ of a reflector prism (figure). Draw its path and write the angle at which it strikes the other two faces. (b) Why is the ray totally reflected at those faces? (c) In which direction does the ray finally leave the prism, and why is this useful on a cycle? (d) In a prism periscope, through what angle does each prism turn the light? (e) If the prisms of a periscope were hollow and filled with water (critical angle 48.6°), would the periscope work? Why?
(5 marks)

💡 Light falling along the normal is not refracted; compare 45° with 42°.

🎬 Step by step
✅ Answer: (a) 45° at both faces   (b) 45° > 42°   (c) back, parallel to the incoming light   (d) 90°   (e) no

(a) The ray enters PQ along the normal (no refraction), strikes the first slanted face at A at 45°, is reflected to the second face at B, again at 45°, is reflected again and leaves through PQ.

(b) The light goes from glass to air (denser → rarer) and 45° is greater than the critical angle 42°, so it is totally internally reflected.

(c) It leaves parallel to the incoming light, going back towards the source – the headlights of the vehicle behind – so its driver sees the cycle at night without any bulb.

(d) Each prism turns the light through 90° (one total internal reflection at the hypotenuse).

(e) No. The light would meet the slanted face at 45°, which is less than the critical angle of water (48.6°). It would refract out instead of being totally reflected, and the image would be very dim.

Marking: (a) 2 (path 1, angles 1), (b) 1, (c) ½ + ½ for the use, (d) ½, (e) ½ with the reason. Total 5.

📖 Review this on Page 11 · Reflector and periscope · Page 15 · Reflector and periscope labs · Page 13 · Let us assess Q6

Marks summary

SectionQuestionsMarks eachTotalTopics
A1–818refractive index, optical density, critical angle, normal incidence, towards/away, twinkling stars, unit of n, Charles K. Kao
B9–13210n = c/v, ray diagram air → water, conditions for total internal reflection, fishing (apparent position), mirror vs total internal reflection
C14–17312speeds and optical density, critical-angle experiment, glass slab ray diagram, mirage
D18–19510refractive-index table and diamond’s sparkle, reflector and periscope prisms
Total40
⏱️ Time management (15 min cool-off + 90 minutes)
🎉 Finished the chapter?

Count your marks: 32 or more – excellent; 24–31 – good, revise the pages linked under the questions you missed; below 24 – go back through pages 5, 6, 7, 10 and 11 with their labs and do the scored quiz again. Then return to the chapter home and tick off what you have learned.