Let Us Assess, Q7–Q11 & Extended Activities – Solutions വിലയിരുത്താം · തുടർപ്രവർത്തനങ്ങൾ
The last five end-of-chapter questions – a critical-angle choice, motorcycle reflectors, Table 1.6 and the speed of light in kerosene, the five ray paths of Fig 1.30, and methanol vs ethanol – followed by the three extended activities: a prism periscope, a critical-angle project and the “can r be 90°?” test. Try each one in your notebook first, then step through the solution.
- Refractive index അപവർത്തനാങ്കം n = c/v, with c = 3 × 10⁸ m/s. It has no unit. Bigger n ⇒ optically denser പ്രകാശികസാന്ദ്രത കൂടിയ ⇒ slower light.
- Rarer → denser obliquely: the ray bends towards the normal ലംബത്തോട് അടുക്കുന്നു. Denser → rarer: it bends away from the normal ലംബത്തിൽ നിന്ന് അകലുന്നു. Along the normal: no bending.
- Critical angle ക്രിട്ടിക്കൽ കോൺ: going denser → rarer, the angle of incidence that gives r = 90°. Glass–air: 42°.
- Total internal reflection പൂർണ്ണാന്തര പ്രതിപതനം needs both: (1) light goes from a denser to a rarer medium, and (2) the angle of incidence is greater than the critical angle.
Question 7 · Which angle gives total internal reflection?
7. ഗ്ലാസിന്റെ ക്രിട്ടിക്കൽ കോൺ 42° ആണ്. തന്നിരിക്കുന്നവയിൽ ഏത് പതനകോണിൽ പതിക്കുമ്പോഴാണ് പ്രകാശത്തിന് പൂർണ്ണാന്തര പ്രതിപതനം സംഭവിക്കുക? a) 40° b) 49° c) 38° d) 42° The critical angle of glass is 42°. At which of the given angles of incidence will light undergo total internal reflection? (a) 40° (b) 49° (c) 38° (d) 42°
Total internal reflection needs an angle of incidence greater than the critical angle – not equal to it. Which option is bigger than 42°?
Inside glass, going towards air (denser → rarer), total internal reflection happens only when the angle of incidence is greater than the critical angle 42°. Of the options only 49° > 42°.
At 40° and 38° (less than 42°) the ray is refracted out into the air, bending away from the normal. At exactly 42° the refracted ray goes along the surface (r = 90°) – that is the critical angle itself, not total internal reflection.
Choosing (d) 42°. At the critical angle the ray just grazes the surface (r = 90°). Total internal reflection starts only when i is more than 42°.
Question 8 · Why are there many small prisms in a motorcycle reflector?
8. മോട്ടോർസൈക്കിളിന്റെ റിഫ്ലക്റ്ററുകളിൽ അനേകം ചെറിയ പ്രിസങ്ങൾ കാണുവാൻ കഴിയും. ഇവ ഉള്ളതുകൊണ്ടുള്ള പ്രയോജനം എന്താണെന്ന് വിശദമാക്കുക. Many small prisms can be seen in the reflectors of a motorcycle. Explain the advantage of having them.
Look again at Fig 1.22 (b). Light from the headlight behind falls normally on face PQ. What is the angle of incidence at A and at B? Compare it with 42°.
The angle at A and B is always 45° (fixed by the prism shape). The trick works only if the material's critical angle is less than 45°.
1. Light from the headlight of a vehicle behind falls normally on the front face (PQ) of each small prism, so it enters without refraction.
2. Inside, it meets the slanted face at A with an angle of incidence of 45°. This is greater than the critical angle of glass (42°) and the light is going from glass to air (denser → rarer), so it is totally internally reflected.
3. At B the angle of incidence is again 45° > 42° – a second total internal reflection. The light now travels back, parallel to the incoming light, and leaves through PQ towards the driver behind.
Advantage: almost all the light is sent back, so the reflector shines brightly without any bulb or battery and the motorcycle is seen from far away at night. Because there are many small prisms, light coming from many directions is returned and the whole reflector glows – safer on the road.
On a dark road from Thrissur to Palakkad, a cycle or bike with a working tail reflector is visible to a bus driver from a long way off – a reflector has no battery to go flat. Keep your reflectors clean and unbroken!
Question 9 · Table 1.6 – speed of light in kerosene
9. പട്ടിക നിരീക്ഷിക്കൂ. Observe the table.
| Medium മാധ്യമം | Refractive index അപവർത്തനാങ്കം |
|---|---|
| air വായു | 1.0003 |
| water ജലം | 1.33 |
| kerosene മണ്ണെണ്ണ | 1.44 |
| turpentine oil ടർപെന്റയിൻ ഓയിൽ | 1.47 |
| crown glass ക്രൗൺ ഗ്ലാസ് | 1.52 |
| diamond വജ്രം | 2.42 |
Table 1.6
a) പട്ടികയിൽ നൽകിയിരിക്കുന്നവയിൽ പ്രകാശവേഗം ഏറ്റവും കുറഞ്ഞ മാധ്യമം ഏത്?b) വായുവിലൂടെയുള്ള പ്രകാശത്തിന്റെ വേഗം 3 × 10⁸ m/s ആണ്. മണ്ണെണ്ണയിലൂടെയുള്ള പ്രകാശവേഗം എത്രയായിരിക്കും?
c) വായുവിൽ നിന്ന് വജ്രത്തിലേക്ക് പ്രകാശരശ്മി ചരിഞ്ഞു പതിക്കുമ്പോൾ അപവർത്തനരശ്മി ലംബത്തോട് അടുക്കുമോ അതോ അകലുമോ? ഉത്തരം സാധൂകരിക്കുക.
- Of the media in the table, in which is the speed of light the lowest?
- The speed of light in air is 3 × 10⁸ m/s. What will the speed of light in kerosene be?
- When a ray of light falls obliquely from air into diamond, does the refracted ray bend towards the normal or away from it? Justify your answer.
n = c/v, so v = c/n: the bigger the refractive index, the smaller the speed. For (b) divide 3 × 10⁸ by 1.44. For (c), compare the refractive indices of air and diamond.
| Medium | n | v = c/n (m/s) |
|---|
Read the last column from top to bottom: as n goes up, v goes down – optical density up, speed down.
(a) Diamond. It has the highest refractive index (2.42), and since v = c/n, the highest n means the lowest speed.
(b) Given: c = 3 × 10⁸ m/s, n (kerosene) = 1.44, v = ?
Formula: n = c/v ⇒ v = c/n
Substitution: v = (3 × 10⁸ m/s) ÷ 1.44
v = 2.08 × 10⁸ m/s (≈ 2.083 × 10⁸ m/s).
(c) It bends towards the normal. Diamond (n = 2.42) has a much higher refractive index than air (n ≈ 1.0003), so diamond is optically denser. A ray going obliquely from a rarer to a denser medium slows down and bends towards the normal (angle of refraction < angle of incidence).
Multiplying instead of dividing: 3 × 10⁸ × 1.44 = 4.32 × 10⁸ m/s – faster than light in vacuum, which is impossible. Light is always slower in a medium, so v must come out less than 3 × 10⁸ m/s.
Question 10 · Fig 1.30 – which ray paths are correct?
10. A, B, C, D എന്നീ മാധ്യമങ്ങളിലൂടെ കടന്നുപോകുന്ന പ്രകാശരശ്മിയുടെ പാത നൽകിയിരിക്കുന്നു. ശരിയായവ തെരഞ്ഞെടുക്കുക. (മാധ്യമങ്ങളുടെ പ്രകാശിക സാന്ദ്രത A < B < C < D എന്ന ക്രമത്തിലാണ്.) The paths of a ray of light passing through the media A, B, C and D are given (Fig 1.30 (a)–(e)). Select the correct ones. (The optical densities of the media are in the order A < B < C < D.)
Check every interface separately. The ray travels downwards. Ask: is the next layer denser or rarer than this one? Denser → the ray must become steeper (closer to the normal). Rarer → it must become flatter (further from the normal). One wrong bend makes the whole figure wrong.
(a) A, B, C, D: A → B bends towards ✓; but B → C (into denser) bends away ✗ and C → D (into denser) bends away ✗ → wrong.
(b) A, B, C, D: every layer is denser than the one above, and the ray gets steeper at every interface ✓ ✓ ✓ → correct.
(c) C, B, D, A: C → B is denser → rarer, so it should bend away, but it bends towards ✗ → wrong (B → D towards ✓ and D → A away ✓ do not save it).
(d) A, C, B, D: A → C towards ✓, C → B away ✓, B → D towards ✓ → correct.
(e) C, B, D, A: C → B away ✓, B → D towards ✓, D → A away ✓ → correct.
Looking only at the last layer. In (a) the ray already goes wrong at the B → C interface. Check every interface, and compare the two layers on either side of it – not the order of the letters in the alphabet.
Snell's law says n₁ sin θ₁ = n₂ sin θ₂ at every interface, so in parallel layers n × sin θ is the same in every layer. Then n is proportional to 1/sin θ. Using the angles measured from the book's figure:
Only in (b), (d) and (e) do the numbers come out in the order A < B < C < D – the same verdict as the towards/away rule.
Question 11 · Methanol or ethanol – which has the lower refractive index?
11. മെത്തനോളിലെ പ്രകാശവേഗത്തേക്കാൾ കുറഞ്ഞ വേഗത്തിലാണ് എത്തനോളിൽ കൂടി പ്രകാശം സഞ്ചരിക്കുന്നത്. ഇവയിൽ അപവർത്തനാങ്കം കുറവ് ഏതു മാധ്യമത്തിന് ആയിരിക്കും? എന്തുകൊണ്ട്? Light travels through ethanol at a lower speed than through methanol. Which of them has the lower refractive index? Why?
n = c/v. c is the same for both. If v is bigger, is c/v bigger or smaller?
Refractive index n = c/v, where c (speed in vacuum) is the same for both. Light is faster in methanol, so v is bigger and c/v is smaller. (Equivalently, methanol is the optically rarer medium; ethanol, where light is slower, is optically denser and has the higher n.)
Measured values: methanol n ≈ 1.33, ethanol n ≈ 1.36. Speeds: 3 × 10⁸ ÷ 1.33 ≈ 2.26 × 10⁸ m/s and 3 × 10⁸ ÷ 1.36 ≈ 2.21 × 10⁸ m/s – ethanol is indeed slower.
Extended activity 1 · Build a prism periscope
1. സമതല ദർപ്പണങ്ങൾക്ക് പകരമായി പ്രിസങ്ങൾ ഉപയോഗിച്ച് പെരിസ്കോപ്പുകൾ നിർമ്മിച്ച് പ്രദർശിപ്പിക്കൂ. Make periscopes പെരിസ്കോപ്പ് using prisms പ്രിസം instead of plane mirrors സമതല ദർപ്പണം and exhibit them.
You need
- 2 right-angled glass (or clear acrylic) prisms with angles 45°–45°–90° (from the school lab, or old binocular prisms)
- a long box of thick card, or a PVC pipe about 40–50 cm long with two elbows (about 4 cm wide)
- black chart paper, glue / tape, a knife or saw (adult help), a ruler and protractor
Steps
- Cut a window near the top of the tube on the front, and one near the bottom on the back (opposite sides).
- Line the inside with black paper so stray light is absorbed.
- Fix the top prism so that light coming in through the window falls normally on one short face and meets the long face (hypotenuse) at 45°. The hypotenuse must be slanted at 45° to the tube.
- Fix the bottom prism the same way, its hypotenuse parallel to the top one, so the light going down is turned towards the eye-hole.
- Look through the bottom window and adjust the prisms slightly until the view over a wall or table is sharp.
Why prisms are better than mirrors: at each hypotenuse the angle of incidence is 45°, greater than the critical angle of glass (42°), so the light is totally reflected – almost no light is lost and the image is brighter. A glass mirror gives a faint second image from its front glass surface and its silvering can tarnish; a prism gives no double image and does not fade. That is why submarine periscopes and binoculars use prisms.
Exhibit tip: show a mirror periscope and a prism periscope side by side and let classmates compare how bright and sharp each view is.
Extended activity 2 · Project – compare critical angles
2. ഗ്ലിസറിൻ, ജലം, വെളിച്ചെണ്ണ, ഗ്ലാസ് തുടങ്ങിയ വ്യത്യസ്ത മാധ്യമങ്ങളുടെ ക്രിട്ടിക്കൽ കോൺ ഒരു പരീക്ഷണ പ്രോജക്ടിലൂടെ കണ്ടെത്തി താരതമ്യം ചെയ്യുക. ഉദ്ദേശ്യം, സാമഗ്രികൾ, പഠനരീതി, പഠനഫലങ്ങൾ തുടങ്ങിയ ഘട്ടങ്ങൾ ഉൾപ്പെടുത്തി പ്രോജക്റ്റ് റിപ്പോർട്ട് തയ്യാറാക്കുക. Find and compare the critical angles of different media such as glycerine, water, coconut oil and glass through an experimental project. Prepare a project report including the stages aim, materials, method and findings.
Aim: To find the critical angle (with air) of glycerine, water, coconut oil and glass, and to compare them.
Materials: a semicircular transparent plastic box (or a thick semicircular glass slab for glass), the liquids, a laser torch, chart paper with a circle and angles marked using two protractors (as in Fig 1.16), a pencil.
Method: (1) Place the semicircular box on the circle with the centre of its flat face exactly on the centre of the circle. (2) Fill it with the liquid. (3) Shine the laser from the curved side towards the centre – it then enters the curved surface along the radius without bending and meets the flat surface at the centre. (4) Increase the angle of incidence slowly until the refracted ray just grazes the flat surface (r = 90°). Read this angle of incidence from the circle – it is the critical angle. (5) Go a little further and see the ray totally reflected. (6) Repeat for each medium, three times, and take the average.
Observations table: medium · trial 1 · trial 2 · trial 3 · average critical angle. Findings: the greater the refractive index, the smaller the critical angle. Use the virtual lab below to know what to expect.
| Medium | n | critical angle (with air) |
|---|---|---|
| Press 📋 to fill. | ||
The book quotes 48.6° for water (n = 1.333); with n = 1.33 the lab gives 48.8°. Glass n = 1.5 gives 41.8° ≈ 42°, the book's value.
At the critical angle r = 90°, so Snell's law n sin C = 1 × sin 90° gives sin C = 1/n, C = sin⁻¹(1/n). Water: sin⁻¹(1/1.33) = 48.8°; glycerine: sin⁻¹(1/1.47) = 42.9°; coconut oil: sin⁻¹(1/1.45) ≈ 43.6°; glass: sin⁻¹(1/1.5) = 41.8°; diamond: sin⁻¹(1/2.42) = 24.4°.
Extended activity 3 · Rarer → denser: can r be 90°?
3. പ്രകാശരശ്മി പ്രകാശികസാന്ദ്രത കുറഞ്ഞ മാധ്യമത്തിൽ നിന്ന് കൂടിയ മാധ്യമത്തിലേക്ക് സഞ്ചരിക്കുമ്പോൾ അപവർത്തനകോൺ 90° ആകുമോ? പ്രവർത്തനം ചെയ്ത് നോക്കി സയൻസ് ഡയറിയിൽ കുറിക്കൂ. When a ray of light travels from an optically rarer medium to a denser one, can the angle of refraction be 90°? Do the activity and note it in your science diary.
Going into a denser medium the ray bends towards the normal, so r is always smaller than i. And i itself can never be more than 90°.
This is the reverse of the critical angle: the grazing ray from air enters glass at exactly the critical angle. That is why total internal reflection can never happen when light goes from air into water or glass.