Unit 2: Optics (ஒளியியல்)

அலகு 2: ஒளியியல் — Tamil Nadu Samacheer Kalvi Book Back Answers & Solutions

100% Complete Official Tamil Nadu Samacheer Kalvi Textbook Solutions covering Refraction of light, Snell's law, Convex and Concave lenses, Image formation rules with ray diagrams, Cartesian sign conventions, Lens formula, Magnification, Power of lens, Human eye defects (Myopia, Hypermetropia, Presbyopia, Astigmatism), Microscopes, Book-Back Numerical Problems, and HOT Questions.

ஒளிவிலகல், ஸ்நெல் விதி, குவி மற்றும் குழி லென்சுகள், கதிர் வரைபட விதிகள், லென்ஸ் சமன்பாடு, லென்சின் திறன், கண் குறைபாடுகள் (கிட்டப்பார்வை, தூரப்பார்வை), கூட்டு நுண்ணோக்கி, பாடப்புத்தகக் கணக்கீடுகள் மற்றும் உயர் சிந்தனை வினாக்களுக்கான (HOT) 100% அதிகாரப்பூர்வ தீர்வுகள்.

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📐 Important Formulas & Laws Quick Revision

  • Snell's Law of Refraction (ஸ்நெல் ஒளிவிலகல் விதி): sin i / sin r = μ₂ / μ₁ = constant Ratio of sine of angle of incidence to sine of angle of refraction equals the ratio of refractive indices of the two media.
  • Refractive Index (μ) (ஒளிவிலகல் எண் (μ)): μ = c / v Ratio of speed of light in vacuum (c = 3 × 10⁸ m/s) to speed of light in the medium (v). It has no unit.
  • Rayleigh's Scattering Law (ராலே ஒளிச்சிதறல் விதி): S ∝ 1 / λ⁴ Amount of scattering (S) is inversely proportional to the fourth power of wavelength (λ).
  • Lens Formula (லென்ஸ் சமன்பாடு): 1/f = 1/v - 1/u f = focal length, v = image distance, u = object distance. Follow Cartesian sign convention strictly.
  • Linear Magnification (m) (நேரியல் உருப்பெருக்கம் (m)): m = h₂ / h₁ = v / u Ratio of height of image (h₂) to height of object (h₁). Negative for real-inverted image; positive for virtual-erect image.
  • Power of a Lens (P) (லென்சின் திறன் (P)): P = 1 / f(in m) = 100 / f(in cm) SI Unit: Dioptre (D). 1 D = 1 m⁻¹. Power is positive for convex lens, negative for concave lens.
  • Myopia Lens Correction Focal Length (கிட்டப்பார்வை திருத்துவதற்கான குவியத்தொலைவு): f = - (x × y) / (x - y) ⟹ When y = ∞, f = -x x = far point of myopic eye, y = desired far point (infinity). Requires a concave lens.
  • Hypermetropia Lens Correction Focal Length (தூரப்பார்வை திருத்துவதற்கான குவியத்தொலைவு): f = (d × D) / (d - D) d = near point of hypermetropic eye, D = normal near point (25 cm). Requires a convex lens.

Part I: Choose the Correct Answer (1 Mark)

1. The refractive index of four substances A, B, C and D are 1.31, 1.43, 1.33, 2.4 respectively. The speed of light is maximum in substance ________.

A, B, C, D என்ற நான்கு பொருட்களின் ஒளிவிலகல் எண்கள் முறையே 1.31, 1.43, 1.33, 2.4 எனில், ஒளியின் திசைவேகம் எதில் பெருமமாக (அதிகமாக) இருக்கும்?

  • (a) A ✓ (Correct Answer)
  • (b) B
  • (c) C
  • (d) D
Explanation: Refractive index μ = c / v ⟹ v = c / μ. Speed of light in a medium is inversely proportional to its refractive index (v ∝ 1/μ). Substance A has the lowest refractive index (1.31), so light travels fastest in substance A.

2. Where should an object be placed so that a real and inverted image of same size is obtained by a convex lens?

பொருளின் அளவிற்கு சமமான, தலைகீழான மெய் பிம்பம் கிடைக்க பொருள் வைக்கப்பட வேண்டிய தொலைவு:

  • (a) f
  • (b) 2f ✓ (Correct Answer)
  • (c) infinity
  • (d) between f and 2f
Explanation: When an object is placed at 2F (centre of curvature C₁) of a convex lens, rays refract to form a real, inverted image of the exact same size at 2F₂ on the opposite side (magnification m = -1).

3. A small bulb is placed at the principal focus of a convex lens. When the bulb is switched on, the lens will produce ________.

ஒரு சிறிய மின்விளக்கு ஒரு குவி லென்சின் முதன்மைக் குவியத்தில் வைக்கப்பட்டுள்ளது. மின்விளக்கு எரியும் போது லென்சானது:

  • (a) a convergent beam of light
  • (b) a divergent beam of light
  • (c) a parallel beam of light ✓ (Correct Answer)
  • (d) a coloured beam of light
Explanation: By the reversibility of light rays, when a point source of light is placed at the principal focus (F) of a convex lens, all rays emerging from the focus refract through the lens and travel parallel to the principal axis as a parallel beam.

4. Magnification of a convex lens is ________.

குவி லென்சின் உருப்பெருக்கமானது:

  • (a) Positive
  • (b) negative
  • (c) either positive or negative ✓ (Correct Answer)
  • (d) zero
Explanation: A convex lens can produce both real (inverted) images where magnification m is negative, and virtual (erect) images when the object is placed between F and optical centre where m is positive. Hence magnification can be either positive or negative.

5. A convex lens forms a real, diminished point sized image at focus. Then the position of the object is at ________.

ஒரு குவி லென்சானது மிகச்சிறிய புள்ளி போன்ற மெய் பிம்பத்தை குவியத்தில் உருவாக்குகிறது எனில் பொருளின் நிலை:

  • (a) focus
  • (b) infinity ✓ (Correct Answer)
  • (c) at 2f
  • (d) between f and 2f
Explanation: Parallel rays of light coming from an object situated at infinity converge at the principal focus after refracting through a convex lens, forming a highly diminished, real, point-sized image at F.

6. Power of a lens is -4D, then its focal length is ________.

ஒரு லென்சின் திறன் -4D எனில் அதன் குவியத்தொலைவு:

  • (a) 4 m
  • (b) -40 m
  • (c) -0.25 m ✓ (Correct Answer)
  • (d) -2.5 m
Explanation: Power of a lens P = 1 / f (in metres) ⟹ f = 1 / P = 1 / (-4 D) = -0.25 m. The negative sign denotes a concave lens.

7. In a myopic eye, the image of the object is formed ________.

கிட்டப்பார்வை குறைபாடு உடைய கண்ணில் பொருளின் பிம்பமானது எங்கு தோற்றுவிக்கப்படுகிறது?

  • (a) behind the retina
  • (b) on the retina
  • (c) in front of the retina ✓ (Correct Answer)
  • (d) on the blind spot
Explanation: In myopia (short-sightedness), due to elongation of the eyeball or increased curvature of the eye lens, light rays converge prematurely and form the image in front of the retina.

8. The eye defect 'presbyopia' can be corrected by ________.

'விழி ஏற்பமைவுத் திறன் குறைபாடு' (Presbyopia) எதன் மூலம் சரிசெய்யப்படுகிறது?

  • (a) convex lens
  • (b) concave lens
  • (c) convex mirror
  • (d) Bi focal lenses ✓ (Correct Answer)
Explanation: Presbyopia occurs due to aging when ciliary muscles lose flexibility. Persons with presbyopia often cannot see nearby or distant objects clearly. It is corrected using Bi-focal lenses (upper part concave for distant vision, lower part convex for reading).

9. Which of the following lens would you prefer to use while reading small letters found in a dictionary?

அகராதியில் உள்ள சிறிய எழுத்துக்களைப் படிப்பதற்கு நீங்கள் எவ்வகை லென்சைப் பயன்படுத்த விரும்புவீர்கள்?

  • (a) A convex lens of focal length 5 cm ✓ (Correct Answer)
  • (b) A concave lens of focal length 5 cm
  • (c) A convex lens of focal length 10 cm
  • (d) A concave lens of focal length 10 cm
Explanation: A magnifying glass is a convex lens. Magnification m = 1 + D/f. Smaller focal length 'f' yields higher magnification. Hence a convex lens with a smaller focal length of 5 cm gives greater magnification (m = 1 + 25/5 = 6) than one with 10 cm (m = 1 + 25/10 = 3.5).

10. If V_B, V_G, V_R be the velocity of blue, green and red light respectively in a glass prism, then which of the following statement gives the correct relation?

ஒரு கண்ணாடி முப்பட்டகத்தில் நீலம், பச்சை மற்றும் சிவப்பு நிற ஒளிகளின் திசைவேகங்கள் முறையே V_B, V_G, V_R எனில் அவற்றுக்கு இடையேயான சரியான தொடர்பு:

  • (a) V_B = V_G = V_R
  • (b) V_B > V_G > V_R
  • (c) V_B < V_G < V_R ✓ (Correct Answer)
  • (d) V_B < V_G > V_R
Explanation: In a material medium like glass, refractive index depends on wavelength (Cauchy's relation: μ = A + B/λ²). Red has the longest wavelength (λ_R > λ_G > λ_B), hence the smallest refractive index (μ_R < μ_G < μ_B). Since velocity v = c / μ, red light travels fastest and blue travels slowest: V_B < V_G < V_R.

Part II: Fill in the Blanks (கோடிட்ட இடங்களை நிரப்புக)

1. The path of the light is called as ________.

ஒளி செல்லும் பாதை ________ எனப்படும்.

Answer: ray of light (ஒளிக்கதிர்)

2. The refractive index of a transparent medium is always greater than ________.

ஒரு ஒளிபுகும் ஊடகத்தின் ஒளிவிலகல் எண் எப்போதும் ________-ஐ விட அதிகமாக இருக்கும்.

Answer: one (or 1) (ஒன்று (1))

3. If the energy of incident beam and the scattered beam are same, then the scattering of light is called as ________ scattering.

படுகதிரின் ஆற்றலும் சிதறலடைந்த கதிரின் ஆற்றலும் சமமாக இருந்தால், அச்சிதறல் ________ சிதறல் எனப்படும்.

Answer: elastic (மீட்சிச்)

4. According to Rayleigh's scattering law, the amount of scattering of light is inversely proportional to the fourth power of its ________.

ராலே ஒளிச்சிதறல் விதிப்படி, சிதறல் அளவானது அதன் ________-ன் நான்கு மடிக்கு எதிர்த்தகவில் அமையும்.

Answer: wavelength (அலைநீளம்)

5. Amount of light entering into the eye is controlled by ________.

கண்ணினுள் நுழையும் ஒளியின் அளவைக் கட்டுப்படுத்துவது ________ ஆகும்.

Answer: iris (ஐரிஸ் (Iris))

Part III: State True or False with Statement Correction (சரியா? தவறா?)

1. Velocity of light is greater in denser medium than in rarer medium.

அடர்வு மிகு ஊடகத்தில் ஒளியின் திசைவேகமானது, அடர்வு குறை ஊடகத்தில் இருப்பதை விட அதிகமாக இருக்கும்.

✗ FALSE (தவறு)

Correct Statement: Velocity of light is greater in a rarer medium than in a denser medium (v_rarer > v_denser).

2. The power of lens depends on the focal length of the lens.

லென்சின் திறன் லென்சின் குவியத்தொலைவைச் சார்ந்தது.

✓ TRUE (சரி)

3. Increase in the converging power of eye lens cause 'hypermetropia'.

விழி லென்சின் குவிக்கும் திறன் அதிகரிப்பதால் 'தூரப்பார்வை' ஏற்படுகிறது.

✗ FALSE (தவறு)

Correct Statement: Increase in the converging power of eye lens causes 'myopia' (short-sightedness). Decrease in converging power causes 'hypermetropia'.

4. The convex lens always gives small virtual image.

குவி லென்சானது எப்போதும் சிறிய மாய பிம்பத்தையே உருவாக்கும்.

✗ FALSE (தவறு)

Correct Statement: The concave lens always gives a small (diminished) virtual image. (A convex lens produces an enlarged virtual image only when object is placed between F and optical centre).

Part IV: Match the Following (பொருத்துக)

  • 1. Retina (1. விழித்திரை) ➔ d. Screen of the eye
  • 2. Pupil (2. கண் பாவை) ➔ a. Path way of light
  • 3. Ciliary muscles (3. சிலியரித் தசைகள்) ➔ f. Power of accommodation
  • 4. Myopia (4. கிட்டப்பார்வை) ➔ b. Far point comes closer
  • 5. Hypermetropia (5. தூரப்பார்வை) ➔ c. Near point moves away

Part V: Assertion and Reason (கூற்றும் காரணமும்)

Question 1:

Assertion (A): If the refractive index of the medium is high (denser medium), the velocity of light in that medium will be small.

Reason (R): Refractive index of the medium is inversely proportional to the velocity of the light.

Correct Answer: (a) If both assertion and reason are true and reason is the correct explanation of assertion.

By definition, refractive index μ = c / v. Thus, μ ∝ 1 / v. A medium with a higher refractive index causes light to travel at a slower speed. Therefore, both assertion and reason are true, and the reason correctly explains the assertion.

Question 2:

Assertion (A): Myopia is due to the increase in the converging power of eye lens.

Reason (R): Myopia can be corrected with the help of concave lens.

Correct Answer: (b) If both assertion and reason are true but reason is not the correct explanation of assertion.

Assertion is true: Myopia occurs when excessive curvature increases the converging power of the eye lens, causing rays to focus before the retina. Reason is also true: A diverging concave lens counteracts this excess converging power to focus the rays on the retina. However, the reason is a statement of correction, not the scientific cause of the assertion.

Part VI: Short Answer Questions — 2 Marks (சுருக்கமான விடையளி)

Q1. What is refractive index? State its unit.

ஒளிவிலகல் எண் என்றால் என்ன? அதன் அலகைக் கூறுக.

Answer:
  • 1. Definition: The ratio of the speed of light in vacuum (c) to the speed of light in a medium (v) is defined as the refractive index (μ) of that medium.
  • 2. Formula: μ = c / v
  • 3. Unit: Since it is a ratio of two similar quantities (velocities), it has NO UNIT and NO DIMENSIONS.

Q2. State Snell’s law.

ஸ்நெல் ஒளிவிலகல் விதியைக் கூறுக.

Answer:
  • First Law: The incident ray, the refracted ray at the point of incidence and the normal to the interface of two transparent media all lie in the same plane.
  • Second Law (Snell's Law): The ratio of the sine of the angle of incidence to the sine of the angle of refraction is equal to the ratio of refractive indices of the two media.
  • Formula: sin i / sin r = μ₂ / μ₁

Q3. Draw a ray diagram to show the image formed by a convex lens when the object is placed between F and 2F.

பொருள் F மற்றும் 2F-க்கு இடையே வைக்கப்படும் போது குவி லென்சினால் தோன்றும் பிம்பத்திற்கான கதிர் வரைபடம் வரைக.

Answer:
  • 1. Nature of Image: Real, inverted, and magnified (enlarged).
  • 2. Position of Image: Formed beyond 2F on the other side of the lens.
  • 3. Ray 1: A ray parallel to the principal axis passes through the principal focus F₂ after refraction.
  • 4. Ray 2: A ray passing through the optical centre (O) goes undeviated.
  • 5. Both rays intersect beyond 2F₂ to produce the enlarged image A'B'.

Q4. Define dispersion of light.

ஒளி நிறப்பிரிகை என்றால் என்ன?

Answer:
  • 1. Definition: When a beam of white light or composite light is passed through an optically transparent medium such as a glass prism, it is split up into its constituent colors (VIBGYOR). This phenomenon is called dispersion of light.
  • 2. Cause: Different colors of light have different wavelengths, so they travel at different speeds and suffer different degrees of deviation through the prism (Violet deviates most; Red deviates least).

Q5. State Rayleigh’s law of scattering.

ராலே ஒளிச்சிதறல் விதியைக் கூறுக.

Answer:
  • 1. Statement: Rayleigh's law of scattering states that the amount of scattering of light is inversely proportional to the fourth power of its wavelength.
  • 2. Mathematical Expression: S ∝ 1 / λ⁴ (where S is the amount of scattering and λ is the wavelength of the incident light).
  • 3. Inference: Shorter wavelengths (blue/violet) are scattered much more strongly than longer wavelengths (red/orange).

Q6. Differentiate convex lens and concave lens.

குவி லென்ஸ் மற்றும் குழி லென்சுகளை வேறுபடுத்துக.

Answer:
  • 1. Shape: Convex lens is thicker at the centre and thinner at the edges. Concave lens is thinner at the centre and thicker at the edges.
  • 2. Nature: Convex lens is a converging lens. Concave lens is a diverging lens.
  • 3. Focal Length: Focal length is positive (+f) for convex lens. Focal length is negative (-f) for concave lens.
  • 4. Image: Convex lens produces mostly real and inverted images (except when object is inside F). Concave lens always produces virtual, erect and diminished images.

Q7. What is power of accommodation of the eye?

விழி ஏற்பமைவுத் திறன் என்றால் என்ன?

Answer:
  • 1. Definition: The ability of the eye lens to adjust its focal length so as to see nearby as well as distant objects clearly is called the power of accommodation of the eye.
  • 2. Mechanism: This is achieved by changing the curvature of the flexible crystalline eye lens with the help of ciliary muscles.
  • 3. For distant vision: Ciliary muscles relax, lens becomes thin, focal length increases.
  • 4. For near vision: Ciliary muscles contract, lens becomes thick, focal length decreases.

Q8. What are the causes of 'Myopia'?

'கிட்டப்பார்வை' குறைபாட்டிற்கான காரணங்கள் யாவை?

Answer:
  • Myopia (Short-sightedness) is caused due to two primary anatomical reasons:
  • 1. Elongation of the eyeball: The distance between the eye lens and retina becomes abnormally large.
  • 2. Excessive curvature of the eye lens: The focal length of the eye lens is shortened, increasing its converging power.
  • Result: Light rays from distant objects converge in front of the retina instead of on the retina.

Q9. Why does the sky appear in blue colour?

வானம் ஏன் நீல நிறமாகத் தோன்றுகிறது?

Answer:
  • 1. According to Rayleigh's law of scattering: S ∝ 1 / λ⁴ (Scattering is inversely proportional to the fourth power of wavelength).
  • 2. Blue light has a much shorter wavelength than red light.
  • 3. When sunlight enters the Earth's atmosphere, blue light is scattered much more strongly by gas molecules in all directions than colours of longer wavelengths.
  • 4. This scattered blue light enters our eyes, making the sky appear blue.

Q10. Why are traffic signals red in colour?

போக்குவரத்து சமிக்ஞை விளக்குகளில் சிவப்பு நிறம் பயன்படுத்தப்படுவது ஏன்?

Answer:
  • 1. Red light has the longest wavelength among all visible colors in the spectrum.
  • 2. According to Rayleigh's law: Scattering S ∝ 1 / λ⁴.
  • 3. Because of its large wavelength, red light is scattered the least by fog, smoke, and air molecules.
  • 4. It can travel the longest distance without being attenuated, making it clearly visible from very far distances to warn drivers.

Q11. While doing an experiment for the determination of focal length of a convex lens, Raja Suddenly dropped the lens. It got broken into two halves along the axis. If he continues his experiment with the same lens, (a) can he get the image? (b) Is there any change in the focal length?

குவி லென்சின் குவியத்தொலைவைக் கண்டறியும் சோதனையின் போது, ராஜா தவறுதலாக லென்சை கீழே தவறவிட்டு விடுகிறார். அது முதன்மை அச்சை ஒட்டி இரு பாதிகளாக உடைகிறது. அவர் அதே உடைந்த லென்சைக் கொண்டு சோதனையைத் தொடர்ந்தால், (அ) அவரால் பிம்பத்தைப் பெற முடியுமா? (ஆ) குவியத்தொலைவில் ஏதேனும் மாற்றம் ஏற்படுமா?

Answer:
  • (a) Yes, he can still obtain the complete image of the object. Every part of a lens, no matter how small, is capable of forming the complete image of the object. However, since the aperture (light-transmitting area) of each broken half is reduced to half, the brightness and intensity of the image formed will be less.
  • (b) No, there is no change in the focal length of the lens. According to Lens Maker's Formula: 1/f = (μ - 1)(1/R₁ - 1/R₂). When the lens is cut or broken along its principal axis, the radius of curvature of each curved surface (R₁ and R₂) and the refractive index (μ) remain unchanged. Therefore, the focal length remains the same (f).

Q12. The eyes of the nocturnal birds like owl are having a large cornea and a large pupil. How does it help them?

ஆந்தை போன்ற இரவு நேரப் பறவைகளின் கண்களில் பெரிய அளவிலான கருவிழி (cornea) மற்றும் கண் பாவை (pupil) காணப்படுகின்றன. இது அவைகளுக்கு எவ்வாறு உதவுகிறது?

Answer:
  • 1. Maximum Light-Gathering Power: A large cornea and a wide pupil provide a broad aperture that allows the maximum amount of light to enter the eyes, which is essential for vision in extremely dim light conditions at night.
  • 2. Rod-Dominated Retina: In addition, their retina contains a large number of rod cells (which are highly sensitive to low intensity and dim light) and very few cone cells.
  • 3. Conclusion: The combination of an enlarged cornea, wide pupil, and rod-rich retina gives nocturnal birds like owls excellent night vision to navigate and hunt effectively in the dark.

Part VII: Long Answer Questions — 5 / 7 Marks (விரிவான விடையளி)

Q1. List any five properties of light.

ஒளியின் ஏதேனும் ஐந்து பண்புகளைக் கூறுக.

1. Form of Energy and Straight-Line Propagation (1 Mark)
  • Light is a form of energy.
  • Light always travels along a straight line path (rectilinear propagation).
2. Medium Independence and Ultimate Speed (1.5 Marks)
  • Light does not require any material medium for its propagation. It can travel even through a vacuum.
  • The speed of light in vacuum or air is c = 3 × 10⁸ m s⁻¹.
3. Wave Nature and Wave Equation (1 Mark)
  • Since light is in the form of electromagnetic waves, it is characterized by a wavelength (λ) and a frequency (ν).
  • Speed, frequency and wavelength are related by the fundamental equation: c = νλ.
4. Spectrum of Colours and Wavelength Order (1.5 Marks)
  • Different coloured lights have different wavelengths and different frequencies.
  • Among visible light, violet light has the lowest wavelength (~400 nm) and red light has the longest wavelength (~700 nm).
  • When light meets the interface between two media, it is partly reflected and partly refracted.

Q2. Explain the rules for obtaining images formed by a convex lens with the help of ray diagram.

குவி லென்சினால் பிம்பங்கள் தோன்றுவதற்கான விதிகளை கதிர் வரைபடங்களுடன் விளக்குக.

Rule-1: Ray Passing Obliquely Through the Optical Centre (1.5 Marks)
  • Rule Statement: When a ray of light strikes the convex lens obliquely at its optical centre (O), it continues to follow its path without any deviation or lateral shift.
  • Explanation: At the exact geometric centre of the lens, the two spherical surfaces are parallel, so the ray emerges along the same straight direction.
Rule-2: Ray Parallel to the Principal Axis (1.5 Marks)
  • Rule Statement: When a ray of light parallel to the principal axis strikes a convex lens, the refracted ray converges and passes through the principal focus (F₂) on the other side of the lens.
  • Explanation: The convex lens is a converging lens; parallel incident rays are refracted towards the optic axis and intersect at the principal focal point.
Rule-3: Ray Passing Through the Principal Focus (2 Marks)
  • Rule Statement: When a ray of light passing through the principal focus (F₁) strikes a convex lens, the refracted ray emerges parallel to the principal axis.
  • Explanation: This rule follows the principle of reversibility of light rays — if an incident ray passes through focus, its refracted ray travels parallel to the axis.

Q3. Differentiate the eye defects: Myopia and Hypermetropia.

கண் குறைபாடுகளான கிட்டப்பார்வை மற்றும் தூரப்பார்வையை வேறுபடுத்துக.

1. Definition and Nature of Defect (1 Mark)
  • Myopia (Short-Sightedness): A person can see nearby objects clearly, but cannot see distant objects clearly.
  • Hypermetropia (Long-Sightedness): A person can see distant objects clearly, but cannot see nearby objects clearly.
2. Anatomical Causes (1.5 Marks)
  • Myopia: (i) Elongation of the eyeball, (ii) Excessive curvature of the eye lens (focal length is shortened, converging power increases).
  • Hypermetropia: (i) Shortening of the eyeball, (ii) Inadequate curvature of the eye lens (focal length is lengthened, converging power decreases).
3. Position of Image Formation (1 Mark)
  • Myopia: The image of a distant object is formed in front of the retina.
  • Hypermetropia: The image of a nearby object is formed behind the retina.
4. Shift in Reference Points (0.5 Mark)
  • Myopia: The far point of the eye comes closer than infinity.
  • Hypermetropia: The near point of the eye moves away beyond 25 cm.
5. Correcting Lens & Formula (1 Mark)
  • Myopia Correction: Corrected by using a Concave Lens (diverging lens) of focal length f = -x (or f = -xy / (x - y)). Power is negative (-D).
  • Hypermetropia Correction: Corrected by using a Convex Lens (converging lens) of focal length f = (d × D) / (d - D). Power is positive (+D).

Q4. Explain the construction and working of a ‘Compound Microscope’.

கூட்டு நுண்ணோக்கியின் அமைப்பையும் செயல்படும் விதத்தையும் படத்துடன் விளக்குக.

1. Construction and Parts (1.5 Marks)
  • A compound microscope consists of two converging convex lenses mounted coaxially in two sliding tubes:
  • (a) Objective Lens: The lens placed near the tiny object. It has a very small focal length (f_o) and small aperture.
  • (b) Eyepiece (Eye lens): The lens placed near the observer's eye. It has a larger focal length (f_e) and larger aperture than the objective.
2. Working Mechanism & Formation of Image (2.5 Marks)
  • Step 1: The tiny object AB is placed slightly beyond the principal focus F_o of the objective lens.
  • Step 2: The objective lens forms a real, inverted, and magnified intermediate image A'B'.
  • Step 3: This intermediate image A'B' acts as an object for the eyepiece.
  • Step 4: The position of the eyepiece is adjusted such that A'B' lies within its principal focus F_e.
  • Step 5: The eyepiece behaves as a simple magnifier and forms a final virtual, erect (with respect to A'B'), and highly magnified image A"B" at the least distance of distinct vision (D = 25 cm).
3. Total Magnification Formula (1 Mark)
  • The total magnification produced by a compound microscope is the product of magnification of the objective (m_o) and eyepiece (m_e):
  • m = m_o × m_e = (v_o / u_o) × (1 + D / f_e)

Part VIII: Solved Problems Corner / Numerical Calculations

Problem 1: An object is placed at a distance 20cm from a convex lens of focal length 10cm. Find the image distance and nature of the image.

10 செ.மீ குவியத்தொலைவு கொண்ட குவி லென்சிலிருந்து 20 செ.மீ தொலைவில் பொருள் வைக்கப்படுகிறது. பிம்பம் தோன்றும் தொலைவு மற்றும் பிம்பத்தின் தன்மையைக் காண்க.

Step-by-Step Working:
Step 1: Write down the Lens Formula:
1/f = 1/v - 1/u ⟹ 1/v = 1/f + 1/u
Step 2: Substitute the given values with Cartesian signs:
1/v = 1/10 + 1/(-20)
1/v = 1/10 - 1/20
Step 3: Find common denominator (LCM = 20):
1/v = (2 - 1) / 20 = 1/20
Step 4: Invert to solve for image distance v:
v = +20 cm.
Step 5: Calculate magnification to determine nature and size:
m = v / u = (+20) / (-20) = -1.
Step 6: Interpretation of Results:
• The positive sign of 'v' (+20 cm) means the image is formed on the other side of the lens.
• Magnification m = -1 indicates that the image is Real, Inverted, and of the Same Size as the object (formed at 2F₂).
Boxed Final Answer: Image distance v = +20 cm; Nature: Real, Inverted and of Same Size as the object.

Problem 2: An object of height 3cm is placed at 10cm from a concave lens of focal length 15cm. Find the size of the image.

15 செ.மீ குவியத்தொலைவு கொண்ட குழி லென்சிலிருந்து 10 செ.மீ தொலைவில் 3 செ.மீ உயரமுள்ள பொருள் வைக்கப்படுகிறது எனில், பிம்பத்தின் அளவைக் காண்க.

Step-by-Step Working:
Step 1: Write down the Lens Formula:
1/f = 1/v - 1/u ⟹ 1/v = 1/f + 1/u
Step 2: Substitute the given values with Cartesian signs:
1/v = 1/(-15) + 1/(-10) = -1/15 - 1/10
Step 3: Find common denominator (LCM of 15 and 10 = 30):
1/v = (-2 - 3) / 30 = -5 / 30 = -1 / 6
Step 4: Solve for image distance v:
v = -6 cm (The negative sign confirms the virtual image is formed on the same side as the object).
Step 5: Apply magnification formula to find image size h₂:
m = h₂ / h₁ = v / u
h₂ = (v / u) × h₁
h₂ = [(-6) / (-10)] × 3
h₂ = 0.6 × 3 = +1.8 cm.
Step 6: Interpretation: The size of the image is 1.8 cm. The positive sign of h₂ confirms the image is virtual and erect.
Boxed Final Answer: Size of the image h₂ = 1.8 cm (Virtual, erect and diminished; formed at v = -6 cm)

Problem Example 1: Light rays travel from vacuum into a glass whose refractive index is 1.5. If the angle of incidence is 30°, calculate the angle of refraction inside the glass.

வெற்றிடத்திலிருந்து 1.5 ஒளிவிலகல் எண் கொண்ட கண்ணாடிக்குள் ஒளிக்கதிர் செல்கிறது. படுகோணம் 30° எனில், கண்ணாடியினுள் ஒளிவிலகல் கோணத்தைக் கணக்கிடுக.

Step-by-Step Working:
Step 1: State Snell's law equation: μ₁ sin i = μ₂ sin r
Step 2: Substitute values: (1.0) × sin 30° = 1.5 × sin r
Step 3: Since sin 30° = 0.5 (or 1/2):
1.0 × 0.5 = 1.5 × sin r
sin r = 0.5 / 1.5 = 1 / 3 = 0.333
Step 4: Calculate r:
r = sin⁻¹(0.333) = 19.45°
Boxed Final Answer: Angle of refraction r = 19.45°

Problem Example 2: A beam of light passing through a diverging lens of focal length 0.3m appear to be focused at a distance 0.2m behind the lens. Find the position of the object.

0.3 மீ குவியத்தொலைவு கொண்ட விரிக்கும் லென்ஸ் வழியே செல்லும் ஒளிக்கற்றை லென்சிற்குப் பின்புறம் 0.2 மீ தொலைவில் குவியப்படுவது போல் தோன்றுகிறது எனில், பொருளின் நிலையைக் காண்க.

Step-by-Step Working:
Step 1: Rearrange lens formula for 1/u:
1/u = 1/v - 1/f
Step 2: Substitute values: v = -0.2 m, f = -0.3 m:
1/u = 1/(-0.2) - [1/(-0.3)] = -10/2 + 10/3 = -5 + 3.33 = -10/6
Step 3: Solve for u:
u = -6/10 = -0.6 m.
Boxed Final Answer: Object distance u = -0.6 m (placed 0.6 m in front of the lens)

Problem Example 3: A person with myopia can see objects placed at a distance of 4m. If he wants to see objects at a distance of 20m, what should be the focal length and power of the concave lens he must wear?

கிட்டப்பார்வை குறைபாடு உடைய ஒருவரால் 4 மீ தொலைவில் உள்ள பொருள்களைக் காண முடிகிறது. அவர் 20 மீ தொலைவில் உள்ள பொருள்களைக் காண விரும்பினால், அவர் அணிய வேண்டிய குழி லென்சின் குவியத்தொலைவு மற்றும் திறன் யாது?

Step-by-Step Working:
Step 1: Formula for focal length: f = (x × y) / (x - y)
Step 2: Substitute x = 4 m and y = 20 m:
f = (4 × 20) / (4 - 20) = 80 / (-16) = -5 m.
Step 3: Calculate power of the lens:
P = 1 / f = 1 / (-5) = -0.2 D.
Boxed Final Answer: Focal length f = -5 m; Power P = -0.2 D (Concave lens)

Problem Example 4: For a person with hypermetropia, the near point has moved to 1.5m. Calculate the focal length of the correction lens in order to make his eyes normal.

தூரப்பார்வை குறைபாடு கொண்ட ஒருவருக்கு அண்மைப் புள்ளி 1.5 மீ என நகர்ந்துள்ளது. அவரது பார்வையை இயல்பாக்கத் தேவையான திருத்தும் லென்சின் குவியத்தொலைவைக் கணக்கிடுக.

Step-by-Step Working:
Step 1: Formula for focal length: f = (d × D) / (d - D)
Step 2: Substitute d = 1.5 m and D = 0.25 m:
f = (1.5 × 0.25) / (1.5 - 0.25)
f = 0.375 / 1.25 = +0.3 m (or +30 cm).
Step 3: Positive sign indicates a Convex lens is required.
Boxed Final Answer: Focal length f = +0.3 m (30 cm, Convex lens)