🔭 Physics · Chapter 2 · Samacheer Kalvi Grade 10

Samacheer Class 10 Science - Optics

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Samacheer Kalvi Class 10 Science Chapter 2 Optics book back questions and answers — complete MCQ, short answer and long answer solutions with step-by-step explanations. This page covers reflection, refraction, Snell's law, total internal reflection, lenses, mirrors, the human eye, compound microscope and telescope as per the TN SSLC syllabus. All answers are aligned with the official Samacheer Kalvi Grade 10 Science textbook.

Optics — key concepts & quick answers

What is the difference between reflection and refraction of light?
Reflection is the bouncing back of light from a surface; refraction is the bending of light as it passes from one medium to another because its speed changes.
What are the laws of reflection?
(i) The angle of incidence equals the angle of reflection. (ii) The incident ray, the reflected ray and the normal at the point of incidence all lie in the same plane.
What is refractive index?
The refractive index of a medium is the ratio of the speed of light in vacuum to its speed in the medium, n = c/v. A higher refractive index means light travels slower in that medium and bends more.
What is the least distance of distinct vision?
It is the closest distance at which the human eye can see an object clearly without strain — about 25 cm for a normal eye.
Which lens corrects myopia and which corrects hypermetropia?
Myopia (short-sightedness) is corrected with a concave (diverging) lens; hypermetropia (long-sightedness) is corrected with a convex (converging) lens.
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Multiple Choice Questions Fill in the Blanks True or False Match the Following Short Answer Questions Numerical Problems Long Answer Questions Higher Order Thinking
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MCQI. Multiple Choice Questions1 mark each
Q.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
✓ Answer: (A) A (n = 1.31) - speed is maximum in the medium with lowest refractive index
Q.2 Where should an object be placed so that a real and inverted image of same size is obtained by a convex lens
✓ Answer: (B) 2f (object at twice the focal length)
Q.3 A small bulb is placed at the principal focus of a convex lens. When the bulb is switched on, the lens will produce
✓ Answer: (C) a parallel beam of light (collimated beam)
Q.4 Magnification of a convex lens is
✓ Answer: (C) either positive or negative
Q.5 A convex lens forms a real, diminished point sized image at focus. Then the position of the object is at
✓ Answer: (B) infinity (object at infinity)
Q.6 Power of a lens is -4 D. What is its focal length?

-0.25 m. Power P = 1/f (in metres), so f = 1/P = 1/(-4) = -0.25 m.

Q.7 In a myopic eye, the image of the object is formed
✓ Answer: (C) in front of the retina
Q.8 The eye defect ‘presbyopia’ can be corrected by
✓ Answer: (D) Bi focal lenses
Q.9 Which of the following lens would you prefer to use while reading small letters found in a dictionary?
✓ Answer: (A) A convex lens of focal length 5 cm
Q.10 If VB, VG, VR be the velocity of blue, green and red light respectively in a glass prism, then which of the following statement gives the correct relation?
✓ Answer: (C) VB < VG < VR
FillII. Fill in the Blanks1 mark each
#Statement (Answer in bold)
1The path of the light is called as ray
2The refractive index of a transparent medium is greater than or equal to one; for ordinary transparent media like water and glass, it is greater than one.
3If the energy of incident beam and the scattered beam are same, then the scattering of light is called as elastic scattering.
4According to Rayleigh's scattering law, the amount of scattering of light is inversely proportional to the fourth power of its wavelength
5Amount of light entering into the eye is controlled by Iris
T/FIII. True or False1 mark each
#StatementAnswerCorrection (if False)
1Velocity of light is greater in denser medium than in rarer mediumFalseVelocity of light is lesser in denser medium than in rarer medium.
2The power of lens depends on the focal length of the lensTruePower of a lens is P = 1/f, where f is measured in metre.
3Increase in the converging power of eye lens cause ‘hypermetropia’FalseAn increase in the converging power of the eye lens causes myopia, because light focuses in front of the retina. A decrease in converging power may cause hypermetropia.
4The convex lens always gives small virtual image.FalseA concave lens always forms a virtual, erect and diminished image for a real object. A convex lens may form real or virtual images depending on object position.
MatchIV. Match the Following1 mark each
Column AColumn B
RetinaScreen of the eye where the image is formed
PupilOpening in the iris that allows light to enter the eye
Ciliary musclesMuscles that change the curvature of the eye lens for accommodation
MyopiaDistant objects are blurred because the image forms in front of the retina; far point comes closer; corrected by a concave lens.
HypermetropiaNearby objects are blurred because the image forms behind the retina; near point moves away; corrected by a convex lens.
ShortVI. Short Answer Questions2 marks each
Q.1 What is refractive index?
✓ Answer
Refractive index is the ratio of the speed of light in vacuum to the speed of light in a medium. It is given by μ = c/v, where c is the speed of light in vacuum and v is the speed of light in the medium. Refractive index has no unit.
Q.2 State Snell’s law.
✓ Answer
Snell's law states that, for a given pair of media, the ratio of sine of the angle of incidence to sine of the angle of refraction is constant, when the angles are measured from the normal to the surface. It can be written as sin i / sin r = n2 / n1, or n1 sin i = n2 sin r, where n1 and n2 are the refractive indices of the two media, i is the angle of incidence and r is the angle of refraction.
Q.3 Draw a ray diagram to show the image formed by a convex lens when the object is placed between F and 2F.
✓ Answer
Ray diagram construction steps: (i) Draw the principal axis and mark the convex lens, F and 2F on both sides. Place the object AB between F and 2F on the left side. (ii) From the top A, draw a ray parallel to the principal axis; after refraction it passes through the focus on the other side. (iii) From A, draw another ray through the optical centre; it goes undeviated. (iv) The refracted rays meet at A' beyond 2F on the right side. Draw A'B' perpendicular to the principal axis. Thus the image is formed beyond 2F, and it is real, inverted and magnified. With Cartesian sign convention, if f = +10 cm and u = -15 cm, then 1/v - 1/u = 1/f gives v = +30 cm. Magnification m = v/u = 30/(-15) = -2; the negative sign shows inversion and the image is twice the size of the object.
Q.4 Define dispersion of light
✓ Answer

Dispersion of light is the phenomenon observed when a beam of white light passes through a transparent medium, such as a glass prism or water droplets, and splits into its constituent colours. This splitting occurs because each colour of light, corresponding to a different wavelength, travels at a slightly different speed within the medium. Consequently, each colour is refracted at a slightly different angle. Violet light, having the shortest wavelength, is deviated the most, while red light, with the longest wavelength, is deviated the least. This results in the formation of a spectrum of colours, typically observed as red, orange, yellow, green, blue, indigo, and violet (ROYGBIV).

Q.5 State Rayleigh’s law of scattering
✓ Answer

Rayleigh's law of scattering describes how light is scattered by particles that are much smaller than the wavelength of the light, such as molecules of gas in the atmosphere. The law states that the intensity of scattered light is inversely proportional to the fourth power of its wavelength. Mathematically, this can be expressed as: Intensity of scattering ∝ 1/λ⁴, where λ is the wavelength of the light. This relationship explains why shorter wavelength colours, such as blue and violet, are scattered much more effectively by atmospheric particles than longer wavelength colours, like red and orange. This is the reason why the sky appears blue during the day, as blue light is scattered in all directions across the sky, and why sunsets and sunrises often appear red, as the blue light has been scattered away over the longer path through the atmosphere.

Q.6 Differentiate convex lens and concave lens.
✓ Answer

A convex lens is a converging lens that is thicker at the center and thinner at the edges. Its surfaces curve outwards. When parallel rays of light pass through a convex lens, they converge at a point called the principal focus, forming a real image. Consequently, a convex lens has a positive focal length. Depending on the position of the object, a convex lens can form either real, inverted images or virtual, erect, and magnified images. In contrast, a concave lens is a diverging lens that is thinner at the center and thicker at the edges, with surfaces curving inwards. When parallel rays of light strike a concave lens, they diverge as if originating from a virtual focus on the same side as the incident light. Therefore, a concave lens has a negative focal length. For a real object, a concave lens typically forms a virtual, erect, and diminished image.

Q.1 A convex lens is thicker in the middle than at edges.
✓ Answer

A convex lens is characterized by its shape, being thicker in the middle and thinner at the edges. This specific curvature causes parallel rays of light incident upon it to converge at a point after passing through the lens. Because it brings light rays together, it is also known as a converging lens. This property makes it useful in various optical instruments where focusing light is essential.

Q.2 It is converging
✓ Answer

A convex lens is termed a converging lens because of its optical behavior. When parallel rays of light strike its surface, the lens refracts these rays inwards, causing them to converge at a specific point called the principal focus. This ability to bring light rays together is the defining characteristic of a converging lens, enabling it to form real images.

Q.3 A convex lens produces mostly real images
✓ Answer

A convex lens is capable of producing both real and virtual images, but it predominantly forms real images. When an object is placed at a distance greater than its focal length from the convex lens, the refracted rays converge to form a real, inverted image on the opposite side of the lens. A virtual, erect, and magnified image is formed only when the object is placed between the optical center and the principal focus of the lens.

Q.4 It is used to treat hypermetropia
✓ Answer

A convex lens is effectively used to correct the vision defect known as hypermetropia, or long-sightedness. In hypermetropia, the eye's focusing power is insufficient, causing the image of nearby objects to be formed behind the retina, making them appear blurred. The convex lens, with its converging power, adds to the eye's refractive power, ensuring that light from nearby objects is focused precisely on the retina, thus restoring clear vision.

Q.1 A concave lens is thinner in the middle than at edges
✓ Answer

A concave lens is identified by its physical structure, which is thinner at the center and thicker at the edges. This shape causes parallel rays of light to diverge, or spread out, after refraction. Consequently, a concave lens is also referred to as a diverging lens. This divergence of light is fundamental to its optical properties and applications.

Q.2 It is diverging
✓ Answer

A concave lens is classified as a diverging lens because of how it interacts with light. When parallel rays of light pass through a concave lens, they are refracted outwards, spreading away from each other. These diverging rays, when traced backward, appear to originate from a point on the same side of the lens as the incident light, known as the principal focus. This spreading of light is the hallmark of a diverging lens.

Q.3 It produces a virtual image
✓ Answer

A concave lens, when used with a real object, consistently produces an image that is virtual, erect, and diminished in size. This means the image cannot be projected onto a screen and appears upright and smaller than the object. The virtual nature arises because the refracted rays diverge and do not actually meet; they only appear to diverge from a point from which the image is formed.

Q.4 It is used to treat myopia
✓ Answer

The vision defect known as myopia, or short-sightedness, is effectively treated using a concave lens. Myopia occurs when the eye focuses distant objects in front of the retina, leading to blurred vision for faraway objects. A concave lens, being a diverging lens, reduces the converging power of the eye's optical system. It causes the incoming parallel rays from distant objects to diverge slightly before entering the eye, ensuring that the final image is formed precisely on the retina, thus correcting the defect.

Q.7 What is power of accommodation of eye?
✓ Answer
Power of accommodation is the ability of the eye lens to change its focal length or optical power so that both nearby and distant objects can be focused clearly on the retina. It is measured as the change in power of the eye lens, in dioptre (D). For a normal eye, the near point is about 25 cm, so the maximum accommodation is approximately 1/0.25 m = 4 D.
Q.8 What are the causes of ‘Myopia’?
✓ Answer

Myopia, or short-sightedness, arises from two primary causes related to the eye's structure and refractive power. Firstly, the eyeball may become abnormally elongated from front to back. Secondly, the eye lens might possess excessive converging power, meaning it bends light too strongly. In either case, parallel rays from distant objects are focused at a point *in front* of the retina, rather than on it. This results in clear vision for nearby objects but blurred vision for distant ones. The condition is typically corrected by using a concave lens, which diverges the light rays before they enter the eye, effectively shifting the focal point back onto the retina.

Q.9 Why does the sky appear in blue colour?
✓ Answer

The phenomenon responsible for the blue color of the sky is known as Rayleigh scattering. When sunlight enters the Earth's atmosphere, it collides with the tiny gas molecules present. These molecules scatter the sunlight in all directions. Blue light, having a shorter wavelength compared to other visible colors like red, is scattered much more effectively than longer wavelengths. As this scattered blue light reaches our eyes from all parts of the sky, the sky appears to be blue. This scattering is more pronounced for shorter wavelengths.

Q.10 Why are traffic signals red in colour?
✓ Answer

Traffic signals are designed to be red primarily because of the physics of light scattering and the psychological impact of color. Red light possesses the longest wavelength within the visible spectrum. This long wavelength means it is scattered the least by atmospheric particles like dust, water droplets, and air molecules. Consequently, red light travels longer distances without significant dispersion, making it clearly visible from afar, even in conditions of fog or haze. Furthermore, red is a color that is easily distinguishable and has been conventionally adopted as a warning or stop signal across many cultures, allowing drivers to quickly and unambiguously recognize its meaning, thus enhancing road safety.

LongVIII. Long Answer Questions5 marks each
Q.1 List any five properties of light
✓ Answer
(i) Light is a form of energy.
(ii) Light travels in a straight line in a homogeneous medium.
(iii) Light can travel through vacuum and does not require a material medium.
(iv) The speed of light in vacuum is c = 3 x 10^8 m s^-1.
(v) Different colours of light have different wavelengths and frequencies.
Q.2 Explain the rules for obtaining images formed by a convex lens with the help of ray diagram.
✓ Answer
Rules for drawing images formed by a convex lens:
(i) A ray parallel to the principal axis passes through the principal focus on the other side after refraction.
(ii) A ray passing through the optical centre of the lens goes undeviated.
(iii) A ray passing through the principal focus on the object side emerges parallel to the principal axis after refraction.
To obtain the image in a ray diagram, draw any two of these rays from the top of the object. The point where the refracted rays actually meet gives a real image; if the refracted rays diverge, extend them backwards to locate the virtual image. For a convex lens, the image may be real and inverted when the object is beyond F, or virtual, erect and magnified when the object is between F and the lens. Important cases are: object beyond 2F gives a diminished image between F and 2F; object at 2F gives same-size image at 2F; object between F and 2F gives a magnified image beyond 2F; object at F gives image at infinity; object within F gives a virtual, erect and enlarged image on the same side.
Q.3 Differentiate the eye defects: Myopia and Hypermetropia
✓ Answer

Myopia, commonly known as short-sightedness, is an eye defect where individuals can see nearby objects clearly but distant objects appear blurred. This occurs either because the eyeball has become too long (elongated) or because the eye lens has too much converging power. As a result, the image of distant objects is focused in front of the retina instead of on it. Myopia is corrected using a concave lens, which diverges the light rays before they enter the eye, ensuring the image is formed on the retina. Hypermetropia, or long-sightedness, is the opposite condition where nearby objects appear blurred, while distant objects are seen clearly. This defect arises when the eyeball is too short or the eye lens has insufficient converging power. Consequently, the image of nearby objects is focused behind the retina. Hypermetropia is corrected using a convex lens, which converges the light rays, helping to focus the image on the retina.

Q.4 Explain the construction and working of a ‘Compound Microscope’.
✓ Answer
A compound microscope is used to see tiny objects. It consists of two convex lenses fixed coaxially in a tube. The lens near the object is called the objective lens; it has a short focal length and relatively large aperture to collect more light. The lens near the eye is called the eyepiece; it has a comparatively longer focal length. The microscope also has a stage to hold the slide, a mirror or light source for illumination, and coarse and fine adjustment screws for focusing.
Working: The object is placed just beyond the focal length of the objective lens. The objective forms a real, inverted and magnified intermediate image inside the tube, near the focal plane of the eyepiece. This intermediate image acts as the object for the eyepiece. The eyepiece works like a simple microscope and produces a further enlarged virtual image either at the near point or at infinity. The final image is enlarged and inverted with respect to the original object.
The total magnification is the product of the magnification by the objective and the magnification by the eyepiece. For final image at the near point, approximately M = (L/f_o) x (1 + D/f_e). For final image at infinity, approximately M = (L/f_o) x (D/f_e). Here L is the tube length, f_o is the focal length of the objective, f_e is the focal length of the eyepiece and D is the least distance of distinct vision, about 25 cm.
NumericalVII. Numerical Problems3 marks each
Q.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.
✓ Answer
Using Cartesian sign convention: f = +10 cm and u = -20 cm because the object is on the left side of the convex lens.
Lens formula: 1/v - 1/u = 1/f.
1/v - 1/(-20) = 1/10
1/v + 1/20 = 1/10
1/v = 1/10 - 1/20 = 1/20, so v = +20 cm.
The positive value of v shows that the image is formed 20 cm on the right side of the lens. Magnification m = v/u = 20/(-20) = -1. Therefore, the image is real, inverted and of the same size as the object.
Q.2 An object of height 3cm is placed at 10cm from a concave lens of focal length 15cm. Find the size of the image.
✓ Answer
Given: u = -10 cm, f = -15 cm, object height h = 3 cm.
Lens formula: 1/v - 1/u = 1/f
=> 1/v - 1/(-10) = 1/(-15)
=> 1/v + 1/10 = -1/15
=> 1/v = -1/15 - 1/10 = - (2+3)/30 = -5/30 = -1/6
=> v = -6 cm (virtual image, 6 cm on object side).
Magnification m = v/u = (-6)/(-10) = 0.6
Image height h' = m h = 0.6 x 3 cm = 1.8 cm.
Therefore: image is virtual, erect, diminished, located 6 cm from the lens on the object side, and its size is 1.8 cm.
HOTIX. Higher Order Thinking3 marks each
Q.1 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?
✓ Answer
(a) Yes, he can still get the image using one half of the broken convex lens.
(b) There is no change in the focal length because the curvature of the lens surfaces does not change. However, the intensity or brightness of the image will be reduced because less light passes through the broken lens.
Q.2 The eyes of the nocturnal birds like owl are having a large cornea and a large pupil. How does it help them?
✓ Answer

The eyes of nocturnal birds like owls are adapted to see in low light conditions. A large cornea, which is the transparent outer layer of the eye, allows a greater amount of light to enter. Similarly, a large pupil, the opening in the center of the iris, can expand significantly to let in even more light. This increased light gathering capacity ensures that more light reaches the retina, where photoreceptor cells convert it into neural signals. Consequently, the bird can form a brighter and clearer image of its surroundings, enabling it to hunt and navigate effectively in the dim light of night.

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