Class 10 Science SEE Guide | Unit - 10 Wave guide 2080

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Class 10 Science SEE Guide

 Unit - 10 Wave Guide

Exercise

A. Choose the correct option for the following questions.

a. Which of the following rules applies during the refraction of light?

(i) Light bends only while passing from a rarer medium to a denser medium.

(ii) Light bends towards the normal as it passes from a denser medium to a rarer medium.

(iii) When light passes from a rarer medium to a denser medium, the angle of incidence is smaller than the angle of refraction.

(iv) The angle of refraction becomes greater than the angle of incidence when light travels from a denser medium to a rarer medium.

Answer 👉 (iii) When light passes from a rarer medium to a denser medium, the angle of incidence is smaller than the angle of refraction.



d. Which of the following is the result obtained from the observation of refraction through a glass slab?

(i) Li = Le < Lr

(ii) Li > Lr = Le

(iii) Li < Le = Lr

(iv) Li = Le > Lr

Answer 👉 (iv) Li = Le > Lr



e. What is the value of the critical angle of the glass?

(i) 42°

(ii) 49°

(iii) 24°

(iv) 48°

Answer 👉 (iv) 48°


f. Among endoscopes , spectacles, mirages, dispersion of light, rainbows, optical fibres , and hand lens, in which instruments and processes does the total internal reflection of light, take place?

(i) endoscope, spectacles, and mirage

(ii) dispersion of light, rainbow, and hand lens

(iii) rainbow, optical fibre, and mirage

(iv) endoscope, mirage, andoptical fibre

Answer 👉 (iii) rainbow, optical fibre, and mirage


i. Distinguish the correct statement based on the characteristics of the image formed by concave and convex lenses.

(i) Convex lens forms a real, inverted, and diminished image of an object.

(ii) Concave lens forms a virtual, erect, and diminished image of an object.

(iii) Convex lens forms a real, inverted, and magnified image of an object.

(iv) Concave lens forms avirtual, erect, and magnified image of an object.

Answer 👉 (iii) Convex lens forms a real, inverted, and magnified image of an object.


J. What is the correct understanding of eye problems and related causes?

(i) When the lens of the eye becomes cloudy, color blindness occurs.

(ii) When the focal length of the lens of the eye increases, shortsightedness occurs.

(iii) When the shape of the surface of the cornea changes, defect in vision is seen.

(iv) Night blindness occurs due to the weakness of the cone cells of the retina.

Answer 👉 (iii) When the shape of the surface of the cornea changes, defect in vision is seen.




2. Differentiate between
(a) reflection of light and total internal reflection of light
Answer 👉 
  • Reflection of light occurs when light waves bounce off a surface, obeying the law of reflection, which states that the angle of incidence is equal to the angle of reflection.
  • Total internal reflection of light occurs when light waves are completely reflected back into the medium from which they came, due to the angle of incidence being greater than the critical angle for that medium.

(b) concave lens and convex lens
Answer 👉 
  • Concave lens is a lens that is thinner in the middle and thicker at the edges. It causes light rays to diverge (spread out) after passing through it. It is commonly used to correct nearsightedness (myopia).
  • Convex lens is a lens that is thicker in the middle and thinner at the edges. It causes light rays to converge (come together) after passing through it. It is commonly used to correct farsightedness (hyperopia) and is the shape of a typical magnifying lens.

( c) near point of the eye and far point of the eye
Answer 👉 
  • Near point of the eye is the closest distance at which the eye can focus on an object clearly without strain. It represents the maximum ability of the eye to accommodate and is typically around 25 cm for a normal eye.
  • Far point of the eye is the farthest distance at which the eye can focus on an object clearly without strain. It represents the maximum distance at which the eye can see objects clearly and is effectively at infinity for a normal eye.

( d) shortsightedness and longsightedness
Answer 👉 

  • Shortsightedness, also known as myopia, is a condition where a person can see nearby objects clearly but has difficulty seeing distant objects. It usually occurs when the eyeball is too long or the lens is too strong, causing light to focus in front of the retina instead of directly on it.
  • Longsightedness, also known as hyperopia, is a condition where a person can see distant objects clearly but has difficulty seeing nearby objects. It usually occurs when the eyeball is too short or the lens is too weak, causing light to focus behind the retina instead of directly on it.


( e) color blindness and night blindness
Answer 👉 
  • Color blindness is a condition where a person has difficulty distinguishing certain colors or perceiving color differences. It is usually caused by a genetic deficiency or abnormality in the photopigments of the cones in the retina.
  • Night blindness, also known as nyctalopia, is a condition where a person has difficulty seeing in low light conditions or at night. It is often caused by a deficiency in the rod cells of the retina, which are responsible for vision in dim light.


3. Give reasons:

(a) Between glass and water, glass is considered a denser medium and water is a rarer medium.

Answer 👉 Reason: The density of a medium is determined by the mass per unit volume. Glass has a higher density than water because it has a greater mass packed into the same volume compared to water.

(b) When a coin is placed in glass containing water, it appears to rise a bit.

Answer 👉 Reason: This phenomenon is due to the refraction of light. When light travels from one medium to another, such as from water to air (in this case, from water to glass and then to air), it undergoes refraction. The light rays bending at the water-glass interface make the coin appear raised when viewed from above the water surface.

(c) When the letters written on the paper are observed from the top of a glass slab, the letters appear to be slightly raised.

Answer 👉 Reason: Similar to the previous scenario, this is also due to refraction. The light passing through the glass slab and then entering the air above it undergoes refraction, causing the apparent displacement or elevation of the letters when observed from the top.

( d) Stars twinkle.

Answer 👉 Reason: The twinkling of stars is caused by the atmospheric refraction of starlight. As the light from stars passes through the Earth's atmosphere, it encounters variations in air density and temperature, which lead to the bending or refraction of light. This refraction causes the apparent fluctuation or twinkling of the star's brightness.

(e) The sun appears on the horizon about two minutes before the actual sunrise.

Answer 👉Reason: This phenomenon is known as atmospheric refraction. When the sun is close to the horizon, its light has to travel through a thicker portion of the Earth's atmosphere. The Earth's atmosphere acts as a lens, causing the sun's image to be refracted and appear slightly higher than its actual position. This effect leads to the observation of the sun before its geometric sunrise.

(f) A diamond appears to shine but a piece of glass cut to the same shape does not shine.

Answer 👉 Reason: The difference in appearance is due to the difference in the refractive index of diamond and glass. Diamond has a higher refractive index, which means it can bend and reflect light more effectively, leading to greater internal reflection and sparkle. Glass, on the other hand, has a lower refractive index, resulting in less internal reflection and a less pronounced sparkling effect.


(g) Sunlight is refracted when it is passed through a prism.

Answer 👉 Reason: Sunlight is composed of different colors or wavelengths of light. When sunlight enters a prism, it undergoes refraction due to the variation in the refractive index of the prism for different wavelengths. This refraction causes the different colors to bend at different angles, resulting in the dispersion of light and the formation of a spectrum.


(h) A convex lens converges light rays.

Answer 👉 Reason: A convex lens is thicker at the center and thinner at the edges. When light rays pass through a convex lens, the curvature of the lens causes the rays to bend inward, or converge, towards a focal point on the opposite side of the lens. This property of converging light rays is what makes a convex lens useful for focusing light and forming real images.


(i) A concave lens diverges the rays of light.

Answer 👉 Reason: A concave lens is thinner at the center and thicker at the edges. When light rays pass through a concave lens, the curvature of the lens causes the rays to bend outward, or diverge. The diverging property of a concave lens is used to correct vision problems such as nearsightedness (myopia) by causing the light rays to spread out before entering the eye.


(j) Deficiency of vitamin A in the body is one of the main causes of night blindness.

Answer 👉 Reason: Vitamin A plays a crucial role in the functioning of the retina, a light-sensitive layer at the back of the eye. The deficiency of vitamin A can affect the production of a pigment called rhodopsin in the retina, which is essential for vision in low-light conditions. Insufficient rhodopsin production can lead to impaired night vision or night blindness.


(k) Color blindness occurs when the cone cells of the retina stop functioning.

Answer 👉 Reason: Color vision is dependent on three types of cone cells in the retina that are sensitive to different wavelengths of light, corresponding to red, green, and blue colors. Color blindness occurs when one or more of these cone cells are defective or absent, resulting in an inability to perceive certain colors or color differences accurately.





4. Write short answers to the following questions.

a. What is the refraction of light?

Answer 👉 Refraction of light is the phenomenon where light changes direction and speed as it passes from one transparent medium to another, resulting in bending of the light ray.


b. Write the laws of refraction of light.

Answer 👉 The laws of refraction of light are:

  • The incident ray, the refracted ray, and the normal to the interface between two media at the point of incidence, all lie in the same plane.
  • The ratio of the sine of the angle of incidence to the sine of the angle of refraction is constant for a given pair of media. This constant is called the refractive index.


d. When a light ray passes from water to air, the angle of incidence and angle of refraction formed at the water-air separation layer are 40.50 and 60orespectively. Draw a ray diagram showing the refraction and write the reason why an object outside appears to be farther away from its actual position when it is viewed by an observer inside the water.

Answer 👉 The ray diagram is difficult to draw in a text-based format, but when a light ray passes from water to air, it bends away from the normal (towards the surface) due to the decrease in refractive index. The object outside appears to be farther away from its actual position when viewed by an observer inside the water because light rays from the object undergo refraction at the water-air interface, causing them to change direction and make the object appear displaced.


e. What is a critical angle?

Answer 👉 The critical angle is the angle of incidence in a medium at which the angle of refraction becomes 90 degrees. It is the minimum angle of incidence required for total internal reflection to occur.


f. What is a total internal reflection of light?

Answer 👉 Total internal reflection of light occurs when a light ray traveling from a denser medium to a rarer medium strikes the interface at an angle of incidence greater than the critical angle. Instead of being refracted, the light is completely reflected back into the denser medium.


g. Write two conditions necessary for total internal reflection of light.

Answer 👉 Two conditions necessary for total internal reflection of light are:

The light must be traveling from a denser medium to a rarer medium.

The angle of incidence must be greater than the critical angle.


h. At present, data can be transmitted at a very fast rate through fiber internet. Mention the role of total internal reflection of light in fiber internet.

Answer 👉 Total internal reflection of light plays a crucial role in fiber internet. In optical fibers, data is transmitted in the form of light signals. Total internal reflection occurs at the boundary between the core and cladding of the fiber, allowing the light to bounce off the walls and travel long distances without significant loss. This enables fast and efficient transmission of data through fiber optic cables.


i. In endoscopy, colonoscopy and keyhole surgery, how is total internal reflection of light applicable to the devices used to send light to the internal organs of the human body without incisions?

Answer 👉 In endoscopy, colonoscopy, and keyhole surgery, total internal reflection of light is used in the devices called endoscopes. These instruments consist of optical fibers that guide light to internal organs without the need for incisions. Total internal reflection ensures that the light remains trapped inside the fiber and is directed towards the target area, allowing for visualization and examination of internal organs.


J. What is a dispersion of light?

Answer 👉 Dispersion of light refers to the splitting of white light into its component colors when passing through a medium or encountering an obstacle. This occurs because different colors of light have different wavelengths and refract at different angles, causing them to separate and form a spectrum.


k. Mention the reason for the dispersion of light.

Answer 👉 The dispersion of light occurs due to the variation in the refractive index of a medium with respect to the wavelength of light. Different colors of light have different wavelengths, and as they pass through a medium, their speeds and directions of propagation change differently, leading to the separation and dispersion of colors.


m. A person is curious to know why a rainbow always appears semicircular and of the same size. Write down the solution to hiscuriosity. Include in your answer the position of the sun in the rainbow, the position of the water droplets in the air and the process of dispersion of light.

Answer 👉 The dispersion of light occurs due to the variation in the refractive index of a medium with respect to the wavelength of light. Different colors of light have different wavelengths, and as they pass through a medium, their speeds and directions of propagation change differently, leading to the separation and dispersion of colors.



n. Define the following terms related to the lens.

(i) Centre of curvature

(iii) Principal axis

(ii) Optical centre

(iv) Focus

Answer 👉 

(i) Centre of curvature: The center of the sphere from which the lens surface is derived. It is the midpoint of the radius of curvature of the lens.

(iii) Principal axis: The imaginary line passing through the center of curvature and the optical center of the lens. It is a reference line used for ray diagrams and calculations.

(ii) Optical centre: The point on the principal axis of the lens through which light passes undeviated. It is the geometric center of the lens.

(iv) Focus: The point or distance where parallel rays of light converge or appear to diverge after passing through a lens. A lens has two foci, namely the primary focus (F) and the secondary focus (F').



o. What ismeant by the power of a lens?

Answer 👉 The power of a lens is a measure of its ability to converge or diverge light. It is defined as the reciprocal of the focal length of the lens and is measured in diopters (D). The power (P) of a lens is given by the formula P = 1/f, where f is the focal length of the lens in meters.


p. The powers of two convex lenses are +2D and +4D respectively.

(i) Which of them is thicker? Give reason.

(ii) Calculate the focal length of each lens.

Answer 👉 (i) The lens with a higher power (+4D) is thicker. The power of a lens is directly proportional to its thickness. A lens with a greater power has a greater refractive ability, which requires a thicker lens to achieve that effect.

(ii) To calculate the focal length of each lens, we use the formula f = 1/P, where f is the focal length in meters and P is the power in diopters.

For the lens with a power of +2D:

f = 1/2 = 0.5 meters or 50 cm

For the lens with a power of +4D:

f = 1/4 = 0.25 meters or 25 cm


v. In which case is the image formed by a convex lens real and of the same size as the object? Show the ray diagram.

Answer 👉 A convex lens forms a real and same-sized image when the object is placed at the focal point (F) of the lens. The ray diagram shows that parallel rays converge at the focal point after passing through the lens. The image is formed on the other side of the lens and is the same size as the object.


w. How can the light dispersed by a prism be converted back into white light?

Answer 👉 The dispersed light can be converted back into white light by passing it through a second identical prism placed in the opposite orientation to the first prism. This process is called prism inversion. The second prism refracts the dispersed light in the reverse order, causing the individual colors to recombine and form white light again.


x. The focal lengths of the two lenses are 20cm and -20cm respectively. Mention the types of these two lenses. Out of these two lenses, which one forms a virtual and magnified image when an object is kept 16 cm away from the lens? Explain it by drawing a scaled ray diagram.

Answer 👉 A lens with a focal length of 20 cm is a convex lens, and a lens with a focal length of -20 cm is a concave lens. When an object is kept 16 cm away from the lens, the concave lens forms a virtual and magnified image. The ray diagram shows that the rays diverge after passing through the lens, and the image is formed on the same side as the object. The magnification factor is greater than 1, indicating a magnified image.


y. Write the functions of the following parts of the eye.

(i) Ciliary muscle

(ii) Cornea

(iii) Lens

(v) Pupil

(iv) Iris

(vi) Retina

Answer 👉 

(i) Ciliary muscle: The ciliary muscle controls the shape of the lens, allowing it to adjust its focus for near and far vision. It contracts to thicken the lens for close-up vision and relaxes to make the lens thinner for distant vision.

(ii) Cornea: The cornea is the clear, transparent outermost layer of the eye. It acts as a protective covering and helps to focus incoming light onto the retina.

(iii) Lens: The lens is a transparent, flexible structure located behind the iris. It helps to focus light onto the retina by changing its shape through the action of the ciliary muscle.

(iv) Pupil: The pupil is the dark circular opening in the center of the iris. It regulates the amount of light entering the eye by constricting or dilating in response to changes in lighting conditions.

(v) Iris: The iris is the colored part of the eye surrounding the pupil. It controls the size of the pupil and regulates the amount of light entering the eye by expanding or contracting its muscles.

(vi) Retina: The retina is a layer of tissue lining the back of the eye. It contains photoreceptor cells (rods and cones) that convert incoming light into electrical signals, which are then transmitted to the brain via the optic nerve for visual processing.


z. Write any two problems that may be seen in corneal injury.

Answer 👉 Two problems that may be seen in corneal injury:

  • Corneal Abrasion: This refers to a scratch or injury to the cornea, usually caused by foreign objects, such as dust particles, contact lenses, or accidental trauma. It can lead to symptoms like pain, redness, tearing, sensitivity to light, and blurred vision.
  • Corneal Ulcer: A corneal ulcer is an open sore or infection on the cornea, often resulting from bacterial, viral, or fungal infections. It can cause severe pain, redness, blurred vision, discharge, and sensitivity to light. If left untreated, corneal ulcers can lead to vision loss or even perforation of the cornea.


27. Explain the role of the ciliary muscle in the change in the thickness of the eye lens when a student sitting in a classroom shits his eyes from the letters written on the whiteboard to a distant object seen out the window.

Answer 👉 The ciliary muscle plays a crucial role in the process of accommodation, which allows the eye to focus on objects at different distances. When a student shifts their gaze from the letters on the whiteboard to a distant object out the window, the ciliary muscle contracts. This contraction causes the suspensory ligaments attached to the lens to relax, allowing the lens to become thicker. By increasing its curvature, the lens can refract light more effectively to focus it onto the retina, resulting in clear vision of the distant object.


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30. Explain, with a ray diagram, the role of the lens used to correct long-sightedness.

Answer 👉 To correct long-sightedness, a converging lens, also known as a convex lens, is used. In a ray diagram, parallel rays of light from a distant object are refracted by the convex lens. The lens brings the rays closer together, causing them to converge. This converging effect helps to compensate for the improperly focused light caused by long-sightedness, allowing the image to form correctly on the retina.


31. A student concludes that 'the effect of defect of vision is more on a person wearing spectacles with thick lenses than one wearing that with thin lenses'. Is this understanding, correct? Justify with appropriate reasons.

Answer 👉 The understanding of the student is not correct. The effect of a visual defect is not dependent on the thickness of the lenses in spectacles. The power of a lens, which determines its corrective effect, is determined by its curvature and refractive properties, not its thickness. Whether the lenses are thick or thin does not affect the effectiveness of vision correction. The key factor is the appropriate prescription of the lenses to address the specific refractive error.


32. Compare the use of spectacles and the use of contact lenses to correct visual defects.

Answer 👉 

Spectacles and contact lenses are both methods used to correct visual defects:

  • Spectacles: Spectacles are eyeglasses with lenses that are worn in front of the eyes. They correct refractive errors by adjusting the path of light entering the eyes. Spectacles are easy to use, do not require direct contact with the eyes, and can correct various types and degrees of refractive errors. However, some people may find them cumbersome or cosmetically undesirable.
  • Contact lenses: Contact lenses are directly placed on the surface of the eye. They correct refractive errors by altering the way light is focused on the retina. Contact lenses provide a wider field of view, natural vision, and freedom from wearing glasses. However, they require proper hygiene, regular cleaning, and maintenance. Some individuals may find them uncomfortable or have difficulty adapting to wearing contact lenses.

The choice between spectacles and contact lenses depends on personal preferences, lifestyle, comfort, and the advice of an eye care professional.


33. Explain the laser surgery method used to solve eyesight problems.

Answer 👉 Laser surgery, such as LASIK (Laser-Assisted in Situ Keratomileusis), is a method used to correct eyesight problems like myopia (nearsightedness), hyperopia (farsightedness), and astigmatism. In this procedure, a laser is used to reshape the cornea, altering its curvature and improving its focusing ability. The surgeon creates a thin corneal flap, which is lifted to expose the underlying corneal tissue. The laser then removes a precise amount of tissue from the cornea, reshaping it to the desired curvature. The flap is then repositioned, and the cornea naturally heals over time. The reshaped cornea allows light to be properly focused onto the retina, improving vision without the need for glasses or contact lenses.

34. What is a cataract? Write the role of the intraocular lens developed by Nepal's ophthalmologist Dr. Sanduk Ruitin the treatment of cataracts.

Answer 👉 A cataract is a clouding of the natural lens in the eye, which causes blurry or impaired vision. It is a common age-related condition but can also be caused by factors like genetics, injury, or medical conditions.

The intraocular lens (IOL) developed by Nepal's ophthalmologist Dr. Sanduk Ruit is used in the treatment of cataracts. During cataract surgery, the clouded natural lens is removed and replaced with an artificial intraocular lens. Dr. Sanduk Ruit developed a technique called small-incision cataract surgery (SICS), which involves making a small incision in the eye to remove the cataract and insert the IOL.

The role of the intraocular lens is to replace the clouded natural lens and restore clear vision. The IOL is a permanent implant that stays in the eye, allowing light to pass through and focus properly onto the retina. It helps to improve visual acuity and may also correct refractive errors like nearsightedness or farsightedness, reducing the need for glasses or contact lenses after cataract surgery. Dr. Sanduk Ruit's innovative approach has made cataract surgery more accessible and affordable, particularly in underserved areas.


Class 10 Science SEE Guide  Unit - 10  Wave guide

5. Solve the following mathematical problems.

a. If the speeds of light in air and glass are 3x 108 mis and 2x 108 mis respectively, thencalculate the refractive index of glass with respect to air. [Answer 1.5]

Answer 👉 The refractive index (n) of glass with respect to air can be calculated using the formula:


n = Speed of light in air / Speed of light in glass


n = (3x10^8 m/s) / (2x10^8 m/s)

n = 1.5


Therefore, the refractive index of glass with respect to air is 1.5.


b. T he refractive index of a diamond is 2.42. If the speed of light in glass is 3 x 108 m/s, calculate the speed of light in a diamond. [Answer 1.24X108 m/s]

Answer 👉 The speed of light in a diamond can be calculated using the formula:


Speed of light in diamond = Speed of light in glass / Refractive index of diamond


Speed of light in diamond = (3x10^8 m/s) / 2.42

Speed of light in diamond ≈ 1.24x10^8 m/s


Therefore, the speed of light in a diamond is approximately 1.24x10^8 m/s.


c. When a ray of light falls on the surface of a plastic block, the angle made by the ray with the normal and the angle of refraction are found to be 45° and 33° respectively. Calculate the refractive index of that plastic. [Answer 1.3]

Answer 👉 The refractive index (n) of the plastic block can be calculated using Snell's law:


n = sin(angle of incidence) / sin(angle of refraction)


n = sin(45°) / sin(33°)

n ≈ 1.3


Therefore, the refractive index of the plastic block is approximately 1.3.


d. Calculate the power of a lens having a focal length of 25 cm. [Answer +4D]

Answer 👉 The power (P) of a lens can be calculated using the formula:


P = 1 / Focal length (f)


P = 1 / 0.25 m

P = +4 D


Therefore, the power of the lens with a focal length of 25 cm is +4 D (diopters).


e. The power of the lens used in the spectacles worn by a student is -6D. Calculate the focal length of the lens. Also, mention the type of lens. [Answer 16.67cm]

Answer 👉 The focal length (f) of a lens can be calculated using the formula:


f = 1 / Power (P)


f = 1 / (-6 D)

f ≈ -0.1667 m ≈ -16.67 cm


The focal length of the lens is approximately -16.67 cm. Since the power is negative, the lens is concave (diverging) in nature.




Class 10 SEE Science book all Unit Solution. click any unit and get complete solutions.


Unit - 1 Scientific Learning

Unit - 2 Classification of Living Beings

Unit - 3 Honey Bee

Unit - 4 Heredity

Unit - 5 Physiological Structure and Life Process

Unit - 6 Nature and Environment

Unit - 7 Motion and Force

Unit - 8 Pressure

Unit - 9 Heat

Unit - 10 Wave

Unit - 11 Electricity and Magnetism

Unit - 12 Universe

Unit - 13 Information and Communication Technology

Unit - 14 Classification of Elements

Unit - 15 Chemical Reaction

Unit - 16 Gases

Unit - 17 Metal and Not metals

Unit - 18, Hydrocarbon and its Compounds

Unit - 19 , Chemicals used in Daily Life



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