CHAPTER 9  LIGHT REFLECTION AND REFRACTION KSEEB SSLC CLASS 10 SCIENCE SOLUTIONS

CHAPTER 9  LIGHT REFLECTION AND REFRACTION KSEEB SSLC CLASS 10 SCIENCE SOLUTIONS

 

CHAPTER 9  LIGHT REFLECTION AND REFRACTION KSEEB SSLC CLASS 10 SCIENCE SOLUTION English medium Karnataka state board,the Answers Are Prepared By Our Teachers Which Are Simple ,Pointwise,Easy To Read And Remember .

 

CHAPTER 9  LIGHT REFLECTION AND REFRACTION KSEEB SSLC CLASS 10 SCIENCE SOLUTIONS
  1. Define the principal focus of a concave mirror.

Definition of the principal focus of a concave mirror:

  • The principal focus of a concave mirror is the point on its principal axis where parallel rays of light converge after reflection.
  • It is represented by the symbol ‘F’.
  • Rays parallel to the principal axis, after reflection, pass through or appear to diverge from the principal focus of a concave mirror.

 

  1. The radius of curvature of a spherical mirror is 20 cm. What is its focal length?

The focal length (f) of a spherical mirror can be determined using the relationship between the focal length and the radius of curvature (R):

    \[ f = \frac{R}{2} \]

Given:

    \[ R = 20 \text{ cm} \]

Calculation:

    \[ f = \frac{20 \text{ cm}}{2} = 10 \text{ cm} \]

Therefore, the focal length of the spherical mirror is 10 cm.

 

  1. Name a mirror that can give an erect and enlarged image of an object.

. Mirror that gives an erect and enlarged image of an object:

  • A concave mirror can produce an erect and enlarged image of an object when the object is placed between its principal focus (F) and the mirror.

 

  1. Why do we prefer a convex mirror as a rear-view mirror in vehicles?

Advantages of convex mirrors as rear-view mirrors:

  • Provide a wider field of view compared to plane mirrors.
  • Always produce an erect image, aiding in accurate perception.
  • Diminish the size of objects, allowing drivers to see a larger area behind the vehicle.
  • Curvature reflects light divergently, minimizing blind spots and providing a broader perspective for safer driving.

 

  1. Find the focal length of a convex mirror whose radius of curvature is 32 cm.

. Focal length of convex mirror:

  • Given radius of curvature (R) = 32 cm.
  • Using the formula f = R/2, where f is the focal length.
  • Substituting R = 32 cm into the formula, we get:
  • f = 32/2 = 16 cm

 

  1. A concave mirror produces three times magnified (enlarged) real image of an object placed at 10 cm in front of it. Where is the image located?

Location of image produced by concave mirror:

  • Given magnification (m) = 3 (enlarged).
  • Object distance (u) = -10 cm (negative as object is placed in front of the mirror).
  • Using the magnification formula m = -v/u, where v is the image distance.
  • Substituting m = 3 and u = -10 cm into the formula, we get:
  • 3 = -v/-10.
  • Solving for v, we find v = 30 cm.
  • Since the image is real and formed on the same side as the object, it is located 30 cm in front of the concave mirror.

 

  1. A ray of light travelling in air enters obliquely into water. Does the light ray bend towards the normal or away from the normal? Why?

When a ray of light travels from air into water obliquely, it bends towards the normal. This happens because water is optically denser than air, causing the light ray to slow down and change direction towards the normal.

  1. Light enters from air to glass having refractive index 1.50. What is the speed of light in the glass? The speed of light in vacuum is 3 × 108 m s–1 .

The speed of light in glass can be found using the refractive index formula n = c/v, where c is the speed of light in vacuum and v is the speed of light in the medium. Given that the refractive index of glass  n  is 1.50 and the speed of light in vacuum  c  is  3 times 10^8 m/s, we can rearrange the formula to find the speed of light in glass: v = c/n = 3 *10^8/1.50= 2 *10^8 m/s

  1. Find out, from Table 10.3, the medium having highest optical density. Also find the medium with lowest optical density.

The medium with the highest optical density can be identified by finding the medium with the highest refractive index from Table 10.3. Similarly, the medium with the lowest optical density can be identified by finding the medium with the lowest refractive index from the same table.

 

  1. You are given kerosene, turpentine and water. In which of these does the light travel fastest? Use the information given in Table 10.3.

To determine in which medium light travels fastest among kerosene, turpentine, and water, we need to compare their refractive indices from Table 10.3. The medium with the highest refractive index has the lowest speed of light, while the medium with the lowest refractive index has the highest speed of light.

 

 

  1. The refractive index of diamond is 2.42. What is the meaning of this statement?

The statement “The refractive index of diamond is 2.42” means that when light enters diamond from air, it slows down to approximately 1/2.42 times its speed in air.

This indicates that diamond has a high optical density compared to air and causes a significant bending of light, making it an excellent material for optical applications like lenses and prisms.

 

  1. Define 1 dioptre of power of a lens.

– One dioptre (1 D) of power of a lens is the power of a lens whose focal length is 1 meter. It indicates the ability of a lens to converge or diverge light rays.

 

  1. A convex lens forms a real and inverted image of a needle at a distance of 50 cm from it. Where is the needle placed in front of the convex lens if the image is equal to the size of the object? Also, find the power of the lens.

The needle is placed at a distance of 25 cm in front of the convex lens if the image is equal to the size of the object. This is because in this case, the object distance (u) is equal to half the image distance (v), as per the lens formula. The power of the lens can be calculated using the formula: Power (P) = 1 / focal length (f). Given that the image distance (v) is 50 cm, the focal length (f) is also 50 cm, resulting in a power of +2 dioptres.

 

  1. Find the power of a concave lens of focal length 2 m.

The power of a concave lens is negative. Therefore, for a concave lens of focal length 2 m, the power (P) can be calculated using the formula: Power (P) = 1 / focal length (f). Substituting the given focal length, we get P = 1 / (-2) = -0.5 dioptres. So, the power of the concave lens is -0.5 dioptres.

 

1. Which one of the following materials cannot be used to make a lens?
– (d) Clay

2. The image formed by a concave mirror is observed to be virtual, erect, and larger than the object. Where should be the position of the object?
– (d) Between the pole of the mirror and its principal focus.

3. Where should an object be placed in front of a convex lens to get a real image of the size of the object?
– (b) At twice the focal length.

4. A spherical mirror and a thin spherical lens have each a focal length of –15cm. The mirror and the lens are likely to be:
– (a) Both concave.

5. No matter how far you stand from a mirror, your image appears erect. The mirror is likely to be:
– (d) Either plane or convex.

6. Which of the following lenses would you prefer to use while reading small letters found in a dictionary?
– (c) A convex lens of focal length 5 cm.

7. We wish to obtain an erect image of an object, using a concave mirror of focal length 15 cm. What should be the range of distance of the object from the mirror? What is the nature of the image? Is the image larger or smaller than the object? Draw a ray diagram to show the image formation in this case.
– The object should be placed within the focal length of the mirror (less than 15 cm from the mirror).
– The image formed will be virtual, erect, and larger than the object.
– [Ray Diagram]

8. Name the type of mirror used in the following situations:

– (a) Headlights of a car:

Concave mirror. (Because it converges light to a point to produce a strong beam.)
– (b) Side/rear-view mirror of a vehicle:

Convex mirror. (Because it provides a wider field of view.)
– (c) Solar furnace:

Concave mirror. (Because it converges sunlight to a single point to generate high temperatures.)

 

9. One-half of a convex lens is covered with black paper. Will this lens produce a complete image of the object? Verify your answer experimentally. Explain your observations.
– Yes, the lens will produce a complete image of the object but with reduced brightness. Each part of the lens forms the complete image, so covering half will still allow the other half to produce the full image, though the intensity of light passing through will be less.

10. An object 5 cm in length is held 25 cm away from a converging lens of focal length 10 cm. Draw the ray diagram and find the position, size, and nature of the image formed.
– Use the lens formula \(\frac{1}{f} = \frac{1}{v} - \frac{1}{u}\) to find the image position.
– Here, \(u = -25\) cm and \(f = 10\) cm.
– Solving, \(\frac{1}{v} = \frac{1}{10} + \frac{1}{25} = \frac{5}{50} + \frac{2}{50} = \frac{7}{50}\) - \(v = \frac{50}{7} \approx 7.14\) cm (positive value indicates real image).
– Magnification \(m = \frac{v}{u} = \frac{7.14}{-25} \approx -0.29\).
– Image size = \(m \times \text{object size} = -0.29 \times 5 \approx -1.45\) cm (image is inverted and smaller).

Ray Diagram:

.

– Intersection of refracted rays gives the position of the image between F and 2F on the right, inverted, and smaller.

11. A concave lens of focal length 15 cm forms an image 10 cm from the lens. How far is the object placed from the lens? Draw the ray diagram.
– Use the lens formula \(\frac{1}{f} = \frac{1}{v} - \frac{1}{u}\).
– Here, \(f = -15\) cm and \(v = -10\) cm (since image is virtual).
– Solving, \(\frac{1}{u} = \frac{1}{-10} - \frac{1}{-15} = -\frac{3}{30} + \frac{2}{30} = -\frac{1}{30}\) - \(u = -30\) cm (negative value indicates object is on the same side as light entering the lens).

Ray Diagram:

– Intersection of the virtual extensions of refracted rays gives the position of the virtual image.

 

12. An object is placed at a distance of 10 cm from a convex mirror of focal length 15 cm. Find the position and nature of the image.
– Using the mirror formula \(\frac{1}{f} = \frac{1}{v} + \frac{1}{u}\).
– Here, \(f = 15\) cm and \(u = -10\) cm.
– Solving, \(\frac{1}{v} = \frac{1}{15} - \frac{1}{10} = \frac{2}{30} - \frac{3}{30} = -\frac{1}{30}\) - \(v = -30\) cm (negative value indicates a virtual image).
– The image is virtual, erect, and smaller than the object.

13. The magnification produced by a plane mirror is +1. What does this mean?
– The magnification of +1 means the image formed by a plane mirror is of the same size as the object and erect.

14. An object 5.0 cm in length is placed at a distance of 20 cm in front of a convex mirror of radius of curvature 30 cm. Find the position of the image, its nature, and size.
– For a convex mirror, the focal length \(f = \frac{R}{2} = 15\) cm.
– Use the mirror formulalatex] \(\frac{1}{f} = \frac{1}{v} + \frac{1}{u}\).[/latex]
– Here, \(u = -20\) cm and \(f = 15\) cm.
– Solving, \(\frac{1}{v} = \frac{1}{15} - \frac{1}{20} = \frac{4}{60} - \frac{3}{60} = \frac{1}{60}\) - \(v = 60\) cm (positive value indicates a virtual image).
– Magnification \(m = \frac{v}{u} = \frac{60}{-20} = -3\).
– Image size = \(m \times \text{object size} = -3 \times 5 = -15\) cm (inverted and magnified).

15. An object of size 7.0 cm is placed at 27 cm in front of a concave mirror of focal length 18 cm. At what distance from the mirror should a screen be placed, so that a sharp-focused image can be obtained? Find the size and the nature of the image.

– Using the mirror formula [lattex]\(\frac{1}{f} = \frac{1}{v} + \frac{1}{u}\).[/latex]
– Here, \(u = -27\) cm and \(f = 18\) cm.
– Solving, \(\frac{1}{v} = \frac{1}{18} - \frac{1}{-27} = \frac{3}{54} + \frac{2}{54} = \frac{5}{54}\) - \(v = \frac{54}{5} = 10.8\) cm (positive value indicates real image).
– Magnification \(m = \frac{v}{u} = \frac{10.8}{-27} \approx -0.4\).
– Image size = \(m \times \text{object size} = -0.4 \times 7 = -2.8\) cm (inverted and smaller).

16. Find the focal length of a lens of power –2.0 D. What type of lens is this?
– The focal length \(f = \frac{1}{P} = \frac{1}{-2} = -0.5\) m or -50 cm.
– This is a concave lens (negative focal length indicates diverging lens).

17. A doctor has prescribed a corrective lens of power +1.5 D. Find the focal length of the lens. Is the prescribed lens diverging or converging?
– The focal length \(f = \frac{1}{P} = \frac{1}{1.5} \approx 0.67\) m or 67 cm.
– This is a convex lens (positive focal length indicates converging lens).

 

 

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