Class 10 Science - Chapter 10: Light - Reflection and Refraction

Here is the Blogger post content up to Page 184 with proper mathematical solutions for Class 10 Science - Chapter 10: Light - Reflection and Refraction.

---

Class 10 Science - Chapter 10: Light - Reflection and Refraction

Textual Questions and Answers

---

Page – 168

Question 1: Define the principal focus of a concave mirror.

Answer: The principal focus of a concave mirror is a point on its principal axis where all parallel light rays close to the axis converge after reflection from the mirror.

---

Question 2: The radius of curvature of a spherical mirror is 20 cm. What is its focal length?

Answer:
We know that the relationship between radius of curvature (R) and focal length (f) of a spherical mirror is:

$$f = \frac{R}{2}$$

Given,
R = 20 cm

$$f = \frac{20}{2} = 10 \text{ cm}$$

Thus, the focal length of the mirror is 10 cm.

---

Question 3: Name a mirror that can give an erect and enlarged image of an object.

Answer: A concave mirror can give an erect and enlarged image when the object is placed between the focus (F) and the pole (P) of the mirror.

---

Question 4: Why do we prefer a convex mirror as a rear-view mirror in vehicles?

Answer: A convex mirror is used as a rear-view mirror in vehicles because:

1. It provides a wider field of view, allowing the driver to see more of the traffic behind.
2. It always forms a virtual, erect, and diminished image, ensuring that objects appear smaller and fit within the mirror’s frame.

---

Page – 171

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

Answer:
Using the formula:

$$f = \frac{R}{2}$$

Given, R = 32 cm

$$f = \frac{32}{2} = 16 \text{ cm}$$

Thus, the focal length of the convex mirror is 16 cm.

---

Question 2: A concave mirror produces a three-times magnified (enlarged) real image of an object placed at 10 cm in front of it. Where is the image located?

Answer:
Given:

• Magnification (m) = –3 (negative for real images)
• Object distance (u) = –10 cm (according to the sign convention)
• Image distance (v) = ?

We use the magnification formula:

$$m = \frac{-v}{u}$$ $$-3 = \frac{-v}{-10}$$ $$v = 3 \times 10$$ $$v = -30 \text{ cm}$$

Thus, the image is located 30 cm in front of the mirror.

---

Page – 176

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

Answer: The light ray bends towards the normal because water is optically denser than air. The speed of light decreases when it enters a denser medium, causing it to bend towards the normal.

---

Question 2: Light enters from air into glass having a refractive index of 1.50. What is the speed of light in glass? (Speed of light in vacuum = 3 × 10⁸ m/s)

Answer:
We use the formula:

$$n = \frac{\text{Speed of light in air}}{\text{Speed of light in medium}}$$

Rearranging,

$$\text{Speed of light in glass} = \frac{\text{Speed of light in air}}{n}$$ $$= \frac{3 \times 10^8}{1.50}$$ $$= 2 \times 10^8 \text{ m/s}$$

Thus, the speed of light in glass is $2 \times 10^8$ m/s.

Question 3: Find out from table 10.3 the medium having highest optical density. Also find the medium with lowest optical density.

Answer: The medium having highest refractive index has the highest optical density. Therefore diamond has the highest optical density.

The medium having lowest refractive index has the lowest optical density. Therefore air has the lowest optical density.

Question 4: You are given kerosene, turpentine and water. In which of these the light travels fastest? Use the information given in Table on page 225.

Answer: For kerosene, n = 1.44 For turpentine, n = 1.47
For water, n = 1.33

Because water has the lowest refractive index, therefore light travels fastest in this optically rarer medium than kerosene and turpentine oil.

Question 5: The refractive index of diamond is 2.42. What is the meaning of this statement?

Answer: The refractive index of diamond is 2.42. This means that the speed of light in diamond will reduce by a factor of 2.42 as compared to its speed in air.
 

---

Page – 184

Question 9: Define 1 dioptre of power of a lens.

Answer:
Dioptre (D) is the SI unit of lens power. 1 dioptre is the power of a lens with a focal length of 1 metre.

---

Question 2: 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 if the image size is equal to the object size? Also, find the power of the lens.

Answer:
Given:

• Image distance (v) = +50 cm
• Object distance (u) = ?

For an equal-sized real image, the object must be placed at 2F (twice the focal length).

Thus,

$$u = -50 \text{ cm}$$

Now, using the lens formula:

$$\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$$ $$\frac{1}{f} = \frac{1}{50} - \frac{1}{-50}$$ $$\frac{1}{f} = \frac{1}{50} + \frac{1}{50}$$ $$\frac{1}{f} = \frac{2}{50}$$ $$f = \frac{50}{2} = 25 \text{ cm} = 0.25 \text{ m}$$

Now, calculating Power (P):

$$P = \frac{1}{f}$$ $$P = \frac{1}{0.25}$$ $$P = +4 D$$

Thus,

• The needle is placed 50 cm in front of the lens.
• The power of the convex lens is +4 dioptres.

---

Question 3: Find the power of a concave lens of focal length 2m.

Answer:
Given: Focal length (f) = -2 m (negative for concave lens)

$$P = \frac{1}{f}$$ $$P = \frac{1}{-2}$$ $$P = -0.5 D$$

Thus, the power of the concave lens is -0.5 dioptres.

---

Class 10 Science - Chapter 10: Light - Reflection and Refraction

Exercises Questions and Answers

---

Question 1: Which one of the following materials cannot be used to make a lens?

(a) Water
(b) Glass
(c) Plastic
(d) Clay

Answer: (d) Clay

---

Question 2: The image formed by a concave mirror is observed to be virtual, erect, and larger than the object. Where should the position of the object be?

(a) Between the principal focus and the centre of curvature
(b) At the centre of curvature
(c) Beyond the centre of curvature
(d) Between the pole of the mirror and its principal focus

Answer: (d) Between the pole of the mirror and its principal focus

---

Question 3: Where should an object be placed in front of a convex lens to get a real image of the same size as the object?

(a) At the principal focus of the lens
(b) At twice the focal length
(c) At infinity
(d) Between the optical centre of the lens and its principal focus

Answer: (b) At twice the focal length

---

Question 4: A spherical mirror and a thin spherical lens have each a focal length of -15 cm. The mirror and the lens are likely to be:

(a) Both concave
(b) Both convex
(c) The mirror is concave, and the lens is convex
(d) The mirror is convex, but the lens is concave

Answer: (a) Both concave

---

Question 5: No matter how far you stand from a mirror, your image always appears erect. The mirror is likely to be:

(a) Plane
(b) Concave
(c) Convex
(d) Either plane or convex

Answer: (d) Either plane or convex

---

Question 6: Which of the following lenses would you prefer to use while reading small letters in a dictionary?

(a) A convex lens of focal length 50 cm
(b) A concave lens of focal length 50 cm
(c) A convex lens of focal length 5 cm
(d) A concave lens of focal length 5 cm

Answer: (c) A convex lens of focal length 5 cm

---

Question 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.

Answer:

• The object should be placed between the pole (P) and the focus (F) of the concave mirror.
• The image formed will be virtual, erect, and magnified.
• The image will be larger than the object.
• Ray diagram: (Students should draw a ray diagram showing the object placed between P and F, forming a virtual, erect, and enlarged image behind the mirror.)

---

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

(a) Headlights of a car
(b) Side/rear-view mirror of a vehicle
(c) Solar furnace

Answer:
(a) Concave mirror – Produces a parallel beam of light after reflection.
(b) Convex mirror – Provides a wide field of view and forms an erect, virtual, and diminished image.
(c) Concave mirror – Concentrates sunlight at a single focal point to produce heat.

---

Question 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.

Answer:
Yes, the convex lens will still produce a complete image of the object, but the brightness of the image will be reduced.

Explanation:

• Even if half of the lens is covered, light rays from the object will still pass through the uncovered part and get refracted to form a complete image.
• However, fewer rays will contribute to the formation of the image, reducing its intensity.

Experimental Verification:

• Place a convex lens in front of a light source and cover half of the lens.
• Observe the image on a screen; it remains complete but dimmer.

---

Question 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.

Answer:

Given that,

• Object distance, u = – 25 cm
• Focal length, f = +10 cm
• Image distance, v = ?

According to the lens formula:

$$\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$$

Substituting the values:

$$\frac{1}{v} - \frac{1}{-25} = \frac{1}{10}$$ $$\frac{1}{v} + \frac{1}{25} = \frac{1}{10}$$ $$\frac{1}{v} = \frac{1}{10} - \frac{1}{25}$$ $$= \frac{5 - 2}{50}$$ $$= \frac{3}{50}$$ $$\therefore v = \frac{50}{3} = +16.67 \text{ cm}$$

Thus, the position of the image is at a distance of 16.67 cm from the lens.
The positive sign of the image distance indicates that the image is formed on the right side of the lens, meaning the image is real and inverted.

Now, calculating magnification:

$$m = \frac{v}{u}$$ $$= \frac{16.67}{-25}$$ $$= -0.66$$

Now, calculating the size of the image:

Using the magnification formula:

$$m = \frac{h'}{h}$$ $$h' = m \times h$$ $$= -0.66 \times 5$$ $$= -3.3 \text{ cm}$$

Thus, the size of the image is 3.3 cm.
The negative sign indicates that the image is inverted.

---

Question 11: A concave lens of focal length 15 cm forms an image 10 cm from the lens. How far is the object from the lens? Draw the ray diagram.

Answer:

Given:

• Focal length (f) = -15 cm (negative for concave lens)
• Image distance (v) = -10 cm
• Object distance (u) = ?

Using the lens formula:

$$\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$$ $$\frac{1}{-15} = \frac{1}{-10} - \frac{1}{u}$$ $$\frac{1}{u} = \frac{1}{-10} - \frac{1}{-15}$$ $$\frac{1}{u} = \frac{-3 + 2}{30}$$ $$\frac{1}{u} = \frac{-1}{30}$$ $$u = -30 \text{ cm}$$

Thus, the object is placed 30 cm in front of the concave lens.

Nature of Image:

• Virtual, erect, and diminished.

(Students should draw a ray diagram showing image formation by a concave lens.)

---

Question 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.

Answer:

Given:

• Object distance (u) = -10 cm (negative as per sign convention)
• Focal length (f) = +15 cm (positive for convex mirrors)
• Image distance (v) = ?

Using the mirror formula:

$$\frac{1}{f} = \frac{1}{v} + \frac{1}{u}$$ $$\frac{1}{15} = \frac{1}{v} + \frac{1}{-10}$$ $$\frac{1}{v} = \frac{1}{15} + \frac{1}{10}$$ $$\frac{1}{v} = \frac{2 + 3}{30}$$ $$\frac{1}{v} = \frac{5}{30}$$ $$v = 6 \text{ cm}$$

Thus, the image is formed at 6 cm behind the convex mirror.

magnification ($m$), we use the magnification formula for mirrors:

$$m = \frac{-v}{u}$$$$m = \frac{-6}{-10}$$ $$m = 0.6$$

 Nature of Image:

• Virtual and erect (as convex mirrors always form virtual images).
• Diminished in size.

---

Question 13: The magnification produced by a plane mirror is +1. What does this mean?

Answer:

Magnification (m) is given by:

$$m = \frac{\text{Image height} (h')} {\text{Object height} (h)}$$

Since m = +1, this means:

• The image height is equal to the object height.
• The image is erect (as the magnification is positive).
• The image is virtual and formed at the same distance behind the mirror as the object is in front.

Thus, a plane mirror always forms a virtual, erect, and same-sized image.

---

Question 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.

Answer:

Given:

• Object height (h) = 5 cm
• Object distance (u) = -20 cm
• Radius of curvature (R) = +30 cm
• Focal length (f) = R/2 = 30/2 = 15 cm
• Image distance (v) = ?

Using the mirror formula:

$$\frac{1}{f} = \frac{1}{v} + \frac{1}{u}$$ $$\frac{1}{15} = \frac{1}{v} + \frac{1}{-20}$$ $$\frac{1}{v} = \frac{1}{15} + \frac{1}{20}$$ $$\frac{1}{v} = \frac{4 + 3}{60}$$ $$\frac{1}{v} = \frac{7}{60}$$ $$v = 8.57 \text{ cm}$$

Thus, the image is formed at 8.57 cm behind the mirror.

Magnification (m):

$$m = \frac{-v}{u} = \frac{-8.57}{-20} = 0.43$$

Image height (h’):

$$h' = m \times h = (0.43) \times 5 = 2.14 \text{ cm}$$

Nature of Image:

• Virtual and erect.
• Diminished in size (2.14 cm).

---

Question 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.

Answer:

Given:

• Object height (h) = 7 cm
• Object distance (u) = -27 cm
• Focal length (f) = -18 cm
• Image distance (v) = ?

Using the mirror formula:

$$\frac{1}{f} = \frac{1}{v} + \frac{1}{u}$$ $$\frac{1}{-18} = \frac{1}{v} + \frac{1}{-27}$$ $$\frac{1}{v} = \frac{1}{-18} + \frac{1}{27}$$ $$\frac{1}{v} = \frac{-3 + 2}{54}$$ $$\frac{1}{v} = \frac{-1}{54}$$ $$v = -54 \text{ cm}$$

Thus, the screen should be placed 54 cm in front of the concave mirror.

Magnification (m):

$$m = \frac{-v}{u} = \frac{-(-54)}{-27} = -2$$

Image height (h’):

$$h' = m \times h = (-2) \times 7 = -14 \text{ cm}$$

Nature of Image:

• Real and inverted.
• Magnified (14 cm in size).

---

Question 16: Find the focal length of a lens of power -2.0 D. What type of lens is this?

Answer:

Given:

• Power (P) = -2.0 D
• Focal length (f) = ?

Using the power formula:

$$P = \frac{1}{f}$$ $$f = \frac{1}{P} = \frac{1}{-2}$$ $$f = -0.5 \text{ m} = -50 \text{ cm}$$

Since the focal length is negative, the lens is a concave lens.

---

Question 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?

Answer:

Given:

• Power (P) = +1.5 D
• Focal length (f) = ?

Using the power formula:

$$P = \frac{1}{f}$$ $$f = \frac{1}{P} = \frac{1}{1.5}$$ $$f = 0.67 \text{ m} = 67 \text{ cm}$$

Since the focal length is positive, the lens is a convex lens (converging lens).

---

Conclusion: