NCERT Solutions For Class 10 Physics: Light Reflection and Refraction

---

Intext Questions (Page 143)

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

Answer-
The principal focus ($\text{F}$) of a concave mirror is a point on the principal axis where a number of rays parallel to the principal axis, after reflection, all meet or intersect.

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

Answer-
For spherical mirrors of small apertures, the relationship between the radius of curvature ($\text{R}$) and the focal length ($\text{f}$) is $\text{R} = 2\text{f}$.
Given $\text{R} = 20 \text{ cm}$.
Therefore, the focal length $\text{f} = \text{R}/2 = 20 \text{ cm} / 2 = \underline{10 \text{ cm}}$.

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 of an object. This occurs only when the object is placed between the pole ($\text{P}$) and the principal focus ($\text{F}$) of the mirror.

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

Answer-
We prefer a convex mirror as a rear-view (wing) mirror in vehicles for two reasons:

1. It always gives an erect image, though the image is diminished.
2. It has a wider field of view as its reflecting surface is curved outwards, enabling the driver to view a much larger area than would be possible with a plane mirror.

---

Intext Questions (Page 146)

Question 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?

Answer-
The light ray will bend towards the normal.
This happens because light is travelling from air (an optically rarer medium, where the speed of light is higher) to water (an optically denser medium, where the speed of light is lower). When light travels from a rarer medium to a denser medium, it slows down and bends towards the normal.

Question 2: 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 × 10$^8$ m s$^{-1}$.

Answer-
The absolute refractive index ($n_m$) of a medium is the ratio of the speed of light in air ($\text{c}$) to the speed of light in the medium ($\text{v}$):
$n_m = \frac{\text{Speed of light in air}}{\text{Speed of light in the medium}} = \frac{c}{v}$
Given: $n_m = 1.50$ and $c = 3 \times 10^8 \text{ m s}^{-1}$.
$v = \frac{c}{n_m} = \frac{3 \times 10^8 \text{ m s}^{-1}}{1.50} = \underline{2 \times 10^8 \text{ m s}^{-1}}$
The speed of light in the glass is $\underline{2 \times 10^8 \text{ m s}^{-1}}$.

Question 3: Find out, from Table 9.3, the medium having highest optical density. Also find the medium with lowest optical density.

Answer-
Optical density is related to the refractive index; the one with the larger refractive index is optically denser.

1. Medium having highest optical density: From Table 9.3, the material with the highest refractive index is $\underline{\text{Diamond}}$, with a value of 2.42.
2. Medium having lowest optical density: The medium with the lowest refractive index in the table is Air, with a value of 1.0003.

Question 4: You are given kerosene, turpentine and water. In which of these does the light travel fastest? Use the information given in Table 9.3.

Answer-
The speed of light is higher in a rarer medium than a denser medium. The rarer medium is the one with the lower refractive index.
From Table 9.3:

• Water: $n = 1.33$
• Turpentine oil: $n = 1.47$
• Kerosene: $n = 1.44$
  Water has the lowest refractive index (1.33), meaning it is the optically rarest medium among the three. Therefore, the light travels fastest in water.

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

Answer-
The refractive index of diamond, $\text{n}_{\text{m}} = 2.42$, means that the ratio of the speed of light in air (or vacuum) to the speed of light in diamond is equal to 2.42. Since the refractive index is greater than 1, it implies that the speed of light in diamond is 2.42 times slower than the speed of light in air/vacuum. Diamond is the optically densest medium listed in the table.

---

Intext Questions (Page 157)

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

Answer-
1 dioptre ($\text{1 D}$) is defined as the power of a lens whose focal length is 1 metre. This is expressed as $1\text{ D} = 1\text{ m}^{-1}$.

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 in front of the convex lens if the image is equal to the size of the object? Also, find the power of the lens.

Answer-

1. Position of the needle (object): For a convex lens to form a real and inverted image of the same size as the object, the object must be placed at $\underline{2\text{F}_1}$ (twice the focal length). In this case, the image is also formed at $\underline{2\text{F}_2}$ on the other side of the lens.
   Since the image distance ($\text{v}$) is $50 \text{ cm}$, the image is located at $2\text{F}_2$. Therefore, the object distance ($\text{u}$) must be equal to the image distance: $\text{u} = 50 \text{ cm}$ (or $-50 \text{ cm}$ using sign convention). The needle is placed $50\text{ cm in front of the convex lens}$.

2. Power of the lens ($\text{P}$):
   Since $2\text{f} = 50 \text{ cm}$, the focal length $\text{f} = 50/2 = 25 \text{ cm}$.
   We must convert $\text{f}$ to metres: $\text{f} = 0.25 \text{ m}$.
   The power $\text{P}$ is the reciprocal of the focal length in metres ($\text{P} = 1/\text{f}$).
   For a convex lens, the focal length is positive.
   $\text{P} = \frac{1}{0.25 \text{ m}} = \underline{+4.0 \text{ D}}$

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

Answer-
For a concave lens, the focal length ($\text{f}$) is negative.
Given focal length $\text{f} = -2 \text{ m}$.
The power $\text{P}$ is the reciprocal of the focal length ($\text{P} = 1/\text{f}$).
$\text{P} = \frac{1}{-2 \text{ m}} = \underline{-0.5 \text{ D}}$
The power of the concave lens is $\underline{-0.5 \text{ D}}$.

---

Exercise Questions (Page 159-160)

Question 1: Which one of the following materials cannot be used to make a lens?
(a) Water (b) Glass (c) Plastic (d) Clay

Answer-
A lens is a transparent material bound by two surfaces, of which one or both are spherical. Clay is an opaque material and therefore cannot be used to make a lens.
The correct option is (d) Clay.

Question 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?
(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-
For a concave mirror, the image is formed as Virtual and erect and Enlarged only when the object is placed Between P and F (the pole and the principal focus).
The correct option is (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 size of 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-
For a convex lens, to obtain a Real and inverted image that is the Same size as the object, the object must be placed At 2F$_1$ (twice the focal length).
The correct option is (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-
The focal length ($\text{f}$) is given as negative (–15 cm).

1. Mirror: According to the New Cartesian Sign Convention, the focal length of a concave mirror is taken as negative (measured to the left of the pole).
2. Lens: According to the sign convention for lenses, the focal length of a concave lens is taken as negative.
   The correct option is (a) both concave.

Question 5: No matter how far you stand from a mirror, your image appears erect. The mirror is likely to be
(a) only plane.
(b) only concave.
(c) only convex.
(d) either plane or convex.

Answer-

1. Plane Mirror: The image formed by a plane mirror is always virtual and erect.
2. Convex Mirror: A convex mirror always gives an erect, though diminished, image, irrespective of the object position.
3. Concave Mirror: A concave mirror gives an erect image only when the object is placed between P and F, otherwise the image is real and inverted.
   Since the image is always erect regardless of distance, the mirror is likely to be (d) either plane or convex.

Question 6: Which of the following lenses would you prefer to use while reading small letters found 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-
To read small letters, a magnifying glass (a simple microscope) is needed. This requires a lens that produces an enlarged, virtual, and erect image.

1. Lens Type: This is achieved by a convex lens when the object is placed between the optical centre ($\text{O}$) and the principal focus ($\text{F}$).
2. Magnification/Focal Length: A lens with a shorter focal length has a higher power and thus causes higher magnification.
   Therefore, the preferred lens is (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-
To obtain an erect image using a concave mirror, the image must be Virtual.

1. Range of Distance: The object must be placed between the pole ($\text{P}$) and the principal focus ($\text{F}$). Since the focal length ($\text{f}$) is $15 \text{ cm}$, the range of distance of the object from the mirror should be between 0 cm and 15 cm.
2. Nature and Size: The image formed in this case is Virtual and erect. The image is Enlarged (magnified).
3. Ray Diagram: (Requires drawing a ray diagram with the object placed between P and F. Two rays are sufficient: a ray parallel to the principal axis reflects through F, and a ray directed towards C reflects back along C. The backward extension of these reflected rays meet behind the mirror, forming an enlarged, virtual, and erect image.)

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.
Support your answer with reason.

Answer-
(a) Concave mirror.

• Reason: Concave mirrors are used in headlights to get powerful parallel beams of light. The light source is placed at the focus ($\text{F}$) of the concave mirror, so that the light rays reflecting off the mirror emerge parallel to the principal axis.

(b) Convex mirror.

• Reason: Convex mirrors are preferred because they always give an erect, though diminished, image, and they have a wider field of view as they are curved outwards, allowing the driver to see a much larger area.

(c) Concave mirror.

• Reason: Large concave mirrors are used to concentrate sunlight to produce heat in solar furnaces. Parallel rays from the Sun are converged at the principal focus ($\text{F}$) of the mirror, generating high heat.

Question 9: One-half of a convex lens is covered with a black paper. Will this lens produce a complete image of the object? Verify your answer experimentally. Explain your observations.

Answer-
Yes, the lens will produce a complete image of the object.

Explanation of Observations:
Image formation, whether by mirrors or lenses, is achieved by considering the intersection of at least two reflected or refracted rays emanating from every point of the object.
When half of the lens is covered, the lens is still capable of refracting light and forming an image. The rays passing through the covered portion are blocked, but the rays passing through the exposed half still refract and converge (or diverge) to form the image. The image will be less bright or diminished in intensity because fewer light rays are incident on the screen, but it will still be complete.

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 the nature of the image formed.

Answer-
A converging lens is a convex lens.
Given: Object height $h = +5.0 \text{ cm}$; Object distance $u = -25 \text{ cm}$; Focal length $f = +10 \text{ cm}$ (positive for convex lens).

Position of the image ($\text{v}$): Using the lens formula $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$:
$\frac{1}{v} = \frac{1}{f} + \frac{1}{u} = \frac{1}{+10} + \frac{1}{-25} = \frac{5-2}{50} = \frac{3}{50}$
$v = \frac{50}{3} \approx \underline{+16.67 \text{ cm}}$
The image is formed $\underline{16.67 \text{ cm on the other side of the lens}}$.

Nature of the image: Since $\text{v}$ is positive, the image is formed on the opposite side of the lens, meaning the image is Real and inverted.

Size of the image ($\text{h'}$): Using magnification $m = \frac{h'}{h} = \frac{v}{u}$:
$h' = h \frac{v}{u} = (+5.0 \text{ cm}) \frac{+16.67 \text{ cm}}{-25 \text{ cm}} \approx \underline{-3.33 \text{ cm}}$
The image size is $3.33 \text{ cm}$. The negative sign confirms the image is inverted and diminished (smaller than the object).

Question 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.

Answer-
A concave lens has a negative focal length and always forms a virtual, erect image on the same side as the object.
Given: Focal length $f = -15 \text{ cm}$; Image distance $v = -10 \text{ cm}$ (same side as object); Object distance $u = ?$

Using the lens formula $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$:
$\frac{1}{u} = \frac{1}{v} - \frac{1}{f} = \frac{1}{-10} - \frac{1}{-15} = -\frac{1}{10} + \frac{1}{15}$
$\frac{1}{u} = \frac{-3 + 2}{30} = -\frac{1}{30}$
$u = \underline{-30 \text{ cm}}$
The object is placed $\underline{30 \text{ cm in front of the concave lens}}$.

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-
For a convex mirror: Object distance $u = -10 \text{ cm}$; Focal length $f = +15 \text{ cm}$ (positive for convex mirror).

Position of the image ($\text{v}$): Using the mirror formula $\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$:
$\frac{1}{v} = \frac{1}{f} - \frac{1}{u} = \frac{1}{+15} - \frac{1}{-10} = \frac{1}{15} + \frac{1}{10}$
$\frac{1}{v} = \frac{2 + 3}{30} = \frac{5}{30} = \frac{1}{6}$
$v = \underline{+6 \text{ cm}}$
The image is formed at $\underline{6 \text{ cm behind the mirror}}$.

Nature of the image: Since $\text{v}$ is positive, the image is formed behind the mirror. A convex mirror always forms a Virtual and erect image, and it will be Diminished.

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

Answer-
Magnification ($\text{m}$) is the ratio of image height ($\text{h'}$) to object height ($\text{h}$).
The magnification $m = +1$ means two things:

1. Size: The magnitude 1 means that the size of the image is equal to that of the object ($h' = h$).
2. Nature: The positive sign indicates that the image is Virtual and erect.

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 $h = +5.0 \text{ cm}$; Object distance $u = -20 \text{ cm}$; Radius of curvature $R = +30 \text{ cm}$.
Focal length $f = R/2 = +30/2 = +15 \text{ cm}$.

Position of the image ($\text{v}$): Using the mirror formula $\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$:
$\frac{1}{v} = \frac{1}{f} - \frac{1}{u} = \frac{1}{+15} - \frac{1}{-20} = \frac{1}{15} + \frac{1}{20}$
$\frac{1}{v} = \frac{4 + 3}{60} = \frac{7}{60}$
$v = \frac{60}{7} \approx \underline{+8.57 \text{ cm}}$
The image is formed at $\underline{8.57 \text{ cm behind the mirror}}$.

Nature of the image: Since $v$ is positive, the image is Virtual and erect.

Size of the image ($\text{h'}$): Using magnification $m = -\frac{v}{u} = \frac{h'}{h}$:
$h' = -h \frac{v}{u} = -(+5.0 \text{ cm}) \frac{+8.57 \text{ cm}}{-20 \text{ cm}} \approx \underline{+2.14 \text{ cm}}$
The image size is $2.14 \text{ cm}$. Since $m$ is positive and $h'$ is positive, the image is erect and diminished (smaller than the object).

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 focussed image can be obtained? Find the size and the nature of the image.

Answer-
A concave mirror can form a sharp image on a screen, meaning the image is Real.
Given: Object height $h = +7.0 \text{ cm}$; Object distance $u = -27 \text{ cm}$; Focal length $f = -18 \text{ cm}$ (negative for concave mirror).
(Note: Since $u = 27 \text{ cm}$ and $f = 18 \text{ cm}$, the object is placed beyond $\text{C}$ since $C = 2f = 36 \text{ cm}$).

Position of the screen ($\text{v}$): Using the mirror formula $\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$:
$\frac{1}{v} = \frac{1}{f} - \frac{1}{u} = \frac{1}{-18} - \frac{1}{-27} = -\frac{1}{18} + \frac{1}{27}$
$\frac{1}{v} = \frac{-3 + 2}{54} = -\frac{1}{54}$
$v = \underline{-54 \text{ cm}}$
The screen should be placed at $\underline{54 \text{ cm in front of the mirror}}$.

Nature of the image: Since $\text{v}$ is negative, the image is formed in front of the mirror, meaning the image is Real and inverted.

Size of the image ($\text{h'}$): Using magnification $m = -\frac{v}{u} = \frac{h'}{h}$:
$h' = -h \frac{v}{u} = -(+7.0 \text{ cm}) \frac{-54 \text{ cm}}{-27 \text{ cm}} = \underline{-14.0 \text{ cm}}$
The image size is $14.0 \text{ cm}$. The negative sign confirms the image is inverted and enlarged.

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

Answer-
Given: Power $P = -2.0 \text{ D}$.
The power of a lens is the reciprocal of its focal length in metres ($\text{P} = 1/\text{f}$).
$f = \frac{1}{P} = \frac{1}{-2.0 \text{ D}} = \underline{-0.5 \text{ m}}$
The focal length is $\underline{-0.5 \text{ m}}$ (or $-50 \text{ cm}$).
Since the power and focal length are negative, the lens is a concave lens (or diverging 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 $P = +1.5 \text{ D}$.
The focal length is calculated as $f = 1/P$:
$f = \frac{1}{+1.5 \text{ D}} \approx \underline{+0.67 \text{ m}}$
The focal length is approximately $\underline{0.67 \text{ m}}$ (or $66.7 \text{ cm}$).
Since the power and focal length are positive, the lens is a convex lens. A convex lens is also called a converging lens.