NCERT Textbook Questions & Answers + Important Extra Questions
Section A: In-Text Questions
Define the principal focus of a concave mirror.
Answer
The principal focus of a concave mirror is the point on the principal axis where light rays that are parallel to the principal axis converge (meet) after reflection from the mirror. It is denoted by F and lies in front of the concave mirror, between the pole and the centre of curvature. The distance between the pole and the principal focus is called the focal length (f), which is equal to half the radius of curvature (f = R/2).
The radius of curvature of a spherical mirror is 20 cm. What is its focal length?
Answer
Focal length = Radius of curvature ÷ 2
f = R/2 = 20/2 = 10 cm
Name a mirror that can give an erect and enlarged image of an object.
Answer
A concave mirror produces an erect and enlarged (magnified) image when the object is placed between the pole (P) and the focus (F). The image formed is virtual, erect, and larger than the object. This property is used in shaving mirrors and dental mirrors.
Why do we prefer a convex mirror as a rear-view (side) mirror in vehicles?
Answer
Convex mirrors are preferred as rear-view mirrors because:
(i) They always form erect (upright) images, making it easier for the driver to judge the position of vehicles behind.
(ii) The images are diminished (smaller), allowing a wider field of view — the driver can see a much larger area behind the vehicle compared to a plane or concave mirror.
(iii) The image is always virtual and on the same side as the object, providing a consistent, clear view regardless of the distance of the object.
Find the focal length of a convex mirror whose radius of curvature is 32 cm.
Answer
f = R/2 = 32/2 = 16 cm
(The focal length of a convex mirror is positive by sign convention.)
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?
Answer
Given: magnification m = –3 (real image, so negative), object distance u = –10 cm
m = –v/u
⇒ –3 = –v/(–10)
⇒ –3 = v/10
⇒ v = –30 cm
The image is located 30 cm in front of the mirror (on the same side as the object).
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 bends towards the normal. This happens because water is optically denser than air (has a higher refractive index). When light passes from a rarer medium (air) to a denser medium (water), it slows down, and the change in speed causes it to bend toward the normal line at the point of incidence. This phenomenon is called refraction.
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.
Answer
Refractive index (n) = Speed of light in vacuum / Speed of light in medium
1.50 = (3 × 10⁸) / v
v = (3 × 10⁸) / 1.50 = 2 × 10⁸ m/s
Find out from Table 10.3 the medium having the highest optical density. Also find the medium with the lowest optical density.
Answer
Highest optical density: Diamond (refractive index = 2.42) — light travels slowest in diamond.
Lowest optical density: Air (refractive index = 1.0003) — light travels fastest in air (closest to its vacuum speed).
You are given kerosene, turpentine and water. In which of these does the light travel fastest? Use the information given in Table 10.3.
Answer
Refractive indices: Water = 1.33, Kerosene = 1.44, Turpentine = 1.47
Light travels fastest in the medium with the lowest refractive index.
Therefore, light travels fastest in water.
The refractive index of diamond is 2.42. What is the meaning of this statement?
Answer
This statement means that the speed of light in diamond is 1/2.42 times the speed of light in vacuum (or air). In other words, light travels 2.42 times slower in diamond compared to vacuum. It also means that the ratio of the sine of the angle of incidence to the sine of the angle of refraction is 2.42 when light enters diamond from air.
Section B: NCERT Exercise Questions
Which one of the following materials cannot be used to make a lens?
Answer
(a) Water (b) Glass (c) Plastic (d) Clay
Ans. (d) Clay. A lens must be made of a transparent material that allows light to pass through it. Water, glass, and plastic are all transparent and can be used to make lenses. Clay is opaque and does not transmit light.
The image formed by a concave mirror is observed to be virtual, erect and larger than the object. Where should the object be placed?
Answer
The object should be placed between the pole (P) and the principal focus (F) of the concave mirror. Only in this position does a concave mirror produce a virtual, erect, and magnified image (formed behind the mirror).
Where should an object be placed in front of a convex lens to get a real image of the size of the object?
Answer
The object should be placed at 2F₁ (at the centre of curvature), i.e., at a distance equal to twice the focal length from the lens. The image formed will be real, inverted, and the same size as the object, located at 2F₂ on the other side of the lens.
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:
Answer
(a) both concave (b) both convex (c) the mirror is concave and the lens is convex (d) the mirror is convex and the lens is concave
Ans. (a) Both concave. By the sign convention, a negative focal length indicates a concave mirror and a concave (diverging) lens.
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: u = –10 cm, f = +15 cm (convex mirror)
Using mirror formula: 1/v + 1/u = 1/f
1/v = 1/f – 1/u = 1/15 – 1/(–10) = 1/15 + 1/10 = (2+3)/30 = 5/30 = 1/6
v = +6 cm
Magnification m = –v/u = –6/(–10) = +0.6
The image is formed 6 cm behind the mirror. It is virtual, erect, and diminished (0.6 times the object size).
The magnification produced by a plane mirror is +1. What does this mean?
Answer
A magnification of +1 means:
• The size of the image is equal to the size of the object (magnification = 1 means same size).
• The positive sign indicates the image is erect (upright).
• The image formed by a plane mirror is always virtual, erect, and of the same size as the object, with left-right lateral inversion.
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: h = 5 cm, u = –20 cm, R = 30 cm
⇒ f = R/2 = +15 cm
Mirror formula: 1/v + 1/u = 1/f
1/v = 1/15 – 1/(–20) = 1/15 + 1/20 = (4+3)/60 = 7/60
v = 60/7 = +8.57 cm
m = –v/u = –(8.57)/(–20) = +0.43
Image height h' = m × h = 0.43 × 5 = 2.15 cm
The image is 8.57 cm behind the mirror, virtual, erect, and diminished (2.15 cm tall).
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 nature of the image.
Answer
Given: h = 7 cm, u = –27 cm, f = –18 cm
1/v = 1/f – 1/u = 1/(–18) – 1/(–27) = –1/18 + 1/27 = (–3+2)/54 = –1/54
v = –54 cm
m = –v/u = –(–54)/(–27) = –2
h' = m × h = –2 × 7 = –14 cm
Screen should be placed 54 cm in front of the mirror. Image is real, inverted, and enlarged (14 cm, twice the object size).
Find the focal length of a lens of power –2.0 D. What type of lens is this?
Answer
Power P = 1/f (where f is in metres)
f = 1/P = 1/(–2.0) = –0.5 m = –50 cm
Since the focal length is negative, the lens is a concave lens (diverging lens).
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
f = 1/P = 1/1.5 = +0.67 m = +66.7 cm
Since the power and focal length are positive, it is a converging (convex) lens. This lens is typically prescribed for hypermetropia (far-sightedness).
Section C: Important Extra Questions
State the laws of reflection.
Answer
First law: The angle of incidence (i) is always equal to the angle of reflection (r). i.e., ∠i = ∠r
Second law: The incident ray, the reflected ray, and the normal at the point of incidence all lie in the same plane.
State the laws of refraction (Snell's Law).
Answer
First law: The incident ray, the refracted ray, and the normal at the point of incidence all lie in the same plane.
Second law (Snell's Law): For a given pair of media, the ratio of the sine of the angle of incidence to the sine of the angle of refraction is constant.
sin i / sin r = n (refractive index) = constant
Distinguish between real and virtual images.
Answer
Real image: Formed when light rays actually converge at a point. Can be obtained on a screen. Always inverted. Formed by concave mirrors and convex lenses.
Virtual image: Formed when light rays appear to diverge from a point (don't actually meet). Cannot be obtained on a screen. Always erect. Formed by plane mirrors, convex mirrors, and concave lenses.
What is the sign convention used for mirrors and lenses?
Answer
The New Cartesian Sign Convention:
(i) All distances are measured from the pole (P) of the mirror or optical centre (O) of the lens.
(ii) Distances measured in the direction of incident light are taken as positive.
(iii) Distances measured against the direction of incident light are taken as negative.
(iv) Heights measured upward (above principal axis) are positive; downward are negative.
(v) Object distance (u) is always negative (object is always placed on the left side).
An object is placed at a distance of 15 cm from a convex lens of focal length 10 cm. Find the position and nature of the image.
Answer
Given: u = –15 cm, f = +10 cm
Lens formula: 1/v – 1/u = 1/f
1/v = 1/f + 1/u = 1/10 + 1/(–15) = 1/10 – 1/15 = (3–2)/30 = 1/30
v = +30 cm
m = v/u = 30/(–15) = –2
Image is at 30 cm on the other side of the lens. It is real, inverted, and magnified (2 times the object size).