Class 10 Science · Chapter 9

Light - Reflection and Refraction

409 Class 10 Science questions on Light - Reflection and Refraction from the NCERT textbook, NCERT Exemplar and CBSE board papers 2017–2026, each with its answer and a step-by-step solution.

NCERT exercises

Exercises (17) Intext Questions (Page 142) (4) Intext Questions (Page 145) (2) Intext Questions (Page 150) (5) Intext Questions (Page 158) (3)

NCERT Exemplar exercises

Multiple Choice Questions (19) Short Answer Questions (10) Long Answer Questions (9)

CBSE Class 10 Science previous year papers, with answers →

Questions · page 13 of 21

Q241 CBSE 2020
The diagram given below shows an object O and its image I. Without actually drawing the ray diagram, state the following :
1. (i)Type of lens (Converging / Diverging)
2. (ii)Name two optical instruments where such an image is obtained.
3. (iii)List three characteristics of the image formed if this lens is replaced by a concave mirror of focal length ‘ $f$ ’ and an object is placed at a distance ‘ $f /2$ ’ in front of the mirror.
Show answer

Converging lens; magnifying glass and microscope; the concave mirror gives a virtual, erect image, magnified to twice the object's size.
(i) Converging lens (a convex lens). (ii) Magnifying glass (simple microscope) and microscope. (iii) The image is virtual, erect and magnified: with $u = - \frac{f}{2}$ the mirror formula gives $v = + f$ (behind the mirror) and $m = + 2$ , so the image is twice the size of the object.

Step-by-step solution →
Q242 CBSE 2020
1. (a)Water has refractive index 1.33 and alcohol has refractive index 1.36. Which of the two medium is optically denser ? Give reason for your answer.
2. (b)Draw a ray diagram to show the path of a ray of light passing obliquely from water to alcohol.
3. (c)State the relationship between angle of incidence and angle of refraction in the above case.
Show answer

Alcohol is optically denser (higher refractive index); the ray bends towards the normal on entering it, so $i > r$ .
(a) Alcohol is optically denser, because its refractive index ( $1.36$ ) is higher than that of water ( $1.33$ ): a medium with a higher refractive index is optically denser, and light travels more slowly in it. (b) In the ray diagram the ray in water meets the water–alcohol surface obliquely at angle $i$ to the normal and, on entering alcohol, bends towards the normal, making a smaller angle $r$ . (c) The angle of incidence is greater than the angle of refraction, $i > r$ ; by Snell’s law $\frac{s i n i}{s i n r}$ is constant, equal to $\frac{1.36}{1.33} \approx 1.023$ .

Step-by-step solution →
Q243 CBSE 2020
1. (a)State Snell’s law of refraction of light.
2. (b)When a ray of light travelling in air enters obliquely into a glass slab, it is observed that the light ray emerges parallel to the incident ray but it is shifted sideways slightly. Draw a labelled ray diagram to illustrate it.
Show answer

$\frac{s i n i}{s i n r}$ is constant; in a glass slab the emergent ray is parallel to the incident ray but laterally shifted.
(a) Snell's law: the ratio of the sine of the angle of incidence to the sine of the angle of refraction is a constant for light of a given colour and for the given pair of media, $\frac{s i n i}{s i n r} = constant$ . This constant is the refractive index of the second medium with respect to the first. (b) Labelled ray diagram (see the figure): the incident ray strikes the top face at O at angle of incidence $i$ and bends towards the normal NN′, with angle of refraction $r < i$ ; the refracted ray OO′ meets the bottom face at O′ and bends away from the normal MM′, leaving as the emergent ray at angle of emergence $e = i$ . So the emergent ray is parallel to the incident ray, and the perpendicular distance $d$ between the incident ray produced and the emergent ray is the lateral displacement.

Step-by-step solution →
Q244 CBSE 2020

The refractive index of a medium ‘x’ with respect to a medium ‘y’ is $2/3$ and the refractive index of medium ‘y’ with respect to medium ‘z’ is $4/3$ . Find the refractive index of medium ‘z’ with respect to medium ‘x’. If the speed of light in medium ‘x’ is $3 \times 1 0^{8} ms^{- 1}$ , calculate the speed of light in medium ‘y’.

Show answer

$n_{z x} = \frac{9}{8}$ and the speed of light in medium y is $2 \times 1 0^{8} m s^{- 1}$ .
The refractive index of z with respect to x is $n_{z x} = n_{z y} \times n_{y x} = \frac{3}{4} \times \frac{3}{2} = \frac{9}{8}$ (that is, $1.125$ ). The speed of light in medium y is $v_{y} = \frac{v _{x}}{n _{y x}} = \frac{3 \times 1 0 ^{8}}{3/2} = 2 \times 1 0^{8} m s^{- 1}$ .

Step-by-step solution →
Q245 CBSE 2020
1. (a)For the same angle of incidence of $4 5^{\circ}$ , the refraction angle in two transparent media P and Q is $2 0^{\circ}$ and $3 0^{\circ}$ respectively. Which of the two is optically denser and why ?
2. (b)Define 1 dioptre power of a lens.
3. (c)Find the focal length of a lens of power $+ 0.5$ D.
Show answer

P is optically denser (smaller angle of refraction); 1 dioptre is the power of a lens of focal length 1 m; $f = + 2$ m.
(a) P is optically denser. For the same angle of incidence, $4 5^{\circ}$ , the ray bends more towards the normal in P ( $r = 2 0^{\circ}$ ) than in Q ( $r = 3 0^{\circ}$ ); the smaller angle of refraction means a larger refractive index ( $n_{P} \approx 2.07$ , $n_{Q} \approx 1.41$ ), so light travels more slowly in P. (b) One dioptre is the power of a lens whose focal length is $1$ metre. (c) $f = \frac{1}{P} = \frac{1}{+ 0.5 D} = + 2$ m, that is $+ 200$ cm: a converging (convex) lens.

Step-by-step solution →
Q246 CBSE 2020
Draw ray diagram in each of the following cases to show what happens after reflection to the incident ray when
1. (a)it is parallel to the principal axis and falling on a convex mirror.
2. (b)it is falling on a concave mirror while passing through its principal focus.
3. (c)it is coming oblique to the principal axis and falling on the pole of a convex mirror.
Show answer

Parallel ray appears to come from F; ray through F goes parallel to the axis; oblique ray at the pole reflects at equal angle.
(a) Convex mirror: the ray parallel to the principal axis, after reflection, appears to diverge from the principal focus F behind the mirror; the reflected ray is drawn going away from the mirror, and its dashed backward extension passes through F. (b) Concave mirror: the ray passing through the principal focus F, after reflection, travels parallel to the principal axis. (c) Convex mirror: the ray striking the pole P obliquely is reflected obliquely on the other side of the principal axis, making an equal angle with it, because the principal axis is the normal at the pole ( $∠ i = ∠ r$ ). Each ray is drawn with an arrow showing its direction.

Step-by-step solution →
Q247 CBSE 2020

A child is standing in front of a magic mirror. She finds the image of her head bigger, the middle portion of her body of the same size and that of the legs smaller. Explain the construction of the magic mirror using different types of mirrors. Also state the reasons in support of your answer.

Show answer

The magic mirror joins a concave mirror at the top, a plane mirror in the middle and a convex mirror at the bottom.
The magic mirror is three mirrors joined one above another. Top part, a concave mirror: the head is within its focal length, where a concave mirror forms a virtual, erect and enlarged image, so the head looks bigger. Middle part, a plane mirror: it forms a virtual, erect image of the same size. Lower part, a convex mirror: it always forms a virtual, erect and diminished image, so the legs look smaller.

Step-by-step solution →
Q248 CBSE 2020
1. (a)To get an enlarged, real and inverted image of an object by a concave mirror, where should the object be placed ? Draw a labelled ray diagram to justify your answer.
2. (b)If an object is placed at the centre of curvature of this mirror, what will be the magnification produced ?
Show answer

Place the object between C and F; the image is real, inverted and enlarged. At C the magnification is $- 1$ .
(a) The object should be placed between the centre of curvature C and the principal focus F. In the ray diagram, a ray from the top of the object parallel to the principal axis passes through F after reflection, and a ray through F is reflected parallel to the axis; the two reflected rays meet beyond C, below the axis, so the image A′B′ is real, inverted and enlarged (for example, $f = - 10$ cm and $u = - 15$ cm give $v = - 30$ cm and $m = - 2$ ). (b) With the object at C, the image is also formed at C, real, inverted and of the same size, so the magnification is $m = - 1$ .

Step-by-step solution →
Q249 CBSE 2020

State the laws of reflection of light. List four characteristics of the images formed by plane mirrors.

Show answer

The angle of incidence equals the angle of reflection; the plane-mirror image is virtual, erect, the same size and laterally inverted.
Laws of reflection: (i) The angle of incidence is equal to the angle of reflection, $∠ i = ∠ r$ . (ii) The incident ray, the normal to the mirror at the point of incidence and the reflected ray all lie in the same plane. Characteristics of the image formed by a plane mirror (any four): it is virtual and cannot be obtained on a screen; it is erect; it is of the same size as the object; it is as far behind the mirror as the object is in front of it; it is laterally inverted.

Step-by-step solution →
Q250 CBSE 2020

An object is kept at a distance of 30 cm in front of a concave mirror of focal length 20 cm. Use mirror formula to determine the position and magnification of the image produced.

Show answer

The image is formed 60 cm in front of the mirror and the magnification is $- 2$ : real, inverted and twice the size.
Using $\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$ with $u = - 30$ cm and $f = - 20$ cm: $\frac{1}{v} = - \frac{1}{20} + \frac{1}{30} = - \frac{1}{60}$ , so $v = - 60$ cm. The image is formed $60$ cm in front of the mirror. Magnification: $m = - \frac{v}{u} = - \frac{- 60}{- 30} = - 2$ , so the image is real, inverted and twice the size of the object.

Step-by-step solution →
Q251 CBSE 2020

An object is kept at a distance of 18 cm in front of a convex lens of focal length 12 cm. Use lens formula to determine the image distance and magnification of the image produced.

Show answer

The image is 36 cm from the lens, on the other side, with $m = - 2$ : real, inverted and twice the size.
Using $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$ with $u = - 18$ cm and $f = + 12$ cm: $\frac{1}{v} = \frac{1}{12} - \frac{1}{18} = \frac{1}{36}$ , so $v = + 36$ cm. The image is formed $36$ cm from the lens, on the other side. Magnification: $m = \frac{v}{u} = \frac{36}{- 18} = - 2$ , so the image is real, inverted and twice the size of the object.

Step-by-step solution →
Q252 CBSE 2020
Name the type of mirror which facilitates
1. (a)Shaving,
2. (b)Observing large images of the teeth of a patient, and
3. (c)Observing the rear view in vehicles. Give reason to justify your answer in each case.
Show answer

A concave mirror is used for shaving and for viewing teeth, and a convex mirror for the rear view in vehicles.
(a) Shaving: concave mirror. When the face is held between the pole and the focus (within the focal length), the mirror forms an erect and enlarged image of the face, so its details are clearly seen. (b) Large images of the teeth: concave mirror. A tooth held within the focal length gives an erect, enlarged image, so the dentist can see it in detail. (c) Rear view in vehicles: convex mirror. It always forms an erect, diminished image, and because it curves outwards it gives a much wider field of view than a plane mirror, so the driver sees a large area of the traffic behind.

Step-by-step solution →
Q253 CBSE 2020
Study the ray diagram given below and answer the questions that follow :
1. (a)Is the type of lens used converging or diverging ?
2. (b)List three characteristics of the image formed.
3. (c)In which position of the object will the magnification be $- 1$ ?
Show answer

A converging lens; the image is real, inverted and enlarged; the magnification is $- 1$ when the object is at $2 F_{1}$ .
(a) The lens is converging (convex): the ray parallel to the principal axis is bent towards the axis and passes through the focus $F_{2}$ after refraction. (b) The image is real, inverted and enlarged (magnified); it is formed beyond $2 F_{2}$ , on the other side of the lens. (c) When the object is placed at $2 F_{1}$ (at a distance $2 f$ from the lens), the image is formed at $2 F_{2}$ , real, inverted and of the same size, so the magnification is $- 1$ .

Step-by-step solution →
Q254 CBSE 2020
Draw ray diagrams for the following cases when a ray of light :
1. (i)passing through centre of curvature of a concave mirror is incident on it.
2. (ii)parallel to principal axis is incident on convex mirror.
3. (iii)is passing through focus of a concave mirror incident on it.
Show answer

Ray through $C$ retraces its path; parallel ray on a convex mirror appears to come from $F$ ; ray through $F$ reflects parallel.
(i) The ray passing through the centre of curvature $C$ strikes the concave mirror along the normal and is reflected back along the same path, through $C$ again. (ii) The ray parallel to the principal axis is reflected by the convex mirror away from the axis; extended backwards as a dashed line, the reflected ray passes through the principal focus $F$ behind the mirror, so it appears to diverge from $F$ . (iii) The ray passing through the principal focus $F$ of the concave mirror is reflected parallel to the principal axis. The three ray diagrams are drawn in the figures.

Step-by-step solution →
Q255 CBSE 2020
Draw a ray diagram in each of the following cases to show the formation of image, when the object is placed :
1. (i)between optical centre and principal focus of a convex lens.
2. (ii)anywhere in front of a concave lens.
3. (iii)at 2F of a convex lens. State the signs and values of magnifications in the above mentioned cases (i) and (ii).
Show answer

Both virtual images have positive magnification: $m > 1$ for the convex lens inside $F_{1}$ and $m < 1$ for the concave lens.
(i) Convex lens, object between O and $F_{1}$ : the refracted rays diverge, and their backward extensions meet on the same side as the object, farther from the lens than the object; the image is virtual, erect and enlarged (first ray diagram). (ii) Concave lens, object anywhere in front of it: the image is formed between $F_{1}$ and O on the same side as the object; it is virtual, erect and diminished (second ray diagram). (iii) Convex lens, object at $2 F_{1}$ : the image is formed at $2 F_{2}$ on the other side of the lens; it is real, inverted and of the same size (third ray diagram). Magnification: in case (i) the sign of $m$ is positive and $m > 1$ (for example $m = + 2.5$ when $f = + 10$ cm and $u = - 6$ cm); in case (ii) the sign of $m$ is positive and $m < 1$ (for example $m = + 0.4$ when $f = - 10$ cm and $u = - 15$ cm).

Step-by-step solution →
Q256 CBSE 2020
1. (a)Draw and complete the following ray diagram on you answer sheet to show lateral displacement.
2. (b)Define absolute refractive index. Give its mathematical expression.
3. (c)Study the following diagrams : I = Image, O = Object Identify Mirror 1 and Mirror 2 and state one use of each.
Show answer

The emergent ray is parallel to the incident ray; $n_{m} = \frac{c}{v}$ ; Mirror 1 is concave (shaving mirror), Mirror 2 convex (rear-view mirror).
(a) At the top face the ray bends towards the normal ( $∠ r < ∠ i$ ); at the bottom face it bends away from the normal and emerges with $∠ e = ∠ i$ , so the emergent ray is parallel to the incident ray but shifted sideways. The perpendicular distance $d$ between the incident ray produced and the emergent ray is the lateral displacement, as shown in the completed diagram. (b) The absolute refractive index of a medium is the ratio of the speed of light in vacuum (or air) to the speed of light in that medium: $n_{m} = \frac{c}{v}$ . (c) Mirror 1 is a concave mirror, since its image is virtual, erect and enlarged; use: shaving mirror or dentist's mirror (also reflectors in torches, searchlights and vehicle headlights, and solar furnaces). Mirror 2 is a convex mirror, since its image is virtual, erect and diminished; use: rear-view mirror in vehicles (also security mirrors).

Step-by-step solution →
Q257 CBSE 2020
1. (a)Complete the following ray diagram :
2. (b)Find the nature, position and size of the image formed.
3. (c)Use lens formula to determine the magnification in this case.
Show answer

Virtual, erect image 180 cm from the lens on the object's side, enlarged with $m = + 10$ .
(a) From the top of the object, a ray parallel to the principal axis is refracted through $F_{2}$ , and a ray through the optical centre O goes on undeviated. The refracted rays diverge; their dashed backward extensions meet on the object's side of the lens, where the dashed image A′B′ is drawn (see the completed diagram). (b) The image is virtual and erect; it is on the same side of the lens as the object, $180$ cm from the lens (beyond $2 F_{1}$ ); it is enlarged, $10$ times the size of the object. (c) With $u = - 18$ cm and $f = + 20$ cm, $\frac{1}{v} = \frac{1}{f} + \frac{1}{u} = \frac{1}{20} - \frac{1}{18} = - \frac{1}{180}$ , so $v = - 180$ cm, and the magnification is $m = \frac{v}{u} = \frac{- 180}{- 18} = + 10$ .

Step-by-step solution →
Q258 CBSE 2020
An object $4.0$ cm in size, is placed $25.0$ cm in front of a concave mirror of focal length $15.0$ cm.
1. (i)At what distance from the mirror should a screen be placed in order to obtain a sharp image?
2. (ii)Find the size of the image.
3. (iii)Draw a ray diagram to show the formation of image in this case.
Show answer

The screen goes 37.5 cm in front of the mirror; the image is real, inverted and 6.0 cm in size ( $h^{'} = - 6.0$ cm).
(i) $v = - 37.5$ cm: the screen should be placed $37.5$ cm in front of the mirror, on the same side as the object. (ii) $h^{'} = - 6.0$ cm: the image is $6.0$ cm in size, and the minus sign shows that it is inverted; it is real and enlarged ( $m = - 1.5$ ). (iii) Ray diagram: with the object between $C$ and $F$ , the ray parallel to the principal axis is reflected through $F$ and the ray through $F$ is reflected parallel to the axis; the reflected rays meet beyond $C$ , $37.5$ cm from the mirror, where a real, inverted, enlarged image $6.0$ cm high is formed on the screen.

Step-by-step solution →
Q259 CBSE 2020
1. (a)Define power of a lens and write its SI unit. Name the type of lens whose power is negative.
2. (b)A convex lens forms a real and inverted image of finite size at a distance of $50$ cm from it. Where is the object placed in front of the lens ? Give all possible positions of the object stating reason in each case.
3. (c)Draw labelled ray diagram for any one position of the object mentioned in (b) above.
Show answer

Power is the reciprocal of focal length (dioptre); the object can be beyond $2 F_{1}$ , at $2 F_{1}$ or between $F_{1}$ and $2 F_{1}$ .
(a) The power of a lens is its ability to converge or diverge the rays of light falling on it; it is the reciprocal of its focal length, $P = \frac{1}{f}$ ( $f$ in metres). Its SI unit is the dioptre (D), $1 D = 1 m^{- 1}$ . A concave (diverging) lens has negative power. (b) A convex lens forms a real, inverted image of finite size only when the object is beyond $F_{1}$ but not at infinity (between $O$ and $F_{1}$ the image is virtual and erect, at $F_{1}$ it is at infinity, and for an object at infinity it is a point at $F_{2}$ ). The focal length is not given, so there are three possible positions: (i) Beyond $2 F_{1}$ : the image lies between $F_{2}$ and $2 F_{2}$ and is diminished; this happens when $f$ is between $25$ cm and $50$ cm. (ii) At $2 F_{1}$ : the image is at $2 F_{2}$ and of the same size; this happens when $f = 25$ cm, the object being $50$ cm from the lens. (iii) Between $F_{1}$ and $2 F_{1}$ : the image lies beyond $2 F_{2}$ and is enlarged; this happens when $f$ is less than $25$ cm. (c) For the object at $2 F_{1}$ with $f = 25$ cm, the ray parallel to the axis refracts through $F_{2}$ , the ray through $O$ goes straight on, and the ray through $F_{1}$ emerges parallel; they meet at $2 F_{2}$ , giving a real, inverted image of the same size, $u = - 50$ cm and $v = + 50$ cm (see the labelled ray diagram).

Step-by-step solution →
Q260 CBSE 2020
1. (a)Complete the following ray diagram :
2. (b)Find the nature, position and size of the image formed.
3. (c)Use mirror formula to determine the magnification in this case.
Show answer

The image is virtual and erect, formed 30 cm behind the mirror, and enlarged with magnification $m = + 2$ .
(a) From the top of the object, a ray parallel to the principal axis is reflected through F, and a ray striking the pole P is reflected at an equal angle on the other side of the axis. The reflected rays diverge; their dashed extensions behind the mirror meet, and the dashed image A′B′ is drawn there (see the completed diagram). (b) The image is virtual and erect; it is formed behind the mirror, $30$ cm from P; it is enlarged, twice the size of the object. (c) With $f = - 30$ cm and $u = - 15$ cm, $\frac{1}{v} = \frac{1}{f} - \frac{1}{u} = - \frac{1}{30} + \frac{1}{15} = \frac{1}{30}$ , so $v = + 30$ cm, and the magnification is $m = - \frac{v}{u} = - \frac{+ 30}{- 15} = + 2$ .

Step-by-step solution →