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.

CBSE Class 10 Science previous year papers, with answers →

Questions · page 10 of 21

1. Q181 CBSE 2024
   Define power of ​a lens. ​The focal ​length of a lens is $+25$ cm. Determine its nature ​(converging/diverging) and calculate its power. If an ​object is placed ​at a distance of ​50 cm from the optical centre of this ​lens, list two properties of the image formed in this case.
   Show answer

   The lens is convex (converging) with power $+4$ D; an object at 50 cm gives a real, inverted image of the same size.

   Power of a lens: the degree of convergence or divergence of light rays achieved by a lens. It is the reciprocal of its focal length in metres, $P=\frac{1}{f}$, and its SI unit is the dioptre (D); $1$ D is the power of a lens of focal length $1$ m. Nature: the focal length is positive, so the lens is convex, that is converging. Power: $P=\frac{1}{+0.25\text{m}}=+4$ D. Image of an object at $50$ cm, that is at $2F_1$: it is formed at $2F_2$, $50$ cm behind the lens ($v=+50$ cm, $m=-1$). Two properties: (1) it is real and inverted; (2) it is of the same size as the object.

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2. Q182 CBSE 2024
   1. (a)Upper ​half of a convex lens is ​covered with a black paper. Draw a ray diagram to ​show ​the formation of image of an object placed at a distance of $2F$ from such ​a lens. Mention the position and nature of the image formed. State the observable difference in the image obtained if the lens is uncovered. Give reason to justify your answer.
   2. (b)An object is placed at a distance of 30 cm from the optical centre of a concave lens of ​focal ​length 15 cm. Use lens formula to determine the distance of the image ​from the optical centre of the lens.
   Show answer

   (a) Image at $2F_2$, real, inverted, same size but dimmer; brighter when uncovered. (b) $v=-10$ cm.

   (a) With the upper half covered, only rays passing through the lower half form the image; in the ray diagram the ray from the top of the object through $F_1$ emerges parallel to the axis and meets a second ray through the lower half at $2F_2$. Position: at $2F_2$, on the other side of the lens. Nature: real, inverted and of the same size as the object; the image is complete but less bright. If the lens is uncovered, the image is formed at the same place with the same size and nature, but it is brighter. Reason: every part of a lens forms a complete image, because rays from each point of the object pass through all parts of the lens and meet at the same image point; covering half the lens only cuts down the amount of light reaching the image. (b) $f=-15$ cm and $u=-30$ cm, so $\frac{1}{v}=\frac{1}{f}+\frac{1}{u}=-\frac{1}{15}-\frac{1}{30}=-\frac{1}{10}$ and $v=-10$ cm. The image is 10 cm from the optical centre, on the same side as the object (virtual, erect and diminished, $m=+\frac{1}{3}$).

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3. Q183 CBSE 2024
   (i) Draw a ray diagram to ​show the path of the refracted ray ​in ​each of the following cases : A ​ray of light incident on a concave lens

   1. (1)parallel to its principal axis, and
   2. (2)is directed towards its principal focus. (ii) A 4 cm tall object is placed ​perpendicular to the principal axis of convex lens of focal length 24 cm. The distance of object from the ​lens is 16 cm. Find the position ​and size of image formed.
   Show answer

   Concave lens: the parallel ray appears to come from $F_1$; the ray towards $F_2$ emerges parallel. Image: $v=-48$ cm, 12 cm tall, virtual, erect.

   (i) (1) The ray parallel to the principal axis, after refraction through the concave lens, diverges and appears to come from the principal focus $F_1$ on the same side as the incident light. (2) The ray directed towards the principal focus $F_2$ on the other side of the lens, after refraction, emerges parallel to the principal axis. The ray diagrams show both paths. (ii) With $u=-16$ cm and $f=+24$ cm, $\frac{1}{v}=\frac{1}{24}-\frac{1}{16}=-\frac{1}{48}$, so $v=-48$ cm: the image is formed 48 cm from the lens on the same side as the object. $m=\frac{v}{u}=\frac{-48}{-16}=+3$, so $h^{\prime }=3\times 4=+12$ cm: the image is 12 cm tall, virtual, erect and magnified three times.

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4. Q184 CBSE 2024
   The variation ​of image distance $(v)$ ​with object distance $(u)$ for a convex lens is ​given in the following observation table. Analyse it ​and answer the questions that follow:

   1. (i)Without calculation, find the focal length of the convex ​lens. Justify your answer.
   2. (ii)Which observation is ​not correct? Why? Draw ray diagram to find the position of ​the image formed for this position of the object.
   3. (iii)​Find the approximate value of magnification for $u=-30$ cm.

   S. No.	Object distance (u) cm	Image distance (v) cm
   1	$-150$	$+30$
   2	$-75$	$+37.5$
   3	$-50$	$+50$
   4	$-37.5$	$+75$
   5	$-30$	$+150$
   6	$-15$	$+37.5$

   Show answer

   $f=25$ cm; observation 6 is wrong as $v$ should be $-37.5$ cm; the magnification for $u=-30$ cm is $-5$.

   (i) $f=25$ cm. In observation 3 the object and the image are at the same distance, 50 cm, on opposite sides of the lens; this happens only when the object is at 2F₁ and the image at 2F₂, so $2f=50$ cm and $f=25$ cm. (ii) Observation 6 ($u=-15$ cm, $v=+37.5$ cm) is not correct. The object at 15 cm is between the optical centre and F₁ ($15\text{cm}<25\text{cm}$), so the lens forms a virtual, erect, magnified image on the same side as the object and $v$ must be negative. By the lens formula the image is at $v=-37.5$ cm and is 2.5 times the size of the object, as the ray diagram shows. (iii) $m=\frac{v}{u}=\frac{+150}{-30}=-5$: the image is real, inverted and 5 times the size of the object.

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5. Q185 CBSE 2024
   1. (i)State laws of reflection of light.
   2. (ii)An object ​of height $5.0$ cm ​is ​placed at $15$ cm in front of a concave mirror of focal length $10$ cm. At what ​distance from ​the mirror ​should ​a ​screen be placed, so that a focussed image is obtained on it ? Find the height of the image.
   Show answer

   Screen 30 cm in front of the mirror ($v=-30$ cm); the image is 10 cm tall, real and inverted.

   (i) (1) The angle of incidence is equal to the angle of reflection. (2) The incident ray, the normal to the mirror at the point of incidence and the reflected ray all lie in the same plane. (ii) The screen should be placed $30$ cm in front of the mirror ($v=-30$ cm). The height of the image is $h^{\prime }=-10$ cm: the image is $10$ cm tall, real and inverted (magnification $m=-2$).

   Step-by-step solution →
6. Q186 CBSE 2024
   (i) Draw a ray diagram to show the path of ​the reflected ray in each of the ​following cases : A ray of light incident on a convex mirror

   1. (1)parallel to its principal axis, and
   2. (2)is directed towards its principal focus (ii) A 1.5 cm tall ​candle flame is ​placed perpendicular to the principal axis of ​a concave mirror of focal length 12 ​cm. If the distance of the flame ​from the pole of the mirror is 18 cm, ​use mirror formula to determine the position and size of the image formed.
   Show answer

   Convex mirror: parallel ray appears to come from F, ray towards F returns parallel. Concave mirror: $v=-36$ cm, real, inverted, 3 cm tall.

   (i) (1) The ray parallel to the principal axis, after reflection from the convex mirror, diverges and appears to come from the principal focus $F$ behind the mirror. (2) The ray directed towards the principal focus $F$ behind the mirror, after reflection, travels parallel to the principal axis. The ray diagrams show both paths. (ii) With $u=-18$ cm and $f=-12$ cm, $\frac{1}{v}=\frac{1}{f}-\frac{1}{u}=-\frac{1}{12}+\frac{1}{18}=-\frac{1}{36}$, so $v=-36$ cm: the image is formed 36 cm in front of the mirror, on the same side as the flame. $m=-\frac{v}{u}=-\frac{-36}{-18}=-2$, so $h^{\prime }=-2\times 1.5=-3$ cm: the image is 3 cm tall, real, inverted and twice the size of the flame.

   Step-by-step solution →
7. Q187 CBSE 2024
   1. (i)​Define ​principal axis of a lens. Draw a ray diagram to show what happens when a ray ​of light parallel to the principal axis of ​a ​concave lens passes through it.
   2. (ii)The focal ​length of a concave lens is 20 cm. At what distance from the lens should a 5 cm tall object be ​placed so that its image is formed at a distance of 15 cm from the lens? ​Also calculate the size of the image formed.
   Show answer

   The object must be 60 cm in front of the lens; the image is 1.25 cm tall, virtual and erect.

   (i) The principal axis of a lens is the imaginary straight line passing through the two centres of curvature of the lens. A ray parallel to the principal axis, after refraction through a concave lens, diverges (bends away from the axis) and appears to come from the principal focus F₁ on the same side of the lens as the incident light, as shown in the ray diagram. (ii) With $f=-20$ cm and $v=-15$ cm, $\frac{1}{u}=\frac{1}{v}-\frac{1}{f}=-\frac{1}{15}+\frac{1}{20}=-\frac{1}{60}$, so $u=-60$ cm: the object must be placed 60 cm in front of the lens. Then $m=\frac{v}{u}=\frac{-15}{-60}=+0.25$ and $h^{\prime }=0.25\times 5=1.25$ cm: the image is 1.25 cm tall, virtual and erect.

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8. Q188 CBSE 2024
   1. (i)State laws of refraction of light.
   2. (ii)Draw a ray ​diagram to show ​refraction of a ​ray of light through a rectangular glass slab. How is the emergent ​ray ​related ​to incident ray ? Mark lateral ​displacement in the diagram.
   Show answer

   The laws: coplanar rays and constant $\frac{\sin i}{\sin r}$; through a slab the emergent ray is parallel to the incident ray but laterally shifted.

   (i) (1) The incident ray, the refracted ray and the normal to the interface of the two media at the point of incidence all lie in the same plane. (2) 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 a given pair of media ($\sin i/\sin r=\text{constant}$, Snell's law). (ii) In the ray diagram the ray bends towards the normal on entering the glass slab and away from the normal on leaving it. The emergent ray is parallel to the incident ray ($e=i$) but shifted sideways; the lateral displacement is the perpendicular distance between the incident ray produced and the emergent ray, marked $d$ in the diagram.

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9. Q189 CBSE 2024
   A ​student has three convex lenses A, B and C of ​different focal ​lengths. He wants ​to observe ​the images ​formed by these lenses on a screen by placing ​a candle flame at different distances as ​given in following table :

   Case No.	Lens	Focal Length	Object Distance
   1	A	50 cm	25 cm
   2	B	20 cm	60 cm
   3	C	15 cm	30 cm

   1. (i)List two properties of the image formed by the lens B in case no. 2.[1]
   2. (ii)In which one of the three cases given above will the lens form a real image of the same size as the object and why ?[1]
   3. (iii)(a)Name the type of lens presented by opticians for the correction of hypermetropia. State the role of such lenses in correcting the vision of the person suffering from this defect.[2]
   4. (iii)(b)OR : Use Lens formula to find the distance of the image formed from the optical centre of the lens A in case no. 1.[2]
   Show answer

   Image is real and inverted; case 3 gives a real, same-size image; hypermetropia needs a convex lens; OR $v=-50$ cm.

   (i) Case 2: the image is real and inverted (it is also diminished, formed between $F_2$ and $2F_2$ at $v=+30$ cm, with $m=-0.5$). (ii) Case 3: the candle is at $30$ cm $=2f$ of lens C, that is, at $2F_1$, so the real, inverted image forms at $2F_2$ and is of the same size ($v=+30$ cm, $m=-1$). (iii)(a) A convex (converging) lens of suitable power. It supplies the additional focusing power, so light from a near object, which the eye alone focuses behind the retina, is focused on the retina. (iii)(b) With $u=-25$ cm and $f=+50$ cm, $\frac{1}{v}=\frac{1}{50}-\frac{1}{25}=-\frac{1}{50}$, so $v=-50$ cm: the image is $50$ cm from the optical centre, on the same side as the candle (virtual and erect, $m=+2$).

   Step-by-step solution →
10. Q190 CBSE 2024
    Study the ​data given ​below showing the focal ​length of ​three concave mirrors A, B and ​C and the ​respective ​distances of objects placed in ​front of the mirrors :

    Case	Mirror	Focal Length (cm)	Object Distance (cm)
    1	A	20	45
    2	B	15	30
    3	C	30	20

    1. (i)In which one of the above cases the mirror will form a diminished image of the object ? Justify your answer.[1]
    2. (ii)List two properties of the image formed in case 2.[1]
    3. (iii)(a)What is the nature and size of the image formed by mirror C ? Draw ray diagram to justify your answer.[2]
    4. (iii)(b)OR : An object is placed at a distance of $18$ cm from the pole of a concave mirror of focal length $12$ cm. Find the position of the image formed in this case.[2]
    Show answer

    Case 1 gives a diminished image; case 2 a real, same-size image at C; mirror C an enlarged virtual image, $m=+3$; or $v=-36$ cm.

    (i) Case 1 (mirror A). Its focal length is $20$ cm, so C is at $40$ cm; the object at $45$ cm is beyond C, and a concave mirror then forms a real, inverted image between F and C that is smaller than the object ($v=-36$ cm, $m=-0.8$). (ii) In case 2 the object is at C ($30$ cm $=2f$). The image is formed at C; it is real and inverted, and of the same size as the object. (iii)(a) For mirror C the object ($20$ cm) is between P and F ($f=30$ cm): the image is virtual and erect, and enlarged, formed behind the mirror ($v=+60$ cm, $m=+3$, three times the size of the object), as the ray diagram shows. (iii)(b) 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 in front of the mirror, on the side of the object; it is real, inverted and magnified ($m=-2$).

    Step-by-step solution →
11. Q191 CBSE 2024
    A highly polished surface such ​as a mirror ​reflects most of the light falling on it. ​In our daily life we use two types of mirrors — ​plane and spherical. The reflecting surface of a ​spherical mirrors may be curved inwards or outwards. In concave mirrors, reflection takes place ​from the ​inner surface, while in convex ​mirrors reflection takes place from the outer surface.

    1. (a)Define the principal axis of a concave mirror.[1]
    2. (b)A ray of light is incident on a concave mirror, parallel to its principal axis. If this ray after reflection from the mirror passes through the principal axis from a point at a distance of 10 cm from the pole of the mirror, find the radius of curvature of the mirror.[1]
    3. (c)(i)An object is placed at a distance of 10 cm from the pole of a convex mirror of focal length 15 cm. Find the position of the image.[2]
    4. (c)(ii)OR : A mirror forms a virtual, erect and diminished image of an object. Identify the type of this mirror. Draw a ray diagram to show the image formation in this case.[2]
    Show answer

    The principal axis joins the pole and centre of curvature; $R=20$ cm; $v=+6$ cm for the convex mirror; or the mirror is convex.

    (a) The principal axis of a concave mirror is the straight line passing through its pole and its centre of curvature; it is normal to the mirror at the pole. (b) The reflected ray crosses the axis at the principal focus, so $f=10$ cm and $R=2f=20$ cm. (c)(i) With $u=-10$ cm and $f=+15$ cm, $\frac{1}{v}=\frac{1}{15}+\frac{1}{10}=\frac{1}{6}$, so $v=+6$ cm: the image is formed $6$ cm behind the mirror (virtual, erect and diminished, $m=+0.6$). (c)(ii) It is a convex mirror. Ray diagram: a ray parallel to the principal axis reflects as if coming from $F$, and a ray directed towards $C$ reflects back along itself; their backward extensions meet behind the mirror between $P$ and $F$, where the virtual, erect, diminished image is formed.

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12. Q192 CBSE 2024
    Under the ​guidance of his ​teacher on a sunny day, a ​student took a concave mirror of large aperture in his hand and directed its reflecting surface towards the sun. After that he directed the light reflected by the concave mirror on a thick sheet of white paper held close to the mirror. He observed a bright circular spot of light on the paper. Then he moved the ​mirror back and forth gradually until he got the brightest ​and smallest spot of light on the paper. The teacher said this sharp spot of light is, in ​fact, the ​real and inverted image of the sun and the distance between the ​mirror and this image is the focal length of this concave mirror.

    1. (a)Define the term principal focus of a concave mirror.[1]
    2. (b)If the distance between the pole and focus of a concave mirror is 12 cm, what is the radius of curvature of the mirror ?[1]
    3. (c)(i)List two properties of the image formed due to a concave mirror of focal length 15 cm in each of the following cases : (1) object distance 20 cm (2) object distance 10 cm[2]
    4. (c)(ii)OR : A 3 cm long candle flame is placed in front of a concave mirror. If the distance between the pole and the candle flame is 40 cm and the image is formed at the same place where the candle flame is located, find the (1) focal length of the mirror, and (2) magnification of the image formed with sign as per the New Cartesian Sign Convention.[2]
    Show answer

    Radius of curvature $R=24$ cm; image real and inverted at 20 cm, virtual and erect at 10 cm; or $f=-20$ cm and $m=-1$.

    (a) The principal focus of a concave mirror is the point on its principal axis at which rays of light incident parallel to the principal axis meet after reflection from the mirror. (b) $R=2f=2\times 12=24$ cm. (c)(i) (1) Object at $20$ cm (between $F$ and $C$): $v=-60$ cm, $m=-3$; the image is real and inverted (also enlarged, formed beyond $C$). (2) Object at $10$ cm (between $P$ and $F$): $v=+30$ cm, $m=+3$; the image is virtual and erect (also enlarged, formed behind the mirror). OR (c)(ii) (1) With $u=v=-40$ cm, $\frac{1}{f}=-\frac{1}{40}-\frac{1}{40}=-\frac{1}{20}$, so $f=-20$ cm (focal length $20$ cm). (2) $m=-\frac{v}{u}=-1$: the image is real, inverted and of the same size ($3$ cm).

    Step-by-step solution →
13. Q193 CBSE 2024
    In our laboratories we use compound microscopes to see the magnified image of a microscopic object. A compound microscope ​is made up of ​two ​lenses. The lens nearest to the object to be viewed, called objective lens, forms real, inverted and magnified image of the object. This image serves as an object ​for the second lens called eyepiece. ​The eyepiece forms virtual, erect and ​magnified image of its object. Thus, the resultant image formed by ​a microscope is virtual, inverted and magnified with respect to the microscopic object ​viewed.

    1. (a)The image of an object formed by a convex lens of focal length $2$ cm is real, inverted and magnified. What is the range of object distance in this case ?[1]
    2. (b)The image of an object formed by a convex lens of focal length $6$ cm is virtual, erect and magnified. What is the range of object distance in this case ?[1]
    3. (c)(i)An object is placed at a distance of $12$ cm from the optical centre of a convex lens of focal length $18$ cm. Draw a labelled ray diagram to show the formation of image in this case.[2]
    4. (c)(ii)OR : The image formed by a convex lens is real, inverted and of the same size as the object. If the distance between the objects and the image is $60$ cm, determine the focal length of the lens. Give justification for your answer.[2]
    Show answer

    A real, magnified image needs an object between $F_1$ and $2F_1$, a virtual one inside $F_1$; for 12 cm, $v=-36$ cm; or $f=15$ cm.

    (a) The object must be between F₁ and 2F₁: its distance from the lens is more than $2$ cm and less than $4$ cm. (b) The object must be between the optical centre O and F₁: its distance from the lens is less than $6$ cm. (c)(i) The object ($u=-12$ cm) lies between O and F₁ of the lens ($f=18$ cm). A ray from the top of the object parallel to the axis refracts through F₂, and a ray through O goes on undeviated; the refracted rays diverge, and their backward extensions meet on the object's side at $v=-36$ cm, at 2F₁. The image is virtual, erect and magnified three times ($m=+3$), as the ray diagram shows. (c)(ii) A real, inverted image of the same size is formed only when the object is at 2F₁, and the image is then at 2F₂. Object and image are therefore $2f+2f=4f=60$ cm apart, so $f=15$ cm; check: $\frac{1}{30}-\frac{1}{-30}=\frac{1}{15}$.

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14. Q194 CBSE 2023
    A student wants to ​obtain an erect ​image of an object using ​a concave ​mirror of 10 cm focal length. What ​will be ​the distance of the ​object from mirror ​?

    1. (a)Less than 10 cm
    2. (b)10 cm
    3. (c)between 10 cm and 20 cm
    4. (d)more than 20 cm
    Show answer

    (a) Less than 10 cm

    Step-by-step solution →
15. Q195 CBSE 2023
    To ​obtain a magnification ​of $+2$ with ​a ​concave ​mirror of radius of ​curvature $60$ cm the object ​distance must be

    1. (a)$-90$ cm
    2. (b)$-45$ cm
    3. (c)$-30$ cm
    4. (d)$-15$ cm
    Show answer

    (d) $-15$ cm

    Step-by-step solution →
16. Q196 CBSE 2023
    In ​torch ​lights ​and head ​lights of vehicles, ​the bulb ​is ​placed

    1. (a)between the pole and the focus of the reflector.
    2. (b)very near to the focus of the reflector.
    3. (c)between the focus and centre of curvature of the reflector.
    4. (d)at the centre of curvature of the reflector.
    Show answer

    (b) very near to the focus of the reflector.

    Step-by-step solution →
17. Q197 CBSE 2023
    When ​an object ​is placed ​beyond $2F$ of ​a convex lens, ​the ​nature of the image ​formed ​is :

    1. (a)Real, inverted and diminished
    2. (b)Real, erect and magnified
    3. (c)Virtual, erect and magnified
    4. (d)Real, inverted and magnified
    Show answer

    (a) Real, inverted and diminished

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18. Q198 CBSE 2023
    An object is ​placed in front ​of a convex mirror at infinity. According to the ​New Cartesian Sign Convention, the sign of the focal length ​and the sign of the ​image distance ​in this ​case ​are respectively :

    1. (a)$+,-$
    2. (b)$-,+$
    3. (c)$-,-$
    4. (d)$+,+$
    Show answer

    (d) $+,+$

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19. Q199 CBSE 2023
    Find ​the magnification of ​the image ​formed by a ​spherical ​mirror from the ​following data : $u=-20$ ​cm, $f=-15$ cm.
    Show answer

    The magnification is $m=-3$: the image is real, inverted and three times the size of the object.

    Using $\frac{1}{v}=\frac{1}{f}-\frac{1}{u}=-\frac{1}{15}+\frac{1}{20}=-\frac{1}{60}$, the image distance is $v=-60$ cm. Hence $m=-\frac{v}{u}=-\frac{-60}{-20}=-3$. The magnification is $-3$: the image is real, inverted and three times the size of the object, formed $60$ cm in front of the concave mirror.

    Step-by-step solution →
20. Q200 CBSE 2023
    Draw ​a labelled ray diagram ​for the image formation by a concave mirror ​when ​an ​object is ​placed between its centre of ​curvature and focus.
    Show answer

    The image is real, inverted and enlarged, formed beyond the centre of curvature C.

    The diagram is drawn with the pole $P$, the principal focus $F$ and the centre of curvature $C$ marked on the principal axis, $F$ lying midway between $P$ and $C$, and the object standing between $C$ and $F$. One ray from the tip of the object is drawn parallel to the principal axis and is reflected through $F$; a second ray from the tip is drawn through $F$ and is reflected parallel to the principal axis. The two reflected rays actually intersect beyond $C$, and the image is drawn there, from the axis to that intersection. The image formed is real, inverted, enlarged (larger than the object) and situated beyond the centre of curvature $C$.

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