Class 10 Science · Chapter 9

Light - Reflection and Refraction

194 Class 10 Science questions on Light - Reflection and Refraction from the NCERT textbook, NCERT Exemplar and CBSE board papers 2023–2026, each with its answer and a step-by-step solution.

CBSE Class 10 Science previous year papers, with answers →

Questions · page 4 of 10

1. Q61 Exemplar · Long Answer Questions
   Draw ray diagrams ​showing the ​image ​formation by a concave mirror when an object is placed

   1. (a)between pole and ​focus of the mirror
   2. (b)​between focus and centre ​of curvature of the mirror
   3. (c)at centre ​of curvature of the mirror
   4. (d)a little beyond centre of curvature of the mirror
   5. (e)at infinity
   Show answer

   The image is real and inverted for all positions beyond $F$, and virtual, erect and enlarged between $P$ and $F$.

   (a) Object between $P$ and $F$ ($u=-6$ cm): the image is virtual, erect and enlarged, $v=+15$ cm behind the mirror, $m=+2.5$. (b) Object between $F$ and $C$ ($u=-15$ cm): real, inverted and enlarged, $v=-30$ cm, beyond $C$, $m=-2$. (c) Object at $C$ ($u=-20$ cm): real, inverted and the same size, $v=-20$ cm at $C$ itself, $m=-1$. (d) Object a little beyond $C$ ($u=-30$ cm): real, inverted and diminished, $v=-15$ cm, between $F$ and $C$, $m=-0.5$. (e) Object at infinity: real, inverted and highly diminished to a point, formed at the focus, $v=f=-10$ cm.

   Step-by-step solution →
2. Q62 Exemplar · Long Answer Questions
   Draw ray diagrams showing the image formation by a convex ​lens when an object is ​placed

   1. (a)between optical centre and focus of the lens
   2. (b)between focus and twice the focal length of the lens
   3. (c)at twice ​the focal ​length ​of the lens
   4. (d)at infinity
   5. (e)at the ​focus ​of the lens
   Show answer

   Virtual, erect and enlarged between O and F; real and inverted in the other cases, a point at F for an object at infinity.

   (a) Object between $O$ and $F$ ($u=-6$ cm): the image is virtual, erect and enlarged, $v=-15$ cm on the same side as the object, $m=+2.5$. (b) Object between $F$ and $2F$ ($u=-15$ cm): real, inverted and enlarged, $v=+30$ cm, beyond $2F$ on the other side, $m=-2$. (c) Object at $2F$ ($u=-20$ cm): real, inverted and the same size, $v=+20$ cm at $2F$ on the other side, $m=-1$. (d) Object at infinity: real, inverted and highly diminished to a point, formed at the focus, $v=f=+10$ cm. (e) Object at the focus ($u=-10$ cm): the emergent rays are parallel, so the image is formed at infinity and is real, inverted and highly enlarged.

   Step-by-step solution →
3. Q63 Exemplar · Long Answer Questions
   Write laws ​of refraction. Explain the same ​with ​the help of ray diagram, when a ray of ​light passes through a rectangular glass ​slab.
   Show answer

   Incident ray, refracted ray and normal are coplanar, and $\sin i/\sin r$ is constant; in a slab the emergent ray is parallel to the incident ray.

   Laws of refraction. First law: the incident ray, the refracted ray and the normal to the interface at the point of incidence all lie in the same plane. Second law (Snell's law): for light of a given colour passing between a given pair of media, $\frac{\sin i}{\sin r}=n_{21}$, a constant equal to the refractive index of the second medium with respect to the first. Applied to a rectangular glass slab with $n=1.5$ and $i=60^{\circ }$: at the first face the ray enters the denser glass and bends towards the normal, $\sin r=\frac{\sin 60^{\circ }}{1.5}=0.5774$, so $r=35.3^{\circ }$. Because the faces are parallel it meets the second face at the same $35.3^{\circ }$, and there it enters the rarer air and bends away from the normal, $\sin e=1.5\sin r=0.8660$, so $e=60^{\circ }=i$. The emergent ray is therefore parallel to the incident ray, displaced sideways by $d=\frac{t\sin (i-r)}{\cos r}=2.05$ cm for a $4$ cm slab.

   Step-by-step solution →
4. Q64 Exemplar · Long Answer Questions
   Draw ray diagrams showing the image formation ​by a ​concave lens when an object is placed

   1. (a)​at the focus of the lens
   2. (b)between focus and twice the focal ​length of the lens
   3. (c)​beyond twice ​the ​focal length of the lens
   Show answer

   In all three positions a concave lens forms a virtual, erect, diminished image on the object's side, between the optical centre and the focus.

   (a) Object at the focus ($u=-10$ cm): the image is virtual, erect and diminished to half, $v=-5$ cm on the same side as the object, $m=+0.5$. (b) Object between $F$ and $2F$ ($u=-15$ cm): virtual, erect and diminished, $v=-6$ cm on the same side, $m=+0.4$. (c) Object beyond $2F$ ($u=-30$ cm): virtual, erect and diminished to a quarter, $v=-7.5$ cm on the same side, $m=+0.25$. In every case the image lies between the optical centre and the focus on the object's own side, and a concave lens can form no other kind of image of a real object.

   Step-by-step solution →
5. Q65 Exemplar · Long Answer Questions
   Draw ray diagrams ​showing the ​image formation by a ​convex mirror when an object is placed

   1. (a)at infinity
   2. (b)​at finite distance ​from the mirror
   Show answer

   At infinity the image is a point at $F$; at a finite distance it is virtual, erect, diminished, between $P$ and $F$.

   (a) Object at infinity: the image is virtual, erect and highly diminished to a point, formed at the focus $F$ behind the mirror, $v=f=+10$ cm, $m\to 0$. (b) Object at a finite distance (taking $u=-30$ cm): the image is virtual, erect and diminished, formed $v=+7.5$ cm behind the mirror, between the pole $P$ and the focus $F$, with $m=+0.25$. Whatever finite distance is chosen, $m=\frac{f}{f-u}$ lies between $0$ and $1$, so a convex mirror always gives an erect, diminished, virtual image between $P$ and $F$.

   Step-by-step solution →
6. Q66 Exemplar · Long Answer Questions
   The image of a ​candle flame formed by a ​lens is obtained on a screen placed on the other side of the ​lens. If the image is three times the size of the flame and the distance between lens and image is $80$ ​cm, at what distance should ​the candle be placed from the lens? What ​is the ​nature of the image at a distance of $80$ ​cm and the lens?
   Show answer

   The candle is 26.7 cm from the lens, which is convex ($f=+20\text{cm}$); the image is real, inverted and magnified three times.

   The candle should be placed $\frac{80}{3}=26.7$ cm from the lens (that is $u=-26.7$ cm, on the side away from the screen). The lens formula then gives $\frac{1}{f}=\frac{1}{80}+\frac{3}{80}=\frac{1}{20}$, so $f=+20$ cm: the lens is a convex (converging) lens. The image obtained $80$ cm from the lens is real, inverted and magnified three times.

   Step-by-step solution →
7. Q67 Exemplar · Long Answer Questions
   Size of image of an object by a mirror having a focal length of $20$ cm is observed to be ​reduced ​to $1/3$rd of ​its size. At what ​distance the object ​has been ​placed ​from ​the mirror? What is the nature of the image and the mirror?
   Show answer

   The object is 80 cm from the mirror; the image is real, inverted and one third the size, and the mirror is concave.

   Taking the mirror as concave, as the NCERT Exemplar answer list does: the object is placed $80$ cm in front of the mirror, $u=-80$ cm (that is $u=4f$, twice the radius of curvature). The image is then at $v=\frac{u}{3}=-26.7$ cm, in front of the mirror between $F$ and $C$, with $m=-\frac{1}{3}$: the image is real, inverted and diminished to one third, and the mirror is a concave mirror. If instead the printed focal length is read as $+20$ cm, the mirror is convex, the one-third reduction needs $u=-2f=-40$ cm and gives a virtual, erect image at $v=+13.3$ cm behind the mirror; the question does not print a sign, and the answer list takes the concave reading.

   Step-by-step solution →
8. Q68 Exemplar · Long Answer Questions
   Define power of ​a lens. ​What is ​its unit? One ​student uses a ​lens of ​focal length $50$ cm and another of $-50$ ​cm. What ​is the nature of the lens and its power used by each of them?
   Show answer

   Power is the reciprocal of focal length in metres, unit dioptre: the first lens is convex, $+2\text{D}$; the second concave, $-2\text{D}$.

   Power of a lens: the degree of convergence or divergence of light rays that a lens produces, measured as the reciprocal of its focal length in metres, $P=\frac{1}{f}$. Its unit is the dioptre (D); $1$ D is the power of a lens of focal length $1$ m, so $1\text{D}=1{\text{m}}^{-1}$. First student: $f=+50$ cm $=+0.5$ m, so $P=\frac{1}{0.5}=+2$ D, and the lens is a convex (converging) lens. Second student: $f=-50$ cm $=-0.5$ m, so $P=\frac{1}{-0.5}=-2$ D, and the lens is a concave (diverging) lens.

   Step-by-step solution →
9. Q69 Exemplar · Long Answer Questions
   A student focussed ​the image of a ​candle flame on a white screen using a convex lens. He noted down the ​position of the candle screen and the lens as under Position of candle = $12.0$ cm Position of convex lens = $50.0$ cm Position of the screen = $88.0$ cm

   1. (i)What is the focal length of the convex lens?
   2. (ii)Where will the image be formed if he shifts the ​candle towards ​the lens at a position of $31.0$ cm?
   3. (iii)​What will be the nature of the image formed if he further shifts the candle towards the lens?
   4. (iv)Draw a ray ​diagram to show the formation of the image in case (iii) as said above.
   Show answer

   $f=+19.0\text{cm}$; the image is at infinity with the candle at the focus, and virtual, erect and magnified when it is closer.

   (i) The object and image distances are $u=12.0-50.0=-38.0$ cm and $v=88.0-50.0=+38.0$ cm, so $\frac{1}{f}=\frac{1}{38.0}+\frac{1}{38.0}=\frac{1}{19.0}$ and the focal length is $f=+19.0$ cm. (ii) At the $31.0$ cm mark the candle is at $u=-19.0$ cm, exactly at the focus, so $\frac{1}{v}=\frac{1}{19.0}-\frac{1}{19.0}=0$ and the image is formed at infinity: the refracted rays leave the lens parallel and no screen at any finite distance can catch it. (iii) Shifting the candle still closer puts it between the optical centre and the focus, and the image then becomes virtual, erect and magnified, formed on the same side of the lens as the candle. (iv) With the candle at $u=-9.5$ cm, for example, $v=-19.0$ cm and $m=+2$: the ray diagram shows the two emergent rays diverging, with their dashed backward extensions meeting $19.0$ cm from the lens to give an erect image twice the size of the flame.

   Step-by-step solution →
10. Q70 CBSE 2026
    A convex lens of focal length $15$ cm, ​is ​forming ​a real image. ​If the size of ​image is same as the size ​of object, then position of object and position of image will be, respectively ​:

    1. (a)$-15$ cm and $-15$ cm from lens
    2. (b)$-15$ cm and $+15$ cm from lens
    3. (c)$-30$ cm and $+30$ cm from lens
    4. (d)$-30$ cm and $-30$ cm from lens
    Show answer

    (c) $-30$ cm and $+30$ cm from lens

    Step-by-step solution →
11. Q71 CBSE 2026
    From ​the ​following ​options choose ​the ​one which ​gives ​negative magnification greater than one ​:

    1. (a)Concave mirror; Nature of image – real
    2. (b)Concave mirror; Nature of image – virtual
    3. (c)Convex mirror; Nature of image – real
    4. (d)Convex mirror; Nature of image – virtual
    Show answer

    (a) Concave mirror; Nature of image – real

    Step-by-step solution →
12. Q72 CBSE 2026
    Rays from the sun converge at a point $25$ cm behind ​a convex ​lens. The ​distance at which ​an object be placed in front of the lens to get ​a ​virtual ​image, ​is :

    1. (a)$20$ cm
    2. (b)$40$ cm
    3. (c)$50$ cm
    4. (d)More than $50$ cm
    Show answer

    (a) $20$ cm

    Step-by-step solution →
13. Q73 CBSE 2026
    Pushkar was studying a ​light phenomenon and wrote down ​the following statements in that context :

    1. (i)​A ​concave mirror can be used as a shaving ​mirror because it can form an enlarged, virtual image when the object is placed between the pole and focus.
    2. (ii)A convex lens always forms real, inverted images, regardless of the ​position of the object.
    3. (iii)​The relationship between object distance ($u$), image distance ($v$) and focal length ($f$) of the lens ​is given by $\frac{1}{f}=\frac{1}{v}-\frac{1}{u}$. Choose the option enlisting all the correct statements :

    1. (a)(i) and (ii)
    2. (b)(i) and (iii)
    3. (c)(ii) and (iii)
    4. (d)(i), (ii) and (iii)
    Show answer

    (b) (i) and (iii)

    Step-by-step solution →
14. Q74 CBSE 2026
    When a convergent beam ​of light is incident on a convex mirror, the ​reflected ​rays appear to be ​coming from a point 20 cm behind the mirror and retrace ​their path. The focal length of the mirror is

    1. (a)20 cm
    2. (b)10 cm
    3. (c)40 cm
    4. (d)30 cm
    Show answer

    (b) 10 cm

    Step-by-step solution →
15. Q75 CBSE 2026
    Choose the correct option :

    1. (a)Concave mirror and concave lens always form virtual image.
    2. (b)Image formed by a plane mirror and a convex mirror has magnification $m=1$.
    3. (c)Concave lens and convex mirror always forms virtual image.
    4. (d)Image formed by a plane mirror and a concave lens has magnification $m=1$.
    Show answer

    (c) Concave lens and convex mirror always forms virtual image.

    Step-by-step solution →
16. Q76 CBSE 2026
    A convex lens and a ​concave lens having focal lengths 10 ​cm and 20 ​cm respectively are placed ​in ​contact ​with each other. Which of the ​following statements is correct about the lens combination ?

    1. (a)It will behave like a plane glass slab.
    2. (b)It will behave like a convex lens.
    3. (c)It will behave like a concave lens.
    4. (d)It will behave like a cylindrical lens.
    Show answer

    (b) It will behave like a convex lens.

    Step-by-step solution →
17. Q77 CBSE 2026
    In the given ray diagram the ​values of the angle ​of incidence and the ​angle of ​reflection ​are ​:

    1. (a)$i=50^{\circ }$, $r=40^{\circ }$
    2. (b)$i=40^{\circ }$, $r=50^{\circ }$
    3. (c)$i=40^{\circ }$, $r=40^{\circ }$
    4. (d)$i=50^{\circ }$, $r=50^{\circ }$
    Show answer

    (d) $i=50^{\circ }$, $r=50^{\circ }$

    Step-by-step solution →
18. Q78 CBSE 2026
    For Question ​number 32, ​two statements are given ​— one labelled as Assertion (A) ​and the other labelled as Reason ​(R). ​Select the correct ​answer to this question from the codes (a), (b), (c) and (d) as given below.

    Assertion (A): For a medium of given refractive index, the value of the angle of refraction corresponding to an angle of incidence is not a constant.

    Reason (R): With change in angle of incidence, the value of angle of refraction changes. Hence, the ratio of sine of angle of incidence to sine of angle of refraction remains constant.

    1. (a)Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).
    2. (b)Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of Assertion (A).
    3. (c)Assertion (A) is true, but Reason (R) is false.
    4. (d)Assertion (A) is false, but Reason (R) is true.
    Show answer

    (d) Assertion (A) is false, but Reason (R) is true.

    Step-by-step solution →
19. Q79 CBSE 2026
    For Question number 32, two statements ​are given — one ​labelled ​as Assertion ​(A) and the ​other labelled as Reason (R). Select the correct answer to ​this question from the ​codes (a), (b), (c) and (d) as given ​below.

    Assertion (A): When light passes from one medium to another medium, the speed of light may increase or decrease.

    Reason (R): Refractive index of one medium with respect to the other medium may be greater or less than 1.

    1. (a)Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).
    2. (b)Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of Assertion (A).
    3. (c)Assertion (A) is true, but Reason (R) is false.
    4. (d)Assertion (A) is false, but Reason (R) is true.
    Show answer

    (a) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).

    Step-by-step solution →
20. Q80 CBSE 2026
    The given figure shows image ​formation by a ​lens. Analyse the figure and answer the following questions :

    1. (a)What ​is ​the type of lens used for image formation in the given ray diagram ​?
    2. (b)If the real image is formed at a distance of 30 cm ​from the lens and the size of image is ​twice ​the size of the object, then where was the object placed ?

    Show answer

    The lens is convex (converging); the object was placed 15 cm in front of the lens.

    (a) A convex (converging) lens. The ray parallel to the principal axis is bent towards the axis and crosses it beyond the lens, and the image formed is virtual, erect and enlarged - an image only a convex lens can form, with the object between its optical centre and principal focus. (b) The object was placed $15$ cm in front of the lens, on the side opposite to the real image: $u=\frac{v}{m}=\frac{+30}{-2}=-15$ cm. (The lens then has $f=+10$ cm, and the object lies between $F_1$ and $2F_1$.)

    Step-by-step solution →