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 12 of 21

1. Q221 CBSE 2023
   1. (i)Define absolute ​refractive index.
   2. (ii)The absolute refractive indices of ​water and glass are $\frac{4}{3}$ ​and $\frac{3}{2}$ respectively. If the speed of light in glass ​is $2\times 10^8$ m/s, find : (I) Speed ​of light in vacuum (II) Speed ​of light in water
   Show answer

   The speed of light in vacuum is $3\times 10^8$ m/s and in water is $2.25\times 10^8$ m/s.

   (i) The absolute refractive index of a medium is the ratio of the speed of light in vacuum to the speed of light in that medium, $n=\frac{c}{v}$. (ii) (I) From glass, $c=n_g{v}_g=\frac{3}{2}\times 2\times 10^8=3\times 10^8$ m/s. (II) For water, $v_w=\frac{c}{n_w}=\frac{3\times 10^8}{4/3}=2.25\times 10^8$ m/s.

   Step-by-step solution →
2. Q222 CBSE 2023
   1. (a)An object is placed in front ​of a convex ​lens of focal length $f$. If the distance of the object from the lens is $2f$, draw ​a ray diagram to show the formation ​of the image. Write the value of magnification in this case.
   2. (b)A student has focussed the image of a candle flame on a white screen using a convex lens. The situation ​is as given below : Length of the flame $=2$ cm Focal length of the lens $=12$ cm Distance of the flame from the lens $=16$ cm If the flame is perpendicular to the principal axis ​of the lens, calculate the values of the ​following : (i) Distance of the image from ​the lens (ii) Length of the image formed
   Show answer

   At $2F_1$ the image is at $2F_2$ with $m=-1$; for the flame, $v=+48$ cm and the image is 6 cm long.

   (a) The image is formed at $2F_2$ on the other side of the lens; it is real, inverted and the same size as the object, and the magnification is $m=-1$ (see the ray diagram). (b) (i) Distance of the image from the lens $v=48$ cm, on the side of the lens opposite the flame. (ii) Magnification $m=\frac{v}{u}=\frac{48}{-16}=-3$, so $h^{\prime }=m\times h=-6$ cm: the image is $6$ cm long and inverted.

   Step-by-step solution →
3. Q223 CBSE 2023
   (a) Name the type of mirror that ​should ​be used to obtain the following types of ​images :

   1. (i)A magnified and virtual image of an object
   2. (ii)A diminished and virtual image of an ​object (b) Draw labelled ray diagrams ​to justify your answers in ​each case mentioned above. (c) ​Which of these mirrors can also be used to obtain a magnified and real image of an object ? ​State the position of the object in this case.
   Show answer

   Concave mirror: magnified virtual image; convex: diminished virtual image; concave also gives a magnified real image for an object between F and C.

   (a) (i) A concave mirror, with the object between its pole P and principal focus F, gives a magnified and virtual image. (ii) A convex mirror gives a diminished and virtual image. (b) In the ray diagram for the concave mirror the reflected rays diverge and their backward extensions meet behind the mirror, giving a virtual, erect, enlarged image; in the ray diagram for the convex mirror the extensions meet behind the mirror between P and F, giving a virtual, erect, diminished image (see the diagrams). (c) The concave mirror. The object must be placed between the principal focus F and the centre of curvature C; a real, inverted and magnified image is then formed beyond C.

   Step-by-step solution →
4. Q224 CBSE 2023
   1. (a)Draw a ray diagram for showing the convergence of a ​parallel ​beam of light by a concave mirror and mark the positions of pole and the centre of curvature in it.
   2. (b)An object $4$ cm in size is placed at $25$ cm ​in front of a ​concave mirror of focal length $15$ cm. At what distance from the ​mirror should a screen ​be placed in order to obtain a sharp image ? Find the nature and size of ​the image.
   3. (c)List ​any two uses of a concave mirror.
   Show answer

   The screen goes 37.5 cm in front of the mirror; the image is real, inverted, magnified and 6 cm in size ($h^{\prime }=-6$ cm).

   (a) See the ray diagram: rays parallel to the principal axis, after reflection from the concave mirror, converge at the principal focus F on the principal axis; the pole P (where the axis meets the mirror) and the centre of curvature C are marked, with F midway between them (for a mirror of small aperture, $f=R/2$). (b) With $u=-25$ cm and $f=-15$ cm, the mirror formula gives $v=-37.5$ cm, so the screen must be placed $37.5$ cm in front of the mirror. The magnification is $m=-\frac{v}{u}=-1.5$, so the image height is $h^{\prime }=-1.5\times 4=-6$ cm: the image is real, inverted and magnified, and $6$ cm in size. (c) Any two: in torches, search-lights and vehicle headlights to get powerful parallel beams of light; as shaving mirrors to see a larger image of the face; by dentists to see large images of the teeth; in solar furnaces to concentrate sunlight.

   Step-by-step solution →
5. Q225 CBSE 2023
   When a ray of light moving in ​a medium enters obliquely in another medium, it bends from its path. This phenomenon is called ​refraction of light. The ability of a medium to refract light is expressed ​in terms of optical density. The term optical density is not the same as mass density. If ​we compare two optical ​media, the one with the larger absolute refractive index ​is an optically denser medium than the other. Also, the speed of light through a given medium is ​inversely proportional to its optical density.

   1. (a)Define the term ‘absolute refractive’ index of a medium.[1]
   2. (b)Refractive indices of glass, water and carbon disulphide are $1.50$, $1.33$ and $1.62$ respectively. In which one of these media is the speed of light (i) highest, and (ii) lowest ?[1]
   3. (c)(i)The absolute refractive indices of glass and water are $3/2$ and $4/3$ respectively. If the speed of light in water is $2.25\times 10^8$ m/s, find the speed of light in (I) vacuum, and (II) glass.[2]
   4. (c)(ii)OR : The absolute refractive indices of glass and water are $3/2$ and $4/3$ respectively. Find the refractive index of water with respect to glass.[2]
   Show answer

   $n=\frac{c}{v}$; light is fastest in water and slowest in carbon disulphide; $c=3.00\times 10^8$ m/s, glass $2.00\times 10^8$ m/s; or $n_{wg}=\frac{8}{9}$.

   (a) The absolute refractive index of a medium is the ratio of the speed of light in vacuum to the speed of light in that medium, $n=c/v$. (b) Since $v=c/n$, the speed is greatest where the index is least: (i) the speed of light is **highest** in water ($n=1.33$), and (ii) it is **lowest** in carbon disulphide ($n=1.62$). (c)(i) Using $n_w=c/v_w$: (I) $c=n_w{v}_w=\frac{4}{3}\times 2.25\times 10^8=3.00\times 10^8$ m/s in vacuum; (II) $v_g=c/n_g=3.00\times 10^8÷\frac{3}{2}=2.00\times 10^8$ m/s in glass. (c)(ii) OR — the refractive index of water with respect to glass is $n_{wg}=\frac{n_w}{n_g}=\frac{4/3}{3/2}=\frac{8}{9}\approx 0.89$, a pure number with no unit.

   Step-by-step solution →
6. Q226 CBSE 2023
   Hold a concave mirror in your hand and direct its reflecting surface towards the sun. Direct the light reflected ​by the mirror on to a white card-board held close to the mirror. Move the card-board back and ​forth gradually until you ​find a bright, sharp spot of light on the board. This spot ​of light is ​the image of the sun on the sheet of paper; ​which is also termed as “Principal Focus” of the concave mirror.

   1. (a)List two applications of concave mirror.[1]
   2. (b)If the distance between the mirror and the principal focus is 15 cm, find the radius of curvature of the mirror.[1]
   3. (c)(i)Draw a ray diagram to show the type of image formed when an object is placed between pole and focus of a concave mirror.[2]
   4. (c)(ii)OR : An object 10 cm in size is placed at 100 cm in front of a concave mirror. If its image is formed at the same point where the object is located, find : (i) focal length of the mirror, and (ii) magnification of the image formed with sign as per Cartesian sign convention.[2]
   Show answer

   Concave mirrors serve as shaving mirrors and torch reflectors; $R=30$ cm; between P and F the image is virtual, erect; or $f=-50$ cm.

   (a) Two applications: concave mirrors are used as shaving mirrors and as dentists' mirrors, because an object held between the pole and the focus gives a magnified, erect, virtual image; and they are used as the reflectors of torches, vehicle head-lights and search-lights, where a bulb placed at the focus sends out a parallel beam. (They are also used in solar furnaces to concentrate sunlight.) (b) $R=2f=2\times 15=30$ cm. (c)(i) The image is virtual, erect, enlarged and formed behind the mirror. A ray from the tip of the object parallel to the principal axis is reflected through F, and a ray striking the pole is reflected at an equal angle on the other side of the axis; the two reflected rays diverge, and their backward extensions meet behind the mirror at the tip of the image (see the ray diagram). OR (c)(ii) Since the image is formed at the object's own position, the object is at the centre of curvature, so $u=v=-100$ cm. (i) From $\frac{1}{v}+\frac{1}{u}=\frac{1}{f}$, $\frac{1}{f}=-\frac{1}{100}-\frac{1}{100}=-\frac{1}{50}$, so $f=-50$ cm. (ii) $m=-\frac{v}{u}=-\frac{-100}{-100}=-1$, so the image is real, inverted and the same size as the object, that is $10$ cm tall.

   Step-by-step solution →
7. Q227 CBSE 2023
   The ability of ​a medium ​to refract light is expressed in terms ​of ​its optical density. Optical density has ​a definite connotation. It is not the same as mass density. On ​comparing ​two media, the one with the large refractive index is optically denser ​medium than the other. The other medium with a lower refractive index is optically rarer. Also the speed of light through a given medium is inversely proportional to its optical density.

   1. (a)Determine the speed of light in diamond if the refractive index of diamond with respect to vacuum is $2.42$. Speed of light in vacuum is $3\times 10^8$ m/s.[1]
   2. (b)Refractive indices of glass, water and carbon disulphide are $1.5$, $1.33$ and $1.62$ respectively. If a ray of light is incident in these media at the same angle (say $\theta$), then write the increasing order of the angle of refraction in these media.[1]
   3. (c)(i)The speed of light in glass is $2\times 10^8$ m/s and in water is $2.25\times 10^8$ m/s. (a) Which one of the two is optically denser and why ? (b) A ray of light is incident normally at the water-glass interface when it enters a thick glass container filled with water. What will happen to the path of the ray after entering the glass ? Give reason.[2]
   4. (c)(ii)OR : The absolute refractive indices of water and glass are $4/3$ and $3/2$ respectively. If the speed of light in glass is $2\times 10^8$ m/s, find the speed of light in (i) vacuum and (ii) water.[2]
   Show answer

   Speed in diamond is $1.24\times 10^8$ m/s; angle of refraction increases from carbon disulphide to glass to water; glass is denser.

   (a) $v=c/n=(3\times 10^8)/2.42=1.24\times 10^8$ m/s. (b) Since $\sin r=\sin \theta /n$, a larger refractive index gives a smaller angle of refraction, so the increasing order is carbon disulphide $(n=1.62)$ $<$ glass $(n=1.5)$ $<$ water $(n=1.33)$. (c)(i)(a) Glass is optically denser, because light travels more slowly in it ($2\times 10^8$ m/s against $2.25\times 10^8$ m/s), and $n=c/v$ therefore makes its refractive index the larger, $1.5$ against $1.33$. (b) The ray goes straight on into the glass without bending, because it is incident along the normal, so $i=0^{\circ }$ and Snell's law gives $r=0^{\circ }$; its speed, however, falls from $2.25\times 10^8$ m/s to $2\times 10^8$ m/s. OR (c)(ii) (i) $c=n_g{v}_g=\frac{3}{2}\times 2\times 10^8=3\times 10^8$ m/s; (ii) $v_w=c/n_w=(3\times 10^8)/(4/3)=2.25\times 10^8$ m/s.

   Step-by-step solution →
8. Q228 CBSE 2023
   Many optical instruments consist of a number of lenses. ​They are combined to increase the ​magnification and sharpness of the image. The net ​power $(P)$ of the ​lenses placed in contact is given by the algebraic sum of ​the powers of the individual lenses $P_1$, $P_2$, $P_3$ ... as $$P=P_1+P_2+P_3\dots$$ This is also termed as the simple additive property of the power of lens, ​widely used to design lens systems of cameras, microscopes and telescopes. These lens systems can have a combination of convex lenses and also ​concave lenses.

   1. (a)What is the nature (convergent / divergent) of the combination of a convex lens of power $+4$ D and a concave lens of power $-2$ D ?[1]
   2. (b)Calculate the focal length of a lens of power $-2.5$ D.[1]
   3. (c)(i)Draw a ray diagram to show the nature and position of an image formed by a convex lens of power $+0.1$ D, when an object is placed at a distance of $20$ cm from its optical centre.[2]
   4. (c)(ii)OR : How is a virtual image formed by a convex lens different from that formed by a concave lens ? Under what conditions do a convex and a concave lens form virtual images ?[2]
   Show answer

   The combination is convergent ($P=+2$ D); $f=-40$ cm; the image of the object at 20 cm is virtual, erect and enlarged.

   (a) $P=P_1+P_2=(+4\text{D})+(-2\text{D})=+2$ D. The power is positive, so the combination is **convergent**; it behaves like a convex lens of focal length $f=1/P=0.5$ m $=50$ cm. (b) $f=\frac{1}{P}=\frac{1}{-2.5}=-0.4$ m $=-40$ cm. The negative sign shows that the lens is concave (diverging). (c)(i) $f=\frac{1}{+0.1}=+10$ m $=+1000$ cm, so the object at $20$ cm lies between the optical centre and the principal focus. A ray from the tip of the object parallel to the axis passes through F' after refraction, and a ray through the optical centre goes straight on; the two emergent rays diverge and their backward extensions meet on the same side as the object. The image is therefore **virtual**, **erect** and **enlarged**, formed on the same side as the object (see the ray diagram). Numerically $\frac{1}{v}=\frac{1}{1000}-\frac{1}{20}$, so $v\approx -20.4$ cm and $m=v/u\approx +1.02$. OR (c)(ii) Both images are virtual and erect, but a convex lens gives an **enlarged** image lying beyond the object on the object's own side, while a concave lens gives a **diminished** image lying between the optical centre and the focus. A convex lens forms a virtual image only when the object is placed between its optical centre and its principal focus (object distance less than the focal length); a concave lens forms a virtual image for every position of the object.

   Step-by-step solution →
9. Q229 CBSE 2023
   A student took three ​concave ​mirrors of different focal lengths and performed ​the experiment to see the image formation by ​placing an object at different distances with these mirrors as shown ​in the ​following ​table. Now answer the following ​questions :

   Object-distance and focal length used in the three cases

   Case No.	Object-distance	Focal length
   I	45 cm	20 cm
   II	30 cm	15 cm
   III	20 cm	30 cm

   1. (a)List two properties of the image formed in Case I.
   2. (b)In which one of the cases given in the table, the mirror will form real image of same size and why ?
   3. (c)(i)Name the type of mirror used by dentists. Give reason why do they use such type of mirrors.
   4. (c)(ii)OR : Look at the table and identify the situation (object distance and focal length) which resembles the situation in which concave mirrors are used as shaving mirrors ? Draw a ray diagram to show the image formation in this case.
   Show answer

   Case I: real, inverted image; Case II: real, same-size image; dentists use concave mirrors; or Case III suits shaving mirrors, $m=+3$.

   (a) In Case I the object is beyond $C$ ($u=45$ cm, $2f=40$ cm), and the mirror formula gives $v=-36$ cm with $m=-0.8$. The image is therefore **real** and **inverted** (and, if more are wanted, diminished and formed between $F$ and $C$). (b) CASE II. A real image of the same size needs $m=-1$, which happens only when $u=2f$, that is when the object is at the centre of curvature. In Case II $u=30$ cm $=2\times 15$ cm $=2f$, and the mirror formula gives $v=-30$ cm with $m=-1$ — a real, inverted image of the same size, formed at $C$. (c)(i) Dentists use a **concave** mirror. The tooth is held nearer to the mirror than its focal length, so the mirror gives an enlarged, erect, virtual image of the tooth; a concave mirror also converges light onto the tooth and lights up the area being examined. (c)(ii) OR : CASE III — object distance $20$ cm with focal length $30$ cm. Since $u<f$ the object is between the pole and the focus, which is how a concave mirror is used as a shaving mirror. The mirror formula gives $v=+60$ cm and $m=+3$: the image is virtual, erect and three times the size, formed behind the mirror (see the ray diagram).

   Step-by-step solution →
10. Q230 CBSE 2023
    A simple magnifier or microscope is a convex lens of small focal length. In order to use ​such a lens as a microscope, the lens ​is held near the object to get an erect, magnified and virtual image ​of the object. ​However, this microscope has a ​maximum limiting magnification. For much larger ​magnification we use a compound microscope ​in which two convex lenses are used, one compounding the effect of the other. The lens nearest the object, called the objective, forms a real, inverted and magnified image of the object. This image serves as the object for the second lens, the eyepiece which functions essentially like a ​simple microscope or magnifier and produces the final image.

    1. (a)You have a compound microscope in which a convex lens of focal length $f$ is used as objective. What should be the distance of the object from the objective in terms of $f$ ?[1]
    2. (b)List two properties of the final image obtained from a compound microscope.[1]
    3. (c)(i)An object is placed in front of a convex lens at a distance of two times the focal length of the lens. List two properties of the image formed and write the value of magnification produced by the lens using the New Cartesian Sign Convention.[2]
    4. (c)(ii)OR : An object is placed in front of a convex lens at a distance more than two times the focal length of the lens. List two properties of the image formed and write the value of magnification produced by the lens using the New Cartesian Sign Convention.[2]
    Show answer

    Object between $f$ and $2f$; final image virtual and magnified; at $2f$, $m=-1$ (real, same size); or beyond $2f$, $-1<m<0$.

    (a) The object must be placed between $f$ and $2f$ from the objective - that is, at a distance greater than the focal length but less than twice the focal length. Only then does a convex lens form the real, inverted and **magnified** image that the objective of a compound microscope has to produce. (b) The final image is (i) virtual - it cannot be taken on a screen - and (ii) highly magnified; it is also inverted with respect to the original object. (Any two of these.) (c)(i) With $u=-2f$: $\frac{1}{v}=\frac{1}{f}-\frac{1}{2f}=\frac{1}{2f}$, so $v=+2f$. The image is real and inverted, it is of the same size as the object, and it is formed at $2F_2$ on the other side of the lens. Magnification $m=\frac{v}{u}=\frac{+2f}{-2f}=-1$. OR (c)(ii) With the object farther than $2f$: the image is real and inverted, it is diminished (smaller than the object), and it is formed between $F_2$ and $2F_2$ on the other side of the lens. Magnification $m=\frac{v}{u}$ is negative and its magnitude is less than $1$, that is $-1<m<0$; for example, for an object at $3f$, $v=1.5f$ and $m=-0.5$.

    Step-by-step solution →
11. Q231 CBSE 2023
    When a ray ​of light moving ​in a medium enters obliquely into another medium, it bends from its path. This phenomenon is called refraction of light. The ability of a medium to refract light is also ​expressed in terms of optical density. It is not the same as mass density. We use the terms ‘rarer medium’ and ‘denser medium’ which actually means ‘optically ​rarer medium’ and ‘optically denser medium’ respectively. When we say that a medium ‘A’ is optically denser than the other medium ‘B’, we mean that the refractive index of ​medium A is more than the refractive index of medium ‘B’. The ​speed of light is higher in a rarer medium than a ​denser medium. Thus a ray of light travelling from a ​rarer medium to a denser medium slows down and bends towards the normal.

    1. (a)Define the term absolute refractive index of a medium.[1]
    2. (b)Absolute refractive indices of water and glass are $\frac{4}{3}$ and $\frac{3}{2}$ respectively. (i) In which one of the two media is the speed of light more ? (ii) If a ray of light enters obliquely from glass to water, will it bend towards the normal or away from the normal ?[1]
    3. (c)(i)The absolute refractive indices of water and glass are $\frac{4}{3}$ and $\frac{3}{2}$ respectively. If the speed of light in glass is $2\times 10^8$ m/s, find the speed of light in (i) water and (ii) vacuum.[2]
    4. (c)(ii)OR : “A ray of light incident on a rectangular glass slab immersed in any medium emerges parallel to itself.” Draw a labelled ray diagram to justify this statement.[2]
    Show answer

    Water has the greater speed and the ray bends away from the normal; $v_w=2.25\times 10^8$ m/s and $c=3\times 10^8$ m/s.

    (a) The absolute refractive index of a medium is the ratio of the speed of light in vacuum to the speed of light in that medium: $n=\frac{c}{v}$. It has no unit. (b) (i) In **water**. Since $v=c/n$, the smaller refractive index gives the greater speed, and $n_w=\frac{4}{3}\approx 1.33$ is less than $n_g=\frac{3}{2}=1.50$. (ii) **Away** from the normal. Glass is optically denser than water, so the ray travels from a denser to a rarer medium, speeds up and bends away from the normal. (c)(i) $c=n_g{v}_g=\frac{3}{2}\times 2\times 10^8=3\times 10^8$ m/s. Then (i) in water, $v_w=\frac{c}{n_w}=\frac{3\times 10^8}{4/3}=2.25\times 10^8$ m/s; (ii) in vacuum, $c=3\times 10^8$ m/s. OR (c)(ii) See the labelled ray diagram. At the first face $n_1\sin i=n_2\sin r$; because the faces are parallel the ray meets the second face at the same angle $r$, so $n_2\sin r=n_1\sin e$. Comparing the two gives $\sin e=\sin i$, hence $e=i$: the emergent ray is parallel to the incident ray and only displaced sideways by the lateral displacement $d=\frac{t\sin (i-r)}{\cos r}$. The index $n_1$ of the surrounding medium cancels, so this is true whatever medium the slab is immersed in.

    Step-by-step solution →
12. Q232 CBSE 2020
    Define pole ​of ​a ​spherical ​mirror.
    Show answer

    The pole of a spherical mirror is the centre of its reflecting surface.

    The pole of a spherical mirror is the centre of its reflecting surface. It lies on the mirror itself and is usually denoted by the letter $P$; the principal axis passes through it and is normal to the mirror there.

    Step-by-step solution →
13. Q233 CBSE 2020
    How can ​a concave mirror ​be identified ​without touching ​?
    Show answer

    Hold it close to the face: an erect, enlarged image shows that the mirror is concave.

    Hold the mirror close to your face and look at the image. If the image is erect and enlarged (larger than the face), the mirror is concave. A plane mirror would give an erect image of the same size, and a convex mirror an erect but smaller image. (Moving the mirror away, the image in a concave mirror turns inverted, which a plane or convex mirror never shows.)

    Step-by-step solution →
14. Q234 CBSE 2020
    The ​laws of ​reflection hold ​true ​for ​:

    1. (a)plane mirrors only
    2. (b)concave mirrors only
    3. (c)convex mirrors only
    4. (d)all reflecting surfaces
    Show answer

    (d) all reflecting surfaces

    Step-by-step solution →
15. Q235 CBSE 2020
    A real image is formed by the ​light rays after reflection or refraction when ​they : (A) actually meet or intersect ​with each ​other. (B) actually converge at a point. (C) appear to meet when they are produced in the ​backward direction. (D) appear to diverge from a point. Which of ​the above statements are ​correct?

    1. (a)(A) and (D)
    2. (b)(B) and (D)
    3. (c)(A) and (B)
    4. (d)(B) and (C)
    Show answer

    (c) (A) and (B)

    Step-by-step solution →
16. Q236 CBSE 2020
    When ​an object is kept within the focus ​of a concave mirror, ​an enlarged image ​is ​formed ​behind the mirror. This ​image is :

    1. (a)real
    2. (b)inverted
    3. (c)virtual and inverted
    4. (d)virtual and erect
    Show answer

    (d) virtual and erect

    Step-by-step solution →
17. Q237 CBSE 2020
    Consider the ​following properties ​of virtual images : (A) cannot be ​projected on the ​screen (B) are formed by both concave and convex ​lens (C) are always erect (D) are ​always inverted The ​correct properties are :

    1. (a)(A) and (D)
    2. (b)(A) and (B)
    3. (c)(A), (B) and (C)
    4. (d)(A), (B) and (D)
    Show answer

    (c) (A), (B) and (C)

    Step-by-step solution →
18. Q238 CBSE 2020
    A concave mirror is used for image formation for different ​positions of an object. What inferences can be drawn about ​the following ​when an object is placed at a distance of 10 cm from the pole of a ​concave mirror ​of focal length 15 cm ?

    1. (a)Position of the image
    2. (b)Size of the image
    3. (c)​Nature of the ​image Draw a labelled ray diagram to justify ​your inferences.
    Show answer

    The image is 30 cm behind the mirror, magnified three times, virtual and erect.

    (a) Position: the image is formed behind the mirror, $30$ cm from the pole ($v=+30$ cm). (b) Size: the image is magnified, three times the size of the object ($m=+3$). (c) Nature: the image is virtual and erect. Ray diagram: a ray from the top of the object parallel to the principal axis is reflected through F, and a ray along the line from C is reflected back along itself. The reflected rays diverge, and their backward extensions meet behind the mirror, giving a virtual, erect and enlarged image, as in the figure.

    Step-by-step solution →
19. Q239 CBSE 2020
    1. (a)Define the principal focus of a concave ​mirror. It is ​desired to obtain an erect image of an object using a concave mirror of ​focal length 12 cm. What should be the range of ​the ​object distance in the above ​case ?
    2. (b)Why do we prefer a convex mirror as a rear-view mirror in vehicles ? ​List two ​reasons.
    Show answer

    Object distance: less than 12 cm; a convex mirror gives an erect image and a wider field of view.

    (a) Principal focus: rays of light parallel to the principal axis, after reflection from a concave mirror, meet at a point on the principal axis; this point is the principal focus of the concave mirror. For an erect image the object must be placed between the pole and the principal focus, that is, at a distance of less than $12$ cm from the mirror ($0<|u|<12$ cm). The image is then virtual, erect and enlarged. (b) (i) A convex mirror always forms an erect, though diminished, image of the vehicles behind. (ii) Being curved outwards, it has a wider field of view, so the driver can see a much larger area of the traffic behind.

    Step-by-step solution →
20. Q240 CBSE 2020
    What happens after refraction, when :

    1. (i)a ray of ​light parallel ​to the principal ​axis passes through a ​concave lens ?
    2. (ii)a ray ​of light falls on a convex lens while passing ​through its principal focus ?
    3. (iii)a ray of light ​passes through the optical centre of a convex lens ​?
    Show answer

    (i) The ray diverges as if from $F_1$; (ii) it emerges parallel to the axis; (iii) it passes undeviated.

    (i) The ray diverges after refraction: it bends away from the principal axis and appears to come from the principal focus $F_1$ on the same side of the lens as the incoming light. (ii) The ray emerges parallel to the principal axis after refraction. (iii) The ray passes through the optical centre without any deviation, continuing in its original direction.

    Step-by-step solution →