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 19 of 21

1. Q361 CBSE 2017
   “A ​lens can form a magnified erect image ​as well as magnified inverted image of an object placed in front of it.” ​State the nature ​of this lens and draw ray diagrams to justify the above ​statement. Mark the positions of $O$, $F$ and $2F$ in the diagram.
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   A convex lens: magnified erect image for an object between F and O, magnified inverted image for an object between F and 2F.

   The lens is a convex (converging) lens. A concave lens always forms a virtual, erect and diminished image of a real object, so only a convex lens can form magnified images. (i) Magnified erect image: the object is placed between $F_1$ and $O$. The ray parallel to the axis is refracted through $F_2$ and the ray through $O$ goes straight on; the refracted rays diverge, and their backward extensions meet on the object's side, giving a virtual, erect, magnified image (first ray diagram). (ii) Magnified inverted image: the object is placed between $F_1$ and $2F_1$. The same two rays meet beyond $2F_2$ on the other side of the lens, giving a real, inverted, magnified image (second ray diagram). $O$, $F_1$, $F_2$, $2F_1$ and $2F_2$ are marked on the principal axis in both diagrams.

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2. Q362 CBSE 2017
   Distinguish between a real ​and a virtual image. Write two characteristics of the image formed ​by a concave mirror ​when an object is placed ​in ​front of the mirror between ​its centre of curvature and focus.
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   A real image can be caught on a screen, a virtual one cannot; between C and F the image is real, inverted, enlarged.

   Real image: formed where the reflected (or refracted) rays actually meet; it can be obtained on a screen; for an object in front of a single mirror or lens it is inverted. Virtual image: formed where the rays only appear to meet when produced backwards; it cannot be obtained on a screen; for an object in front of a single mirror or lens it is erect. Object between $C$ and $F$ of a concave mirror: the image is formed beyond $C$, and it is (1) real and inverted, (2) enlarged, that is, larger than the object.

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3. Q363 CBSE 2017
   State the laws that are ​followed when ​light is ​reflected by spherical mirrors. Draw a ​ray diagram to ​show the formation of image of an object placed infront of a convex mirror. List two characteristics of the ​image formed. Briefly explain one ​use of convex ​mirrors.
   Show answer

   Equal angles and a common plane; a convex mirror gives a virtual, erect image and serves as a wide-view rear-view mirror.

   Laws of reflection, which hold for spherical mirrors as for plane mirrors: (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. At a spherical mirror the normal at any point is the radius through that point, which passes through the centre of curvature. Ray diagram: from the top $A$ of the object, a ray parallel to the principal axis is reflected so that it appears to come from the principal focus $F$ behind the mirror, and a ray directed towards the centre of curvature $C$ is reflected back along its own path. The reflected rays diverge; produced backwards, they meet behind the mirror between $P$ and $F$, where the image $A^{\prime }{B}^{\prime }$ is formed (see the figure). Characteristics: the image is virtual and erect. (It is also diminished, and it lies behind the mirror between $P$ and $F$.) Use: convex mirrors are used as rear-view (wing) mirrors in vehicles. They always give an erect, though diminished, image, and being curved outwards they have a wider field of view than a plane mirror, so the driver can see a much larger area of the traffic behind.

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4. Q364 CBSE 2017
   1. (a)​Define the term ​“magnification” ​in the context of spherical mirrors. Express it in terms of object-distance ($u$) and image-distance ($v$). ​Besides size and position of image, what other two informations it provides about the image ?
   2. (b)Draw a ray diagram to show the formation of image by a concave mirror when ​the object is placed in front of it between its centre of curvature and focus. What is the ​nature of ​the image formed ?
   Show answer

   $m=\frac{h^{\prime }}{h}=-\frac{v}{u}$, and its sign shows whether the image is real or virtual, erect or inverted; (b) real, inverted, enlarged image beyond $C$.

   (a) The magnification produced by a spherical mirror is the ratio of the height of the image ($h^{\prime }$) to the height of the object ($h$): $m=\frac{h^{\prime }}{h}$. In terms of the object distance and the image distance, $m=-\frac{v}{u}$. Besides the size and position of the image, the magnification tells us (1) whether the image is real or virtual and (2) whether it is erect or inverted: a negative value of $m$ means the image is real and inverted, and a positive value means it is virtual and erect. (b) Ray diagram: from the top A of the object AB placed between C and F, a ray parallel to the principal axis is reflected through F, and a ray through F is reflected parallel to the principal axis; the reflected rays meet beyond C, where the image A′B′ is formed. Nature of the image: real, inverted and enlarged (magnified), formed beyond C.

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5. Q365 CBSE 2017
   (a) Draw labelled ray diagrams for each of the following cases to show the position, nature and size of ​the ​image formed by a convex lens ​when the object ​is placed

   1. (i)between its optical centre (O) and ​principal focus (F)
   2. (ii)between ​F and 2F (b) How will the nature and size of the image formed in the ​above two cases, (i) and (ii) change, if the convex lens is replaced by a concave lens of same ​focal length ?
   Show answer

   Convex lens: virtual, erect, enlarged, then real, inverted, enlarged; a concave lens gives a virtual, erect, diminished image in both cases.

   (a) (i) Object between $O$ and $F_1$: the image $A^{\prime }{B}^{\prime }$ is formed on the same side of the lens as the object, farther from the lens than the object; it is virtual and erect, and enlarged. (The refracted rays diverge; their backward extensions meet at $A^{\prime }$.) (ii) Object between $F_1$ and $2F_1$: the image is formed beyond $2F_2$ on the other side of the lens; it is real and inverted, and enlarged. (The refracted rays actually meet at $A^{\prime }$.) (b) A concave lens of the same focal length gives, in both cases, a virtual, erect and diminished image, formed between the focus $F_1$ and $O$ on the same side as the object. In case (i) the nature stays the same (virtual and erect) but the size changes from enlarged to diminished. In case (ii) the nature changes from real and inverted to virtual and erect, and the size changes from enlarged to diminished.

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6. Q366 CBSE 2017
   1. (a)If the image formed by ​a mirror ​for all positions of the object ​placed ​in front of it is always erect and diminished, what type of mirror is it ? Draw a ​ray diagram in support of your answer. Where and why do we generally use this type of mirror ?
   2. (b)The image of an object placed ​in front of a mirror at a distance of ​20 cm is formed ​behind the mirror at a distance of 10 cm from it. Find the nature and focal length of the mirror.
   Show answer

   (a) A convex mirror, used as a rear-view mirror for its wide field of view; (b) $f=+20$ cm, so the mirror is convex.

   (a) It is a convex mirror: for every position of the object in front of it, a convex mirror forms a virtual, erect and diminished image, between its pole P and focus F behind the mirror (see the ray diagrams for an object far from and close to the mirror). Convex mirrors are generally used as rear-view (wing) mirrors in vehicles, because they always give an erect, though diminished, image and, being curved outwards, have a wider field of view, so the driver can see a much larger area of the traffic behind. (b) With $u=-20$ cm and $v=+10$ cm, $\frac{1}{f}=\frac{1}{v}+\frac{1}{u}=\frac{1}{10}-\frac{1}{20}=\frac{1}{20}$, so $f=+20$ cm. The focal length is positive, so the mirror is convex, and its focal length is $20$ cm. (The image is virtual, erect and half the size of the object: $m=+0.5$.)

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7. Q367 CBSE 2017
   1. (a)What is meant by power of a ​lens and define its SI unit ?
   2. (b)Three ​friends ‘A’, ‘B’ and ‘C’ have ​corrective lenses of focal ​lengths $+50$ cm, $+100$ cm and $-100$ cm ​respectively. Find the ​nature and power of each ​of these lenses.
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   Lens A is convex with $+2$ D, lens B is convex with $+1$ D and lens C is concave with $-1$ D.

   (a) The power of a lens is the degree of convergence or divergence of light rays achieved by it. It is the reciprocal of the focal length expressed in metres, $P=\frac{1}{f}$. Its SI unit is the dioptre (D): 1 dioptre is the power of a lens whose focal length is 1 metre ($1\text{D}=1{\text{m}}^{-1}$). (b) Lens of A: $f=+50$ cm $=+0.50$ m, so $P=\frac{1}{0.50}=+2$ D; the lens is convex (converging). Lens of B: $f=+100$ cm $=+1.00$ m, so $P=+1$ D; the lens is convex (converging). Lens of C: $f=-100$ cm $=-1.00$ m, so $P=-1$ D; the lens is concave (diverging).

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8. Q368 CBSE 2017
   Analyse the following observation table showing variation of ​image distance ($v$) with ​object ​distance ($u$) in case of a convex lens and answer the questions that follow, without doing any calculations :

   1. (a)What ​is the ​focal length of the convex lens ? Give reason in support of your answer.
   2. (b)Write the serial ​number of that observation ​which is not correct. How did you ​arrive at this conclusion ?
   3. (c)Take an appropriate scale to draw ray diagram for the observation at S. No. 4 and find the approximate value of magnification.

   S. No.	Object distance $u$ (cm)	Image distance $v$ (cm)
   1	−90	+18
   2	−60	+20
   3	−30	+30
   4	−20	+60
   5	−18	+90
   6	−10	+100

   Show answer

   $f=+15$ cm from observation 3; observation 6 is incorrect (object inside $f$ cannot give a real image); $m\approx -3$ for observation 4.

   (a) The focal length is $15$ cm ($f=+15$ cm). In observation 3 the object ($u=-30$ cm) and the image ($v=+30$ cm) are at equal distances on opposite sides of the lens. A convex lens does this only when the object is at $2F_1$, the image then forming at $2F_2$, so $2f=30$ cm and $f=15$ cm. (b) Observation 6 is not correct. Its object distance, $10$ cm, is less than the focal length, $15$ cm, so the object lies between the optical centre and $F_1$. A convex lens then forms a virtual, erect, enlarged image on the same side as the object, with a negative image distance; so with the object at $10$ cm it cannot form a real image at $+100$ cm on the other side. (c) Drawn to the scale 1 cm = $5$ cm with the object at $20$ cm (between $F_1$ and $2F_1$): the ray parallel to the axis refracts through $F_2$ and the ray through O goes straight on; they meet about $60$ cm behind the lens, where the image is real, inverted and about three times the height of the object (see the diagram). The magnification is approximately $-3$.

   Step-by-step solution →
9. Q369 CBSE 2017
   Analyse the following observation table showing variation of image-distance ($v$) with object-distance ($u$) in case of a convex ​lens and answer the questions that follow without doing ​any calculations :

   1. (a)What is the focal ​length of the convex lens ? State reason for your answer.
   2. (b)For what ​object-distance ($u$) is the corresponding image-distance ($v$) not correct ? ​How did you arrive at this conclusion ?
   3. (c)Choose an appropriate scale to draw a ​ray diagram ​for the observation at ​S. No. 4 and find the approximate value of magnification.

   S. No.	Object-Distance $u$ (cm)	Image-Distance $v$ (cm)
   1	$-60$	$+12$
   2	$-30$	$+15$
   3	$-20$	$+20$
   4	$-15$	$+30$
   5	$-12$	$+60$
   6	$-9$	$+90$

   Show answer

   $f=+10$ cm; the image distance for $u=-9$ cm is wrong; the magnification for S. No. 4 is about $-2$.

   (a) The focal length is $10$ cm ($f=+10$ cm). Reason: at S. No. 3 the object and the image are both 20 cm from the lens. A convex lens forms a real image as far from the lens as the object only when the object is at $2F_1$ (the image then forms at $2F_2$), so $2f=20$ cm and $f=10$ cm. (b) The image distance is not correct for $u=-9$ cm (S. No. 6). With $f=10$ cm, an object 9 cm from the lens lies between the optical centre and $F_1$. A convex lens then forms a virtual, erect, magnified image on the same side as the object, so $v$ must be negative; a real image at $v=+90$ cm is not possible. (c) With a scale of 1 cm to 5 cm the object is drawn 3 cm from the lens, the foci 2 cm from it, and the rays meet 6 cm (30 cm to scale) beyond the lens. The image is real, inverted and about twice as tall as the object, so the magnification is approximately $-2$.

   Step-by-step solution →
10. Q370 CBSE 2017
    Analyse ​the ​following observation ​table ​showing variation of image-distance ($v$) with object-distance ($u$) in case of a convex lens and answer the questions ​that follow without doing any calculations :

    1. (a)What is the focal length of the convex ​lens ? Give reason to justify your answer.
    2. (b)Write the serial number of the observation which is not correct. On what basis have you arrived at this conclusion ​?
    3. (c)Select an appropriate scale and draw a ​ray diagram for the observation at S.No.2. Also find the approximate value of magnification.

    Object distance and image distance for a convex lens

    S.No.	Object-Distance u (cm)	Image-Distance v (cm)
    1	−100	+25
    2	−60	+30
    3	−40	+40
    4	−30	+60
    5	−25	+100
    6	−15	+120

    Show answer

    (a) $f=+20$ cm; (b) S.No. 6 is incorrect; (c) the image is real, inverted and half the size, $m\approx -0.5$.

    (a) $f=+20$ cm. Reason: in S.No. 3 the object and image distances are equal ($u=-40$ cm, $v=+40$ cm), which happens only when the object is at $2F_1$ and the image at $2F_2$; so $2f=40$ cm and $f=20$ cm. (b) S.No. 6 is not correct. Its object distance, $15$ cm, is less than the focal length, $20$ cm, so the object is between $F_1$ and $O$; a convex lens then forms a virtual, erect image on the same side as the object, so $v$ must be negative, not $+120$ cm. (c) With the scale $1$ cm $=10$ cm (object at $6$ cm, $F$ at $2$ cm), the rays meet $3$ cm from the lens, that is $v=+30$ cm, between $F_2$ and $2F_2$; the image is real, inverted and about half the size of the object, so the magnification is approximately $m=-0.5$.

    Step-by-step solution →
11. Q371 CBSE 2017
    1. (a)A student has focused on the screen the image of a 2 cm high candle flame placed at a distance ​of 24 cm from a convex lens of focal length ​16 cm. ​Use lens formula to calculate the distance of the screen from the lens and length of the image formed.
    2. (b)If the distance between the lens and the flame is increased to 32 ​cm, where would ​the image be formed ​and what would ​be its size ? ​Give reason to support your answer.
    Show answer

    (a) Screen 48 cm from the lens, image 4 cm long; (b) image at 32 cm ($2F_2$), 2 cm long, as the flame is at $2F_1$.

    (a) With $u=-24$ cm and $f=+16$ cm, $\frac{1}{v}=\frac{1}{16}-\frac{1}{24}=\frac{1}{48}$, so $v=+48$ cm: the screen is $48$ cm from the lens, on the other side from the flame. $m=\frac{v}{u}=\frac{48}{-24}=-2$, so $h^{\prime }=-2\times 2=-4$ cm: the image is $4$ cm long, inverted, real and magnified. (b) With $u=-32$ cm, $\frac{1}{v}=\frac{1}{16}-\frac{1}{32}=\frac{1}{32}$, so $v=+32$ cm: the image is formed $32$ cm from the lens on the other side, at $2F_2$. $m=\frac{32}{-32}=-1$, so $h^{\prime }=-2$ cm: the image is $2$ cm long, the same size as the flame, inverted and real. Reason: $32$ cm is twice the focal length, so the flame is at $2F_1$, and a convex lens forms the image of an object at $2F_1$ at $2F_2$, real, inverted and of the same size.

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12. Q372 CBSE 2017
    (a) ​Explain with the help of diagrams the following terms in the context of spherical mirrors :

    1. (i)Pole
    2. (ii)Centre of curvature
    3. (iii)Principal axis (b) ​Consider ​the following diagram. In this M is a mirror, P is an object and Q is ​the image of the object formed by the mirror : (i) ​State the ​type of mirror M. (ii) If the ​numeric value of focal length of the mirror is 15 cm, what must be the range of distance of the object from ​the mirror ? (iii) Draw a ray diagram to show the formation of image in this case.

    Show answer

    M is a concave mirror; the object must be less than 15 cm from it, between the pole and the focus.

    (a) (i) Pole: the centre of the reflecting surface of a spherical mirror, a point on the mirror, written P. (ii) Centre of curvature: the reflecting surface is part of a sphere, and the centre of that sphere is the centre of curvature C; it lies in front of a concave mirror and behind a convex mirror. (iii) Principal axis: the straight line passing through the pole and the centre of curvature; it is normal to the mirror at its pole (see the diagrams). (b) (i) M is a concave mirror: the image Q is behind the mirror (virtual), erect and larger than the object, and only a concave mirror forms such an image. (ii) The object must lie between the pole and the principal focus, that is, at a distance of less than 15 cm from the mirror (0 to 15 cm). (iii) In the ray diagram a ray parallel to the axis is reflected through F and a ray along the line from C returns along itself; the reflected rays diverge, and their backward extensions meet behind the mirror to give a virtual, erect, enlarged image (for an object at 10 cm, the image is 30 cm behind the mirror and three times larger).

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13. Q373 CBSE 2017
    1. (a)Two lenses ‘A’ and ‘B’ of ​focal length $+15$ cm and $-15$ cm respectively were given to a ​student. He fixed these lenses on separate optical benches and placed a burning candle in front of each at a distance of 20 cm from the optical centre of ​each lens. He adjusted the screen placed on other ​side of each lens to obtain the ​image of the flame. What ​will he observe on the screen in each case ? Give reason for your answer. Explain with the help of ray diagrams for each.
    2. (b)The radius of curvature of a convex mirror used to see the rear view in ​a car is 2.0 m. If a bus is located at a distance of ​3.0 m from this mirror, find position and nature of the image seen in the mirror.
    Show answer

    Lens A gives a real, inverted, enlarged image on the screen; lens B gives none; the bus's image is 0.75 m behind the mirror.

    (a) Lens A (convex, $f=+15$ cm): the candle, 20 cm away, is between F₁ and 2F₁, so the student obtains a sharp, real, inverted and enlarged image of the flame on the screen when the screen is 60 cm from the lens on the other side ($v=+60$ cm, $m=-3$: three times the size of the flame). The refracted rays actually meet there, so the image can be caught on the screen. Lens B (concave, $f=-15$ cm): he obtains no image of the flame on the screen at any position. A concave lens makes the rays spread out; they only appear to come from a virtual, erect, diminished image on the same side as the candle ($v=-\frac{60}{7}\approx -8.6$ cm, $m=+\frac{3}{7}\approx +0.43$), and a virtual image cannot be caught on a screen. Ray diagrams: for lens A, a ray from the tip of the flame parallel to the axis passes through F₂ and a ray through O goes straight on; they meet beyond 2F₂. For lens B, the parallel ray spreads out as if from F₁ and the ray through O goes straight on; the two diverge, and their backward extensions meet between F₁ and O. (b) $f=\frac{R}{2}=+1.0$ m and $u=-3.0$ m, so $\frac{1}{v}=\frac{1}{1.0}+\frac{1}{3.0}=\frac{4}{3}$ and $v=+0.75$ m. The image is 0.75 m behind the mirror (between P and F); it is virtual, erect and diminished, with $m=+0.25$, one-quarter of the size of the bus.

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14. Q374 CBSE 2017
    1. (a)To construct a ray diagram we use two rays which are so chosen ​that ​it is easy to know their directions after reflection ​from the mirror. List two such rays and state the path of ​these rays after reflection in case of concave mirrors. Use these two rays and draw ray diagram to locate the image of an object placed between pole and focus of a concave mirror.
    2. (b)A concave mirror produces ​three times magnified image on a screen. If the object is placed $20$ cm in ​front ​of ​the mirror, how far is the screen from the object ?
    Show answer

    (a) The image is virtual, erect and enlarged; (b) the screen is 60 cm from the mirror, 40 cm from the object.

    (a) Two convenient rays: (i) a ray parallel to the principal axis, which after reflection from a concave mirror passes through the principal focus $F$; (ii) a ray passing through the centre of curvature $C$, which strikes the mirror along the normal and is reflected back along the same path. For an object placed between $P$ and $F$, the two reflected rays diverge; when produced backwards they meet behind the mirror. The image is formed behind the mirror and is virtual, erect and enlarged (see the ray diagram). (b) An image on a screen is real, hence inverted, so $m=-3$. With $u=-20$ cm, $m=-\frac{v}{u}$ gives $v=-60$ cm: the screen is $60$ cm in front of the mirror. The screen and the object are on the same side of the mirror, so the screen is $60-20=40$ cm from the object.

    Step-by-step solution →
15. Q375 CBSE 2017
    (a) Define the following terms in the context of spherical mirrors :

    1. (i)Pole
    2. (ii)Centre of curvature
    3. (iii)Radius of ​curvature
    4. (iv)​Principal axis (b) Draw ray diagrams ​to show the principal focus ​of (i) a concave ​mirror, and (ii) a convex mirror. (c) ​In the following diagram, MM′ is a concave mirror and AB is an object. Draw on your answer-sheet a ray diagram ​to show the formation of image of ​this object.

    Show answer

    An object between $F$ and $P$ of a concave mirror gives a virtual, erect, enlarged image behind the mirror.

    (a)(i) Pole: the centre of the reflecting surface of a spherical mirror. It is a point on the surface of the mirror, usually written $P$. (a)(ii) Centre of curvature: the centre of the sphere of which the reflecting surface of the mirror is a part, written $C$. It is not a part of the mirror; it lies in front of a concave mirror and behind a convex mirror. (a)(iii) Radius of curvature: the radius of the sphere of which the reflecting surface of the mirror is a part, written $R$; it is equal to the distance $PC$. (a)(iv) Principal axis: the straight line passing through the pole and the centre of curvature of a spherical mirror; it is normal to the mirror at its pole. (b)(i) Concave mirror: rays parallel to the principal axis, after reflection, converge and actually pass through a point $F$ on the principal axis in front of the mirror. This is the principal focus; it is real. (b)(ii) Convex mirror: rays parallel to the principal axis, after reflection, diverge and appear to come from a point $F$ on the principal axis behind the mirror. This is the principal focus; it is virtual. For both mirrors of small aperture $F$ lies midway between $P$ and $C$ ($f=R/2$). (c) The object lies between $F$ and the pole. A ray from $A$ parallel to the principal axis is reflected through $F$, and a ray from $A$ along the line from $C$ is reflected back along itself. The reflected rays diverge; their backward extensions meet behind the mirror at $A^{\prime }$. The image $A^{\prime }{B}^{\prime }$ is formed behind the mirror and is virtual, erect and enlarged.

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16. Q376 CBSE 2017
    1. (a)If the image formed by a mirror for all positions ​of the object placed in front of it ​is always diminished, erect and virtual, state the ​type of the ​mirror and also ​draw a ray diagram to justify your answer. Write one use such mirrors are put to and why.
    2. (b)Define the radius ​of curvature of ​spherical mirrors. Find the nature and focal length of a spherical mirror whose radius of curvature is ​+ 24 cm.
    Show answer

    A convex mirror, used as a rear-view mirror for its wide field of view; $R=+24$ cm means a convex mirror with $f=+12$ cm.

    (a) The mirror is a convex mirror: for every position of the object it forms a virtual, erect and diminished image behind the mirror, between $P$ and $F$ (see the ray diagram). Use: convex mirrors are used as rear-view (wing) mirrors in vehicles, because they always give an erect, though diminished, image and, being curved outwards, have a wider field of view, so the driver sees a larger area behind. (b) The radius of curvature of a spherical mirror is the radius of the sphere of which its reflecting surface is a part; it is the distance between the pole and the centre of curvature. For $R=+24$ cm the centre of curvature lies behind the mirror, so the mirror is convex, and its focal length is $f=\frac{R}{2}=+12$ cm.

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17. Q377 CBSE 2017
    Write any four ​main points of the New Cartesian sign ​convention for reflection by spherical mirrors. If according of this convention, the radius of curvature of a spherical mirror is 30 cm, write the name and focal length of the ​mirror. An object is placed at a distance of 30 cm from a concave mirror of focal ​length 20 cm. At what distance from ​the mirror should a screen be placed in order to obtain a ​sharp image of the ​object on it.
    Show answer

    Convex mirror with $f=+15$ cm; the screen is 60 cm in front of the concave mirror ($v=-60$ cm).

    Four points of the New Cartesian sign convention (pole $P$ as origin, principal axis as the $x$-axis): (1) The object is always placed to the left of the mirror, so that light from the object falls on the mirror from the left. (2) All distances parallel to the principal axis are measured from the pole. (3) Distances measured to the right of the pole, in the direction of the incident light, are positive; those measured to the left are negative. (4) Distances measured perpendicular to the principal axis are positive above it and negative below it. Radius of curvature $+30$ cm: the centre of curvature lies behind the mirror, so the mirror is a convex mirror; its focal length is $f=\frac{R}{2}=+15$ cm. Screen: with $u=-30$ cm and $f=-20$ cm, $\frac{1}{v}=\frac{1}{f}-\frac{1}{u}=-\frac{1}{20}+\frac{1}{30}=-\frac{1}{60}$, so $v=-60$ cm. The screen should be placed $60$ cm from the mirror, in front of it, on the same side as the object; the image there is real, inverted and twice the size of the object.

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18. Q378 CBSE 2017
    State the laws that ​are followed when light is refracted by lenses. Draw a ray diagram to show formation of ​image by a ​concave ​lens. List any two characteristics of the image formed. The magnification produced by a concave lens ​is +0.5. ​What does this mean ​?
    Show answer

    Refraction obeys the two laws; the concave lens gives a virtual, erect, diminished image; $m=+0.5$ means an erect image half the object's size.

    Laws of refraction (obeyed at each surface of the lens): (1) The incident ray, the refracted ray and the normal to the interface 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 constant for light of a given colour and a given pair of media: $\frac{\sin i}{\sin r}=\text{constant}$ (Snell's law). Ray diagram: a ray from the top $A$ of the object, parallel to the principal axis, diverges after refraction as if it came from the focus $F_1$; a ray through the optical centre $O$ goes on undeviated. The backward extension of the first ray meets the second ray at $A^{\prime }$, between $F_1$ and $O$, giving the image $A^{\prime }{B}^{\prime }$ (see the figure). Characteristics (any two): the image is virtual and erect; it is diminished; it is formed between $O$ and $F_1$, on the same side of the lens as the object. Magnification $+0.5$: the positive sign means that the image is erect (and virtual); $0.5$ means that the image is half the size of the object, $h^{\prime }=0.5h$.

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19. Q379 CBSE 2017
    1. (a)Name ​and state the law of refraction of light that ​explains the refractive ​index of one medium with ​respect to the other. Express it ​mathematically. How is refractive index related ​to ​velocity of light in one medium with respect to the second ? Write an expression for the absolute ​refractive index of a medium.
    2. (b)Refractive index of water is 1.33 and that of turpentine oil is 1.47. Draw a ray diagram to show the path of a ray passing from turpentine oil into water.
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    Snell's law: $\frac{\sin i}{\sin r}=n_{21}=\frac{v_1}{v_2}$, with $n_m=\frac{c}{v}$; a ray from turpentine oil into water bends away from the normal.

    (a) The law is Snell’s law of refraction (the second law of refraction): 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. Mathematically, $\frac{\sin i}{\sin r}=n_{21}$, where the constant $n_{21}$ is the refractive index of the second medium with respect to the first. In terms of speeds, $n_{21}=\frac{v_1}{v_2}$: the speed of light in medium 1 divided by the speed of light in medium 2. The absolute refractive index of a medium is $n_m=\frac{c}{v}$, where $c$ is the speed of light in vacuum and $v$ its speed in the medium. (b) Turpentine oil ($n=1.47$) is optically denser than water ($n=1.33$), so a ray passing from turpentine oil into water bends away from the normal. In the ray diagram the refracted ray in the water makes a larger angle with the normal than the incident ray in the oil ($r>i$; for $i=40^{\circ }$, $\sin r=\frac{1.47}{1.33}\sin 40^{\circ }$ gives $r\approx 45.3^{\circ }$).

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20. Q380 CBSE 2017
    A student is performing the experiment of determining the focal length of a ​given ​concave mirror by focussing a distant ​tree on a ​screen. Which one of the following ​kinds of images he is ​likely to ​obtain on the screen ​?

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

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