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 15 of 21

1. Q281 CBSE 2019
   Light enters from air to ​glass having refractive index 1.50. Calculate the ​speed of light in ​the glass. Given : The ​speed of light in vacuum ​is = $3\times 10^8$ m/s.
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   The speed of light in the glass is $2\times 10^8$ m/s.

   Speed of light in the glass: $v_g=\frac{c}{n_g}=\frac{3\times 10^8}{1.50}=2\times 10^8$ m/s.

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2. Q282 CBSE 2019
   The refractive index ​of a medium ​Y with respect to medium X ​is 1.6 and ​the absolute refractive ​index ​of X is 1.5. Find out the speed of light in the medium Y. The speed ​of light in vacuum is $3\times 10^8{\text{ms}}^{-1}$.
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   The speed of light in medium Y is $1.25\times 10^8$ m/s.

   The speed of light in medium Y is $v_Y=1.25\times 10^8{\text{m s}}^{-1}$. (Absolute refractive index of Y: $n_Y=n_{YX}\times n_X=1.6\times 1.5=2.4$, so $v_Y=\frac{c}{n_Y}=\frac{3\times 10^8}{2.4}=1.25\times 10^8$ m s${}^{-1}$.)

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3. Q283 CBSE 2019
   How is refractive index of medium 2 with respect to medium ​1 represented in terms ​of speed of light ​in the ​two media ? ​If the refractive index of medium 2 with ​respect ​to medium 1 is 1.6 and the speed of light in medium ​1 is $2\times 10^8{\text{ms}}^{-1}$, find the speed of light in medium 2.
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   $n_{21}=\frac{v_1}{v_2}$; the speed of light in medium 2 is $v_2=1.25\times 10^8$ m/s.

   The refractive index of medium 2 with respect to medium 1 is the ratio of the speed of light in medium 1 to the speed of light in medium 2: $n_{21}=\frac{v_1}{v_2}$. Speed of light in medium 2: $v_2=\frac{v_1}{n_{21}}=\frac{2\times 10^8}{1.6}=1.25\times 10^8{\text{m s}}^{-1}$.

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4. Q284 CBSE 2019
   A concave mirror has a focal ​length of ​20 cm. At what distance from the ​mirror should a 4 cm tall object be placed so that it forms ​an image ​at ​a distance of 30 cm from the mirror ? Also ​calculate the size of the image formed.
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   The object should be 60 cm in front of the mirror; the image is 2 cm tall, real and inverted.

   The object must be placed at $u=-60$ cm, that is 60 cm in front of the mirror. The image size is $h^{\prime }=-2$ cm: the image is 2 cm tall and inverted ($m=-\frac{1}{2}$), real and diminished, formed 30 cm in front of the mirror between $F$ and $C$.

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5. Q285 CBSE 2019
   A ​real image $2/3^{\text{rd}}$ of the size of an ​object ​is formed by ​a ​convex lens when ​the ​object is at a distance of 12 cm from it. Find the focal length of the lens.
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   The focal length is $f=+4.8$ cm (image distance $v=+8$ cm).

   The image is real, so it is inverted: $m=-\frac{2}{3}$. With $u=-12$ cm, $v=m\times u=+8$ cm, and $\frac{1}{f}=\frac{1}{v}-\frac{1}{u}=\frac{1}{8}+\frac{1}{12}=\frac{5}{24}$, so the focal length of the lens is $f=\frac{24}{5}=+4.8$ cm.

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6. Q286 CBSE 2019
   How far should an object be placed ​from a ​convex ​lens of ​focal length 20 cm ​to obtain its real image at a distance of 30 cm from ​the lens ? Determine the height of the image if the ​object is 4 ​cm tall.
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   The object is placed 60 cm in front of the lens; the image is real, inverted and 2 cm tall.

   The object should be placed $60$ cm in front of the lens ($u=-60$ cm). The image is $2$ cm tall and inverted ($h^{\prime }=-2$ cm, $m=-0.5$).

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7. Q287 CBSE 2019
   Where should an object be placed ​in ​front of a concave mirror of focal length 20 ​cm ​so ​as to obtain a two times magnified virtual image of the object ​?
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   The object should be placed 10 cm in front of the mirror, between the pole and the focus ($u=-10$ cm).

   The object should be placed $10$ cm in front of the mirror ($u=-10$ cm), between its pole and its focus. The virtual, erect image is then formed $20$ cm behind the mirror ($v=+20$ cm), twice the size of the object.

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8. Q288 CBSE 2019
   A student holding a ​mirror in his hand, directed the reflecting surface of the mirror towards the Sun. He then ​directed the reflected light on to a sheet of paper ​held close to the mirror.

   1. (a)What should he do to burn the paper ?
   2. (b)Which type of mirror does he have ?
   3. (c)Will he be able to determine the approximate value of focal length of ​this mirror from this activity ? ​Give reason and ​draw ray ​diagram to justify your answer in ​this case.
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   He focuses sunlight into a sharp spot and holds it steady; the mirror is concave; the distance $PF$ gives its approximate focal length.

   (a) He should move the sheet of paper (or the mirror) back and forth until the reflected sunlight is focused into the smallest, sharpest bright spot on the paper, and then hold the mirror and the paper steady in that position for a few minutes. The sunlight collected by the whole mirror is concentrated on that spot, and the paper begins to smoke and catches fire. (b) A concave (converging) mirror. (c) Yes. The Sun is very far away, so its rays reach the mirror parallel to the principal axis, and after reflection from a concave mirror they meet at the principal focus F. The sharp bright spot on the paper, the image of the Sun, is therefore formed at F, and the distance between the mirror and the paper (PF) gives the approximate focal length of the mirror. Ray diagram: rays from the Sun parallel to the principal axis are reflected by the concave mirror and all pass through F on the axis, where the paper is held; PF $=f$.

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9. Q289 CBSE 2019
   A ​10 cm tall object ​is placed ​perpendicular ​to the principal axis of a convex lens of focal length 12 cm. The distance of the object ​from the lens ​is 18 cm. Find the nature, position and size ​of ​the image formed.
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   The image is real and inverted, 36 cm from the lens on the other side, and 20 cm tall (twice the object).

   Position: $v=+36$ cm, so the image is formed $36$ cm from the lens on the side opposite to the object (beyond $2F_2$). Nature: real and inverted ($v$ is positive and $m=\frac{v}{u}=\frac{36}{-18}=-2$), and magnified. Size: $h^{\prime }=m\times h=-2\times 10=-20$ cm, so the image is $20$ cm tall, twice the height of the object, and inverted.

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10. Q290 CBSE 2019
    1. (a)​List four characteristics of the image formed by a convex lens when an object is ​placed between its ​optical centre and ​principal focus.
    2. (b)Size of the image of ​an object ​by ​a concave lens of focal length 20 cm is observed to be reduced to ${\frac{1}{3}}^{\text{rd}}$ of its size. Find the distance of the object from the ​lens.
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    The image is virtual, erect, enlarged and on the object's side; the object is 40 cm from the concave lens ($u=-40$ cm).

    (a) When an object is placed between the optical centre and the principal focus of a convex lens, the image formed is: (i) virtual; (ii) erect; (iii) enlarged (magnified); and (iv) on the same side of the lens as the object, farther from the lens than the object. (b) For a concave lens $f=-20$ cm, and its image is virtual and erect, so $m=\frac{v}{u}=+\frac{1}{3}$, which gives $v=\frac{u}{3}$. By the lens formula, $\frac{1}{v}-\frac{1}{u}=\frac{1}{f}$: $\frac{3}{u}-\frac{1}{u}=-\frac{1}{20}$, so $\frac{2}{u}=-\frac{1}{20}$ and $u=-40$ cm. The object is $40$ cm from the lens, on the side from which the light comes (the image is at $v=-\frac{40}{3}\approx -13.3$ cm).

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11. Q291 CBSE 2019
    A concave ​lens of focal length 60 cm is used to ​form an image of an ​object of length 9 cm kept at ​a distance of 30 cm ​from it. Use lens formula to determine ​the nature, position and length of the image formed. Also draw labelled ray diagram to show the image formation in the above case.
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    $v=-20$ cm and $h^{\prime }=+6$ cm: the image is virtual, erect and diminished, 20 cm from the lens on the object's side.

    Using the lens formula with $f=-60$ cm, $u=-30$ cm and $h=+9$ cm: $\frac{1}{v}=\frac{1}{f}+\frac{1}{u}=-\frac{1}{60}-\frac{1}{30}=-\frac{1}{20}$, so $v=-20$ cm. $m=\frac{v}{u}=\frac{-20}{-30}=+\frac{2}{3}$, so $h^{\prime }=m\times h=\frac{2}{3}\times 9=+6$ cm. Nature: virtual, erect and diminished. Position: $20$ cm from the lens on the same side as the object (between the optical centre and the focus $F_1$). Length: $6$ cm. Ray diagram (see the figure): a ray from the top of the object parallel to the axis diverges after the lens as if coming from $F_1$, and a ray through the optical centre goes straight on; the backward extension of the first ray meets the second at the top of the image $A^{\prime }{B}^{\prime }$.

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12. Q292 CBSE 2019
    A 6 cm ​tall object is ​placed perpendicular to the principal axis ​of a concave mirror of focal length 30 ​cm. The distance of the object ​from the mirror is 45 cm. Use mirror formula to determine the position, nature and size ​of the image formed. ​Also ​draw labelled ray diagram to show the image formation in this case.
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    $v=-90$ cm: the image is real, inverted and enlarged, formed 90 cm in front of the mirror, 12 cm tall.

    Position: $v=-90$ cm, that is, the image is formed $90$ cm in front of the mirror, beyond C. Nature: real and inverted (and enlarged). Size: $m=-\frac{v}{u}=-2$, so $h^{\prime }=-2\times 6=-12$ cm: the image is $12$ cm tall, twice the size of the object. The labelled ray diagram shows the two rays from B meeting at B′, beyond C.

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13. Q293 CBSE 2019
    An object 6 cm in ​size ​is placed at 50 cm in front of a convex lens ​of focal length 30 cm. At what distance from ​the ​lens should a screen be placed in ​order to obtain a sharp image of the object ? Find the nature and size of the image. Also draw labelled ray diagram to show the image formation in ​this case.
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    The screen goes 75 cm from the lens, opposite the object; the image is real, inverted, enlarged and 9 cm tall ($m=-1.5$).

    The screen should be placed $75$ cm from the lens, on the side opposite to the object ($v=+75$ cm). Nature: real and inverted (and enlarged). Size: $m=\frac{v}{u}=\frac{75}{-50}=-1.5$, so $h^{\prime }=-1.5\times 6=-9$ cm: the image is $9$ cm tall. The labelled ray diagram shows the image formed beyond $2F_2$ and caught on the screen.

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14. Q294 CBSE 2019
    1. (a)A 5 cm tall object is ​placed perpendicular ​to the ​principal axis of a convex lens of focal length 20 cm. The distance of the ​object from the lens is 30 cm. Find ​the position, nature and size of the image ​formed.
    2. (b)Draw ​a ​labelled ray diagram showing object distance, image distance and focal length in the above case.
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    $v=+60$ cm: the image is real, inverted and magnified, 10 cm tall, on the other side of the lens.

    (a) Using $\frac{1}{v}-\frac{1}{u}=\frac{1}{f}$ with $u=-30$ cm and $f=+20$ cm: $\frac{1}{v}=\frac{1}{20}-\frac{1}{30}=\frac{1}{60}$, so $v=+60$ cm. Position: the image is formed $60$ cm from the lens, on the other side of the lens (beyond $2F_2$). Magnification $m=\frac{v}{u}=\frac{60}{-30}=-2$, so $h^{\prime }=m\times h=-2\times 5=-10$ cm. Nature: real, inverted and magnified. Size: $10$ cm tall, twice the size of the object. (b) In the ray diagram, a ray from the top B of the object parallel to the principal axis passes through $F_2$ after refraction, and a ray through the optical centre O goes straight on; they meet at B′, $60$ cm behind the lens, where the real, inverted image A′B′ forms. The object distance $\text{OA}=30$ cm, the image distance ${\text{OA}}^{\prime }=60$ cm and the focal length ${\text{OF}}_2=20$ cm are marked on it.

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15. Q295 CBSE 2019
    An object is placed at a distance of ​30 cm from a concave lens of focal ​length 30 cm.

    1. (i)Use lens formula to determine the distance of the image from ​the lens.
    2. (ii)​List four characteristics of the image (nature position, size, ​erect/inverted) ​in this case.
    3. (iii)Draw a labelled diagram to justify ​your answer of part ​(ii)
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    $v=-15$ cm: the image is virtual, erect and diminished ($m=+0.5$), 15 cm from the lens on the same side as the object.

    (i) With $u=-30$ cm and $f=-30$ cm, $\frac{1}{v}=\frac{1}{f}+\frac{1}{u}=-\frac{1}{30}-\frac{1}{30}=-\frac{1}{15}$, so $v=-15$ cm: the image is $15$ cm from the lens, on the same side as the object. (ii) Nature: virtual. Position: $15$ cm from the lens on the same side as the object, between the optical centre and the focus $F_1$. Size: diminished, half the size of the object ($m=v/u=+0.5$). Orientation: erect. (iii) Ray diagram: a ray from the top of the object parallel to the principal axis diverges after refraction so that its backward extension passes through $F_1$, and a ray through the optical centre goes on undeviated. They appear to meet $15$ cm from the lens, which gives a virtual, erect, diminished image, as shown in the figure.

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16. Q296 CBSE 2019
    1. (a)List ​four characteristics of the image ​formed by a concave lens ​of focal length 20 cm when the object is placed at ​a distance of 40 ​cm from its optical centre.
    2. (b)The size of image of ​an object by a convex lens of focal length 20 cm is ​observed to be reduced to ${\frac{1}{3}}^{\text{rd}}$ of its size. Find the distance of the object ​from the optical centre of the lens.
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    (a) The image is virtual, erect, diminished and on the object's side; (b) the object is 80 cm from the convex lens.

    (a) The image is (1) virtual, (2) erect, (3) diminished - one-third the size of the object - and (4) formed on the same side of the lens as the object, between the optical centre and the focus, $13.3$ cm from the lens ($\frac{1}{v}=-\frac{1}{20}-\frac{1}{40}$ gives $v=-13.3$ cm and $m=+\frac{1}{3}$). (b) A diminished image formed by a convex lens is real and inverted, so $m=\frac{v}{u}=-\frac{1}{3}$. With $u=-x$ and $v=\frac{x}{3}$, the lens formula gives $\frac{3}{x}+\frac{1}{x}=\frac{1}{20}$, so $x=80$ cm. The object is $80$ cm from the optical centre ($u=-80$ cm; the image forms at $v=+26.7$ cm).

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17. Q297 CBSE 2019
    An object ​is ​placed at a distance of 60 ​cm from ​a concave lens of focal length 30 cm.

    1. (i)Use lens ​formula to find the distance of the image ​from the lens.
    2. (ii)List four characteristics ​of the image (nature, position, size, erect/inverted) formed by the lens in this case.
    3. (iii)Draw ray diagram to justify your answer of part (ii).
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    $v=-20$ cm: the image is virtual, erect and diminished (one third), formed 20 cm from the lens on the object's side.

    (i) By the sign convention $u=-60$ cm and $f=-30$ cm. From $\frac{1}{v}=\frac{1}{f}+\frac{1}{u}=-\frac{1}{30}-\frac{1}{60}=-\frac{1}{20}$, $v=-20$ cm: the image is 20 cm from the lens, on the same side as the object. The magnification is $m=\frac{v}{u}=\frac{-20}{-60}=+\frac{1}{3}$. (ii) Nature: virtual. Position: 20 cm from the lens on the same side as the object, between O and $F_1$. Size: diminished (one third of the object). Erect. (iii) Ray diagram (see the figure): a ray from the top of the object parallel to the axis spreads out after the lens as if coming from $F_1$, and a ray through the optical centre O goes on undeviated; the backward extension of the first ray meets the second ray 20 cm in front of the lens, giving a virtual, erect and diminished image.

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18. Q298 CBSE 2019
    1. (a)List four characteristics of ​the images formed ​by plane mirrors.
    2. (b)A 5 cm tall ​object is placed at a distance of 20 cm ​from a concave mirror of focal length 30 cm. Use mirror formula to ​determine the position ​and size ​of the image formed.
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    Plane-mirror image: virtual, erect, same size, laterally inverted; concave mirror: $v=+60$ cm, virtual, erect, 15 cm tall.

    (a) The image formed by a plane mirror is (1) virtual and erect - it cannot be obtained on a screen; (2) of the same size as the object; (3) as far behind the mirror as the object is in front of it; (4) laterally inverted. (b) With the sign convention, $h=+5$ cm, $u=-20$ cm, $f=-30$ cm. Mirror formula: $\frac{1}{v}=\frac{1}{f}-\frac{1}{u}=-\frac{1}{30}+\frac{1}{20}=\frac{1}{60}$, so $v=+60$ cm: the image is formed $60$ cm behind the mirror. Magnification $m=-\frac{v}{u}=-\frac{60}{-20}=+3$, so $h^{\prime }=m\times h=3\times 5=+15$ cm. The image is $15$ cm tall - virtual, erect and enlarged (three times the size of the object).

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19. Q299 CBSE 2019
    A teacher gives a ​convex ​lens and a concave mirror of focal length ​of 20 cm each to his student and asks him to find their ​focal lengths by obtaining the image of a distant object. The student uses a distant ​tree as the object and obtains its sharp image, ​one by one, on a screen. The distances $d_1$ and $d_2$ between the lens/mirror and the screen in the two cases and the nature of their respective sharp images are ​likely to be Give reason for your answer.

    1. (a)(20 cm, 40 cm) and (erect and erect)
    2. (b)(20 cm, 40 cm) and (inverted and inverted)
    3. (c)(20 cm, 20 cm) and (inverted and inverted)
    4. (d)(20 cm, 40 cm) and (erect and inverted)
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    (c) (20 cm, 20 cm) and (inverted and inverted)

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20. Q300 CBSE 2019
    Study the ​following ray diagram and list two mistakes committed ​by the ​student while tracing it. Rectify these mistakes by drawing the correct ray diagram to show ​the ​real position and size of the image corresponding to ​the position of the object AB.

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    The foci were unequally spaced and the image misplaced; correctly, the image is real, inverted, magnified and beyond $2F_2$.

    Mistakes (any two): (1) $F_1$ and $F_2$ are not marked at equal distances from the optical centre: $O{F}_1\ne O{F}_2$ (and likewise $O2F_1\ne O2F_2$), although the two foci of a lens are equidistant from O. (2) The image is shown between $F_2$ and $2F_2$; for an object placed between $F_1$ and $2F_1$ the image must be formed beyond $2F_2$. (3) The image should be magnified; the image drawn is only slightly larger than AB. Correct ray diagram (see the figure): with $F_1$, $F_2$ equidistant from O and $2F_1$, $2F_2$ at twice that distance, the ray from A parallel to the axis passes through $F_2$ after refraction and the ray through O goes undeviated. They meet beyond $2F_2$, giving a real, inverted and magnified image A′B′ (for example, $u=-1.5f$ gives $v=+3f$ and $m=-2$).

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