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NCERT Class 10 Science Light – Reflection and Refraction: Exercise Solutions

Step-by-step working for every question of End-of-chapter exercises in NCERT's Class 10 Science book, chapter “Light – Reflection and Refraction”, written by Shishya. Each question is in our own short words; open the NCERT chapter for the exact wording and figures.

1. Exercises Q1

   Which cannot be used to make a lens: (a) water (b) glass (c) plastic (d) clay?

   1. A lens must be made of a transparent material that allows light to pass through it.
   2. Water, glass, and plastic are all transparent materials.
   3. Clay is an opaque material that does not allow light to pass through.
   4. Therefore, clay cannot be used to make a lens.

   Answer: (d) clay

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2. Exercises Q2

   A concave mirror forms a virtual, erect, enlarged image. Where is the object: (a) between F and C (b) at C (c) beyond C (d) between the pole and F?

   1. A concave mirror forms a virtual, erect, and enlarged image only when the object is placed between the pole and the focus.
   2. If the object is between F and C, the image is real, inverted, and enlarged.
   3. If the object is at C, the image is real, inverted, and of the same size.
   4. If the object is beyond C, the image is real, inverted, and diminished.
   5. Therefore, the object must be between the pole and F.

   Answer: (d) between the pole and F

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3. Exercises Q3

   Where must an object be placed before a convex lens for a real image of the same size: (a) at F (b) at 2F (c) at infinity (d) between the optical centre and F?

   1. A convex lens forms a real image of the same size as the object when the object is placed at twice the focal length.
   2. This position is known as 2F or the centre of curvature.
   3. At this distance, the image is also formed at 2F on the other side of the lens.
   4. The image is real, inverted, and equal in size to the object.
   5. Therefore, the correct option is (b).

   Answer: (b) at 2F

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4. Exercises Q4

   A spherical mirror and a thin lens each have focal length −15 cm. They are likely (a) both concave (b) both convex (c) concave mirror, convex lens (d) convex mirror, concave lens.

   1. A negative focal length for a spherical mirror indicates it is a concave mirror.
   2. A negative focal length for a thin lens indicates it is a concave lens.
   3. Therefore, the mirror is concave and the lens is concave.

   Answer: (a) both concave

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5. Exercises Q5

   Your image in a mirror is erect however far you stand. The mirror is (a) only plane (b) only concave (c) only convex (d) plane or convex.

   1. A plane mirror always forms an erect image regardless of the object's distance.
   2. A convex mirror always forms an erect and diminished image for any object position.
   3. A concave mirror forms an inverted image when the object is placed beyond the focal point.
   4. Since the image is erect no matter how far you stand, the mirror must be either plane or convex.
   5. Therefore, the correct option is (d).

   Answer: (d) plane or convex

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6. Exercises Q6

   Which lens would you use to read small dictionary letters: (a) convex, f = 50 cm (b) concave, f = 50 cm (c) convex, f = 5 cm (d) concave, f = 5 cm?

   1. A concave lens always forms a diminished virtual image, so options (b) and (d) are not suitable for magnification.
      why?

      Magnification m = height of image ÷ height of object (= −v/u for a mirror, v/u for a lens). A negative m means a real, inverted image; a positive m a virtual, erect one.

   2. A convex lens acts as a magnifying glass when the object is placed within its focal length.
   3. The magnifying power of a simple magnifier is given by the formula M = 1 + d/f, where d is the least distance of distinct vision (25 cm) and f is the focal length.
   4. To get maximum magnification, we need the smallest possible focal length f.
      why?

      Magnification m = height of image ÷ height of object (= −v/u for a mirror, v/u for a lens). A negative m means a real, inverted image; a positive m a virtual, erect one.

   5. Comparing the convex lenses, f = 5 cm gives a much higher magnification than f = 50 cm.
      why?

      Magnification m = height of image ÷ height of object (= −v/u for a mirror, v/u for a lens). A negative m means a real, inverted image; a positive m a virtual, erect one.

   6. Therefore, the convex lens with a focal length of 5 cm is the best choice to read small letters.

   Answer: (c) convex, f = 5 cm

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7. Exercises Q7

   For an erect image with a concave mirror of focal length 15 cm, what range of object distance is needed? What is the image like, and is it larger or smaller? Describe the ray diagram.

   1. A concave mirror forms an erect image only when the object is placed between the pole and the focus.
   2. The focal length of the mirror is given as 15 cm.
   3. Therefore, the object distance must be less than 15 cm.
   4. In this position, the rays appear to diverge from a point behind the mirror.
   5. The image formed is virtual and erect.
   6. The image is larger than the object.
   7. To draw the ray diagram, place the object between the pole and the focus.
   8. Draw a ray parallel to the principal axis which reflects through the focus.
   9. Draw a ray passing through the focus which reflects parallel to the principal axis.
   10. Extend the reflected rays backwards to find the point where they appear to meet behind the mirror.

   Answer: Object distance is less than 15 cm; image is virtual, erect, and larger

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8. Exercises Q8

   Which mirror is used, and why, in (a) car headlights (b) side/rear-view mirrors (c) a solar furnace?

   1. (a) A concave mirror is used in car headlights.
   2. (a) The bulb is placed at the focus of the concave mirror.
   3. (a) This arrangement produces a parallel beam of light that travels a long distance.
   4. (b) A convex mirror is used as a side or rear-view mirror.
   5. (b) A convex mirror provides a wider field of view than a plane mirror.
   6. (b) It forms erect images of vehicles behind, which helps in safe driving.
   7. (c) A concave mirror is used as the reflector in a solar furnace.
   8. (c) The concave mirror converges parallel rays of sunlight to a single point.
   9. (c) This concentration of light produces very high temperatures at the focus.

   Answer: (a) Concave mirror, to produce a parallel beam; (b) Convex mirror, for a wider field of view; (c) Concave mirror, to concentrate sunlight at the focus.

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9. Exercises Q9

   Half of a convex lens is covered with black paper. Does it still form a complete image? Explain.

   1. Covering half of the convex lens blocks light rays passing through that specific half.
   2. However, the uncovered half of the lens still allows light rays from the object to pass through.
   3. These remaining rays refract and converge at the same focal point as before.
   4. Since all rays from a point on the object still meet at the corresponding image point, the entire image is formed.
   5. The only difference is that fewer rays contribute to the image, making it dimmer but not incomplete.

   Answer: Yes, it forms a complete image, but the image will be dimmer.

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10. Exercises Q10

    A 5 cm object is 25 cm from a converging lens of focal length 10 cm. Find the position, size and nature of the image.

    1. Given: object height h_o = 5 cm, object distance u = -25 cm, focal length f = 10 cm.
    2. Using the lens formula 1/v - 1/u = 1/f, substitute the values: 1/v - 1/(-25) = 1/10.
       why?

       Lens formula 1/v − 1/u = 1/f, with distances measured from the optical centre using the New Cartesian sign convention.

    3. Simplify the equation: 1/v + 1/25 = 1/10.
    4. Find a common denominator to solve for 1/v: 1/v = 1/10 - 1/25 = (5 - 2)/50 = 3/50.
    5. Calculate v: v = 50/3 = 16.67 cm.
    6. The positive sign of v indicates the image is formed on the opposite side of the lens, so it is real.
    7. Use the magnification formula m = v/u to find the size ratio: m = 16.67 / (-25).
       why?

       Magnification m = height of image ÷ height of object (= −v/u for a mirror, v/u for a lens). A negative m means a real, inverted image; a positive m a virtual, erect one.

    8. Calculate magnification: m = -0.667.
       why?

       Magnification m = height of image ÷ height of object (= −v/u for a mirror, v/u for a lens). A negative m means a real, inverted image; a positive m a virtual, erect one.

    9. Use m = h_i / h_o to find image height: -0.667 = h_i / 5.
       why?

       Magnification m = height of image ÷ height of object (= −v/u for a mirror, v/u for a lens). A negative m means a real, inverted image; a positive m a virtual, erect one.

    10. Calculate h_i: h_i = -0.667 × 5 = -3.33 cm.
    11. The negative sign of h_i indicates the image is inverted.
    12. Since the image is real and inverted, it can be projected on a screen.

    Answer: v = 16.67 cm, -3.33 cm, real, inverted

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11. Exercises Q11

    A concave lens of focal length 15 cm forms an image 10 cm from the lens. How far is the object?

    1. Given focal length f = -15 cm and image distance v = -10 cm.
    2. Use the lens formula 1/f = 1/v - 1/u.
       why?

       Lens formula 1/v − 1/u = 1/f, with distances measured from the optical centre using the New Cartesian sign convention.

    3. Substitute the values: 1/(-15) = 1/(-10) - 1/u.
    4. Rearrange to find 1/u = 1/(-10) - 1/(-15).
    5. Simplify the right side: 1/u = -1/10 + 1/15.
    6. Find a common denominator: 1/u = -3/30 + 2/30.
    7. Calculate the result: 1/u = -1/30.
    8. Invert to find u: u = -30 cm.

    Answer: u = -30 cm

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12. Exercises Q12

    An object is 10 cm from a convex mirror of focal length 15 cm. Find the position and nature of the image.

    1. Given object distance u = -10 cm, focal length f = +15 cm for a convex mirror.
    2. Using the mirror formula 1/f = 1/v + 1/u.
       why?

       Mirror formula 1/v + 1/u = 1/f, with the New Cartesian sign convention: distances are measured from the pole, positive along the direction of the incident light and negative against it.

    3. Substituting values: 1/15 = 1/v + 1/(-10).
    4. Rearranging to find 1/v: 1/v = 1/15 + 1/10.
    5. Finding common denominator: 1/v = 2/30 + 3/30 = 5/30.
    6. Simplifying to get 1/v = 1/6, so v = 6 cm.
    7. Since v is positive, the image is formed behind the mirror.
    8. Calculating magnification m = -v/u = -6/(-10) = 0.6.
       why?

       Magnification m = height of image ÷ height of object (= −v/u for a mirror, v/u for a lens). A negative m means a real, inverted image; a positive m a virtual, erect one.

    9. Since m is positive, the image is erect and virtual.
    10. The image is diminished as |m| < 1.

    Answer: v = 6 cm, virtual, erect

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13. Exercises Q13

    A plane mirror gives magnification +1. What does this mean?

    1. A plane mirror always forms an image that is the same size as the object.
    2. The magnification of a mirror is defined as the ratio of the height of the image to the height of the object.
       why?

       Magnification m = height of image ÷ height of object (= −v/u for a mirror, v/u for a lens). A negative m means a real, inverted image; a positive m a virtual, erect one.

    3. A magnification value of +1 means the image height is equal to the object height.
       why?

       Magnification m = height of image ÷ height of object (= −v/u for a mirror, v/u for a lens). A negative m means a real, inverted image; a positive m a virtual, erect one.

    4. The positive sign indicates that the image is erect, meaning it is not inverted.
    5. Therefore, the image formed is virtual, erect, and of the same size as the object.

    Answer: The image is erect and of the same size as the object.

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14. Exercises Q14

    A 5.0 cm object is 20 cm in front of a convex mirror of radius of curvature 30 cm. Find the image position, nature and size.

    1. Given: object height h_o = 5.0 cm, object distance u = -20 cm, radius of curvature R = +30 cm.
       why?

       For a spherical mirror the focal length is half the radius of curvature: f = R/2.

    2. Calculate the focal length f using the formula f = R/2, which gives f = 30/2 = 15 cm.
       why?

       For a spherical mirror the focal length is half the radius of curvature: f = R/2.

    3. Use the mirror formula 1/v + 1/u = 1/f to find the image position v.
       why?

       Mirror formula 1/v + 1/u = 1/f, with the New Cartesian sign convention: distances are measured from the pole, positive along the direction of the incident light and negative against it.

    4. Substitute the values: 1/v + 1/(-20) = 1/15, so 1/v = 1/15 + 1/20.
    5. Find the common denominator for 15 and 20, which is 60, so 1/v = 4/60 + 3/60 = 7/60.
    6. Solve for v: v = 60/7 ≈ 8.57 cm.
    7. Since v is positive, the image is formed behind the mirror, making it virtual and erect.
    8. Use the magnification formula m = -v/u to find the magnification.
       why?

       Magnification m = height of image ÷ height of object (= −v/u for a mirror, v/u for a lens). A negative m means a real, inverted image; a positive m a virtual, erect one.

    9. Substitute the values: m = -8.57 / (-20) = 0.4285.
    10. Calculate the image height h_i using h_i = m × h_o, so h_i = 0.4285 × 5.0 ≈ 2.14 cm.

    Answer: v = 8.57 cm, virtual, erect, 2.14 cm

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15. Exercises Q15

    A 7.0 cm object is 27 cm in front of a concave mirror of focal length 18 cm. Where should the screen be for a sharp image? Find the size and nature of the image.

    1. Given: object height h_o = 7.0 cm, object distance u = -27 cm, focal length f = -18 cm.
    2. Using the mirror formula 1/v + 1/u = 1/f, we substitute the values: 1/v - 1/27 = -1/18.
       why?

       Mirror formula 1/v + 1/u = 1/f, with the New Cartesian sign convention: distances are measured from the pole, positive along the direction of the incident light and negative against it.

    3. Solving for 1/v gives 1/v = -1/18 + 1/27 = (-3 + 2)/54 = -1/54.
    4. Therefore, the image distance v = -54 cm, so the screen should be placed 54 cm in front of the mirror.
    5. Now, using the magnification formula m = -v/u, we get m = -(-54)/(-27) = -2.
       why?

       Magnification m = height of image ÷ height of object (= −v/u for a mirror, v/u for a lens). A negative m means a real, inverted image; a positive m a virtual, erect one.

    6. The negative sign indicates the image is inverted, and the magnitude 2 means it is enlarged.
    7. Using m = h_i/h_o, we find h_i = m × h_o = -2 × 7.0 = -14 cm.
       why?

       Magnification m = height of image ÷ height of object (= −v/u for a mirror, v/u for a lens). A negative m means a real, inverted image; a positive m a virtual, erect one.

    8. The negative height confirms the image is inverted and its size is 14 cm.
    9. Since the image is formed on the same side as the object, it is real.

    Answer: v = -54 cm; image size = -14 cm; nature: real, inverted, enlarged

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16. Exercises Q16

    Find the focal length of a lens of power −2.0 D. What type of lens is it?

    1. The power of the lens is given as P = −2.0 D.
       why?

       Power of a lens P = 1/f with f in metres; the unit is the dioptre (D). A convex lens has positive power and a concave lens negative.

    2. The relationship between power and focal length is P = 1/f, where f is in metres.
       why?

       Power of a lens P = 1/f with f in metres; the unit is the dioptre (D). A convex lens has positive power and a concave lens negative.

    3. Rearranging the formula gives f = 1/P.
    4. Substituting the value, f = 1 / (−2.0).
    5. Calculating the result gives f = −0.5 m.
    6. Since the focal length is negative, the lens is a concave lens.

    Answer: f = -0.5 m, concave

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17. Exercises Q17

    A corrective lens has power +1.5 D. Find its focal length. Is it diverging or converging?

    1. The power of the lens is given as P = +1.5 D.
       why?

       Power of a lens P = 1/f with f in metres; the unit is the dioptre (D). A convex lens has positive power and a concave lens negative.

    2. The relationship between power and focal length is P = 1/f, where f is in metres.
       why?

       Power of a lens P = 1/f with f in metres; the unit is the dioptre (D). A convex lens has positive power and a concave lens negative.

    3. Rearranging the formula to find focal length gives f = 1/P.
    4. Substituting the value of power, we get f = 1 / 1.5.
    5. Calculating the division, f = 0.666... m.
    6. Rounding to two decimal places, f = 0.67 m.
    7. Since the power is positive, the lens is converging.

    Answer: f = 0.67 m; converging

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