Class 10 - Physics Chp 9 Light

1. Which one of the following materials cannot be used to make a lens?

Answer: (d) Clay
Explanation: A lens must be transparent. Clay is opaque and does not allow light to pass through.

2. The image formed by a concave mirror is virtual, erect and larger than the object. Where should be the position of the object?

Answer: (d) Between the pole of the mirror and its principal focus
Explanation: Only in this position does a concave mirror form a virtual, erect, and magnified image.

3. Where should an object be placed in front of a convex lens to get a real image of the same size as the object?

Answer: (b) At twice the focal length
Explanation: When an object is at 2F in a convex lens, the image is real, inverted, and the same size.

4. A spherical mirror and a thin spherical lens have each a focal length of –15 cm. The mirror and the lens are likely to be:

Answer: (a) Both concave
Explanation: A negative focal length means both the mirror and lens are concave.

5. No matter how far you stand from a mirror, your image appears erect. The mirror is likely to be:

Answer: (d) Either plane or convex
Explanation: Both plane and convex mirrors always produce erect images.

6. Which of the following lenses would you prefer to use while reading small letters?

Answer: (c) A convex lens of focal length 5 cm
Explanation: Convex lenses magnify objects. A shorter focal length gives more magnification.

7. Concave mirror, erect image, focal length 15 cm — what should be the object range?

  • Answer: The object should be placed between the pole and the focus (0 cm < u < 15 cm).

  • Nature of image: Virtual, erect, magnified

  • Ray Diagram:

    • One ray parallel to axis reflects through focus.

    • One ray aimed at center reflects back on itself.

    • Virtual rays appear to diverge from behind the mirror.

8. Mirror types:

  • (a) Concave mirror – Used in car headlights because it converges light to form a powerful beam.

  • (b) Convex mirror – Used for side mirrors due to wider field of view.

  • (c) Concave mirror – Concentrates sunlight at a point (focus) in solar furnaces.

9. Half of a convex lens covered – will image still form?

Answer: Yes, a complete image is still formed, but it is dimmer. Explanation: Every part of the lens contributes to the whole image; however, brightness reduces.

10. Object 5 cm, 25 cm from convex lens (f = 10 cm):

  • Use lens formula:
    1v−1u=1f\frac{1}{v} - \frac{1}{u} = \frac{1}{f}
    1v=110+125=750⇒v=7.14 cm (approx.)\frac{1}{v} = \frac{1}{10} + \frac{1}{25} = \frac{7}{50}\Rightarrow v = 7.14\text{ cm (approx.)}

  • Image position: 7.14 cm (on opposite side)

  • Magnification:
    m=vu=7.1425≈0.29m = \frac{v}{u} = \frac{7.14}{25} \approx 0.29

  • Image size:
    5×0.29=1.45 cm (inverted)5 \times 0.29 = 1.45\text{ cm (inverted)}

  • Nature: Real, inverted, smaller.

11. Concave lens, f = –15 cm, image distance v = –10 cm

  • Use lens formula:
    1v−1u=1f\frac{1}{v} - \frac{1}{u} = \frac{1}{f}
    1−10−1u=1−15⇒1u=1−10+115=−130\frac{1}{-10} - \frac{1}{u} = \frac{1}{-15} \Rightarrow \frac{1}{u} = \frac{1}{-10} + \frac{1}{15} = -\frac{1}{30}
    u=−30 cmu = -30 \text{ cm}

  • Object distance: 30 cm

  • Ray Diagram: Shows virtual, erect, and diminished image on same side.

12. Convex mirror, f = 15 cm, u = –10 cm

  • Use mirror formula:
    1v+1u=1f⇒1v=115−110=−130⇒v=−30 cm\frac{1}{v} + \frac{1}{u} = \frac{1}{f} \Rightarrow \frac{1}{v} = \frac{1}{15} - \frac{1}{10} = -\frac{1}{30} \Rightarrow v = -30 \text{ cm}

  • Nature: Virtual, erect, diminished image

13. Magnification by plane mirror is +1 — what does it mean?

Answer: Image is same size as object, virtual, erect, and laterally inverted.

14. Convex mirror, u = –20 cm, R = 30 cm ⇒ f = 15 cm

  • Use mirror formula:
    1v=115+120=760⇒v=8.57 cm (approx)\frac{1}{v} = \frac{1}{15} + \frac{1}{20} = \frac{7}{60} \Rightarrow v = 8.57\text{ cm (approx)}

  • Image distance: 8.57 cm (behind mirror)

  • Magnification:
    m=vu=8.57−20≈−0.43m = \frac{v}{u} = \frac{8.57}{-20} \approx -0.43

  • Image size:
    5×0.43≈2.15 cm (erect)5 \times 0.43 \approx 2.15 \text{ cm (erect)}

15. Concave mirror, u = –27 cm, f = –18 cm, object size = 7 cm

  • Use mirror formula:
    1v=1f−1u=1−18−1−27=−1+254=154⇒v=54 cm\frac{1}{v} = \frac{1}{f} - \frac{1}{u} = \frac{1}{-18} - \frac{1}{-27} = \frac{-1+2}{54} = \frac{1}{54} \Rightarrow v = 54 \text{ cm}

  • Screen at: 54 cm from mirror

  • Magnification:
    m=−vu=−54−27=2m = \frac{-v}{u} = \frac{-54}{-27} = 2

  • Image size: 14 cm, real and inverted

16. Power = –2.0 D ⇒ Find focal length

  • f=100P=100−2.0=−50 cmf = \frac{100}{P} = \frac{100}{-2.0} = -50 \text{ cm}

  • Type: Concave (diverging) lens

17. Power = +1.5 D

  • f=100P=1001.5=66.67 cmf = \frac{100}{P} = \frac{100}{1.5} = 66.67 \text{ cm}

  • Type: Convex (converging) lens


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Maths - IFP 1 - 3

Maths - IFP 1 -3