Light - Reflection and Refraction - Solutions

Multiple Choice Questions
Short Answer Questions
Long Answer Questions

Multiple Choice Questions

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

Answer: (d) Clay

Explanation: Clay is opaque and cannot transmit light, which is essential for a lens. Lenses require transparent materials like water, glass, or plastic to allow light to pass through and focus it.

2. The image formed by a concave mirror is observed to be 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: When an object is placed between the pole (P) and the principal focus (F) of a concave mirror, the image formed is virtual, erect, and larger than the object.

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

Answer: (b) At twice the focal length

Explanation: When an object is placed at 2F (twice the focal length) in front of a convex lens, the image formed is real, inverted, and of the same size as the object.

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 indicates a concave mirror and a concave (diverging) lens.

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 mirrors and convex mirrors always form erect images regardless of the object's position.

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

Answer: (c) A convex lens of focal length 5 cm.

Explanation: A convex lens with a shorter focal length (5 cm) has higher converging power and provides greater magnification, making it suitable for reading small text.

Short Answer Questions

7. We wish to obtain an erect image of an object, using a concave mirror of focal length 15 cm. What should be the range of distance of the object from the mirror? What is the nature of the image? Is the image larger or smaller than the object? Draw a ray diagram to show the image formation in this case.

Answer:

To obtain an erect image using a concave mirror, the object must be placed between the pole (P) and the principal focus (F).

Range of object distance: 0 cm to 15 cm (between P and F)

Nature of image: Virtual and erect

Size of image: Larger than the object

Ray diagram: The ray diagram would show the object placed between P and F, with the reflected rays appearing to diverge from behind the mirror, forming a virtual, erect, and enlarged image.

8. Name the type of mirror used in the following situations.

Answer:

(a) Headlights of a car: Concave mirror

Reason: Concave mirrors are used in headlights to produce a powerful parallel beam of light when the bulb is placed at its focus.

(b) Side/rear-view mirror of a vehicle: Convex mirror

Reason: Convex mirrors provide a wider field of view and always give an erect, though diminished, image, allowing drivers to see more traffic behind them.

(c) Solar furnace: Concave mirror

Reason: Concave mirrors are used in solar furnaces to concentrate sunlight at their focus, producing high temperatures.

9. One-half of a convex lens is covered with a black paper. Will this lens produce a complete image of the object? Verify your answer experimentally. Explain your observations.

Answer:

Yes, the lens will still produce a complete image of the object, but the image will be less bright.

Experimental verification: When we cover half of a convex lens with black paper and place an object in front of it, we still get a complete image on the screen, though it appears dimmer.

Explanation: Each part of the lens is capable of forming a complete image. When we cover half the lens, we block half the light rays, reducing the intensity (brightness) of the image, but the remaining half still forms a complete image.

13. The magnification produced by a plane mirror is +1. What does this mean?

Answer:

The magnification produced by a plane mirror is +1. This means:

  • The image is of the same size as the object (since |m| = 1)
  • The image is virtual and erect (since m is positive)
  • The image is formed at the same distance behind the mirror as the object is in front of it
16. Find the focal length of a lens of power -2.0 D. What type of lens is this?

Answer:

Given: Power (P) = -2.0 D

Using the formula: P = 1/f (where f is in meters)

So, f = 1/P = 1/(-2.0) = -0.5 m = -50 cm

Since the focal length is negative, this is a concave (diverging) lens.

17. A doctor has prescribed a corrective lens of power +1.5 D. Find the focal length of the lens. Is the prescribed lens diverging or converging?

Answer:

Given: Power (P) = +1.5 D

Using the formula: P = 1/f (where f is in meters)

So, f = 1/P = 1/(+1.5) = +0.67 m = +67 cm

Since the focal length is positive, this is a convex (converging) lens.

Long Answer Questions

10. An object 5 cm in length is held 25 cm away from a converging lens of focal length 10 cm. Draw the ray diagram and find the position, size and the nature of the image formed.

Answer:

Given:

Height of object (h) = +5 cm

Object distance (u) = -25 cm (negative as per sign convention)

Focal length (f) = +10 cm (positive for convex lens)

Using lens formula: 1/v - 1/u = 1/f

1/v = 1/f + 1/u = 1/10 + 1/(-25) = 0.1 - 0.04 = 0.06

v = 1/0.06 = +16.67 cm

Using magnification formula: m = h'/h = v/u

h' = h × (v/u) = 5 × (16.67/(-25)) = -3.33 cm

Results:

  • Position of image: 16.67 cm on the opposite side of the lens
  • Size of image: 3.33 cm (inverted, as indicated by negative sign)
  • Nature of image: Real and inverted

Ray diagram: The ray diagram would show the object placed beyond 2F (since 25 cm > 20 cm), with the image formed between F and 2F on the other side, real, inverted, and diminished.

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

Answer:

Given:

Focal length (f) = -15 cm (negative for concave lens)

Image distance (v) = -10 cm (negative as image is virtual and on the same side as object)

Object distance (u) = ?

Using lens formula: 1/v - 1/u = 1/f

1/u = 1/v - 1/f = 1/(-10) - 1/(-15) = -0.1 + 0.067 = -0.033

u = 1/(-0.033) = -30 cm

The object is placed 30 cm in front of the lens.

Ray diagram: The ray diagram would show the object placed in front of the concave lens, with the image formed on the same side as the object, virtual, erect, and diminished.

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

Answer:

Given:

Object distance (u) = -10 cm (negative as per sign convention)

Focal length (f) = +15 cm (positive for convex mirror)

Image distance (v) = ?

Using mirror formula: 1/v + 1/u = 1/f

1/v = 1/f - 1/u = 1/15 - 1/(-10) = 0.067 + 0.1 = 0.167

v = 1/0.167 = +6 cm

Results:

  • Position of image: 6 cm behind the mirror (positive v indicates virtual image behind mirror)
  • Nature of image: Virtual and erect (always for convex mirror)
  • Size of image: Diminished (as v < u)
14. An object 5.0 cm in length is placed at a distance of 20 cm in front of a convex mirror of radius of curvature 30 cm. Find the position of the image, its nature and size.

Answer:

Given:

Height of object (h) = +5 cm

Object distance (u) = -20 cm

Radius of curvature (R) = +30 cm (positive for convex mirror)

Focal length (f) = R/2 = +15 cm

Using mirror formula: 1/v + 1/u = 1/f

1/v = 1/f - 1/u = 1/15 - 1/(-20) = 0.067 + 0.05 = 0.117

v = 1/0.117 = +8.55 cm

Using magnification formula: m = h'/h = -v/u

h' = h × (-v/u) = 5 × (-8.55/(-20)) = 5 × 0.4275 = +2.14 cm

Results:

  • Position of image: 8.55 cm behind the mirror
  • Nature of image: Virtual and erect (positive h')
  • Size of image: 2.14 cm (diminished)
15. An object of size 7.0 cm is placed at 27 cm in front of a concave mirror of focal length 18 cm. At what distance from the mirror should a screen be placed, so that a sharp focussed image can be obtained? Find the size and the nature of the image.

Answer:

Given:

Height of object (h) = +7 cm

Object distance (u) = -27 cm

Focal length (f) = -18 cm (negative for concave mirror)

Using mirror formula: 1/v + 1/u = 1/f

1/v = 1/f - 1/u = 1/(-18) - 1/(-27) = -0.0556 + 0.037 = -0.0186

v = 1/(-0.0186) = -53.76 cm

Using magnification formula: m = h'/h = -v/u

h' = h × (-v/u) = 7 × (-(-53.76)/(-27)) = 7 × (-1.99) = -13.93 cm

Results:

  • Position of image: 53.76 cm in front of the mirror (screen should be placed here)
  • Size of image: 13.93 cm (enlarged)
  • Nature of image: Real and inverted (negative h')