Tag: CHAPTER 9 Light – Reflection and Refraction

Q1. Which one of the following materials cannot be used to make a lens?

(a) Water
(b) Glass
(c) Plastic
(d) Clay

Answer: (d) Clay

Explanation:

A material used to make a lens must be transparent to allow light to pass and refract through it.

• Water, glass and plastic are transparent materials and can be shaped into lenses.

• Clay is opaque, so it cannot be used to make a lens.

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Q2. 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?

(a) Between the principal focus and the centre of curvature
(b) At the centre of curvature
(c) Beyond the centre of curvature
(d) Between the pole of the mirror and its principal focus.

Answer: (d) Between the pole of the mirror and its principal focus

Explanation:

A concave mirror forms a virtual, erect, and magnified image only when the object is placed between P and F.
In all other positions, the image formed is real and inverted.

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Q3. Where should an object be placed in front of a convex lens to get a real image of the size of the object?

(a) At the principal focus of the lens
(b) At twice the focal length
(c) At infinity
(d) Between the optical centre and its principal focus

Answer: (b) At twice the focal length

Explanation:

For a convex lens:

• When the object is at 2F, the image is real, inverted, and same size formed at 2F on the other side.

Q4. 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

(a) both concave
(b) both convex
(c) the mirror is concave and the lens is convex
(d) the mirror is convex, but the lens is concave

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Answer: (d) the mirror is convex, but the lens is concave

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Explanation:

According to the sign conventions:

• Focal length of a concave mirror is negative

• Focal length of a convex mirror is positive

• Focal length of a convex lens is positive

• Focal length of a concave lens is negative

Q5. No matter how far you stand from a mirror, your image appears erect. The mirror is likely to be

(a) only plane.
(b) only concave.
(c) only convex.
(d) either plane or convex.

Answer: (d) either plane or convex

Explanation:

• A plane mirror always forms an erect and same-sized image at all distances.

• A convex mirror always forms an erect and diminished virtual image no matter where the object is placed.

• A concave mirror can form inverted images when the object is far away.

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Q6. Which of the following lenses would you prefer to use while reading small letters found in a dictionary?

(a) A convex lens of focal length 50 cm.
(b) A concave lens of focal length 50 cm.
(c) A convex lens of focal length 5 cm.
(d) A concave lens of focal length 5 cm.

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

Explanation:

To magnify tiny letters, we use a magnifying glass, which is a convex lens with short focal length so that it produces a large, virtual, and erect image.

• Convex lens helps in magnification.

• Short focal length = greater magnifying power.

Q 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.

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Answer:

Range of distance

To get an erect image using a concave mirror, the object must be placed:

$$\mathbf{\text{Between}}\mathbf{\text{the}}\mathbf{\text{pole}}\mathbf{\text{(P)}}\mathbf{\text{and}}\mathbf{\text{the}}\mathbf{\text{principal}}\mathbf{\text{focus}}\mathbf{\text{(F)}}$$

Given:

$$f=15\text{ cm}$$

So, the object distance must be:

$$\mathbf{\text{Less}}\mathbf{\text{than}}\mathbf{\text{15}}\mathbf{\text{cm}}\mathbf{\text{from}}\mathbf{\text{the}}\mathbf{\text{mirror}}$$

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Nature of the Image

• Virtual

• Erect

• Magnified (larger than the object)

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Size of the Image

The image is larger (magnified) when the object is placed between P and F.

Q8. Name the type of mirror used in the following situations.

(a) Headlights of a car.
(b) Side/rear-view mirror of a vehicle.
(c) Solar furnace.
Support your answer with reason.

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Answer:

(a) Headlights of a car

Mirror used: Concave mirror

Reason:
A concave mirror, when the bulb is placed at its focus, reflects light as a powerful parallel beam of light. This helps the driver to see long distances clearly at night.

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

Mirror used: Convex mirror

Reason:
A convex mirror:

• Always forms an erect image

• Forms a diminished image, allowing a wide field of view
  Thus, it helps the driver see more traffic area behind the vehicle.

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(c) Solar furnace

Mirror used: Concave mirror

Reason:
A concave mirror converges parallel rays of sunlight to its focus, concentrating solar energy at a single point, producing very high temperature for heating or melting materials.

Q9. 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.

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Answer:

Yes, the lens will still produce a complete image of the object, even if one-half of the convex lens is covered with black paper.

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Explanation:

Every part of the lens refracts light rays from all parts of the object.
So even when half of the lens is covered, the remaining half still allows light rays from the whole object to pass and form the complete image.

However:

• The image becomes dimmer (less bright),

• Because lesser light passes through the uncovered portion.

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Experimental Verification:

Materials needed:

Convex lens, candle (object), screen, black paper.

Procedure:

1. Place a convex lens on a stand.

2. Place a burning candle in front of the lens and place a screen behind the lens.

3. Adjust until you get a clear sharp image.

4. Now cover the upper half of the lens with black paper.

5. Observe the image on the screen.

Observation:

• The complete image of the candle is still formed.

• The image becomes less bright compared to before covering the lens.

Conclusion:

Even a small portion of a lens contains the ability to refract light from all parts of an object, forming a complete imagebut with reduced brightness.

Q10. Solution

Question

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.

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Given Data

• Height of object, h = 5 cm

• Object distance, u = –25 cm

• Focal length, f = +10 cm (positive because converging/convex lens)

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Using Lens Formula

$$\frac{1}{v}-\frac{1}{u}=\frac{1}{f}$$
$$\frac{1}{v}-\frac{1}{-25}=\frac{1}{10}$$
$$\frac{1}{v}+\frac{1}{25}=\frac{1}{10}$$
$$\frac{1}{v}=\frac{1}{10}-\frac{1}{25}=\frac{5-2}{50}=\frac{3}{50}$$
$$v=\frac{50}{3}=16.67\text{ cm (approx)}$$

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Magnification

$$m=\frac{v}{u}=\frac{16.67}{-25}=-0.67$$
$$h^{\mathrm{\prime }}=m\times h=-0.67\times 5=-3.33\text{ cm}$$

Q11. Solution

Question

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.

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Given

• Focal length of concave lens, f = –15 cm

• Image distance, v = –10 cm (negative for virtual image on same side as object)

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Using Lens Formula

$$\frac{1}{v}-\frac{1}{u}=\frac{1}{f}$$
$$\frac{1}{-10}-\frac{1}{u}=\frac{1}{-15}$$
$$-\frac{1}{10}+\frac{1}{15}=\frac{1}{u}$$
$$\frac{-3+2}{30}=\frac{1}{u}$$
$$\frac{-1}{30}=\frac{1}{u}$$
$$u=-30\text{ cm}$$

Final Answer Summary

 

Q12. 

Question

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.

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Solution:

Given

• Object distance: u = –10 cm (negative as per sign convention, object placed in front)

• Focal length of convex mirror: f = +15 cm
  (Convex mirror focal length is positive because its focus lies behind the mirror)

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Using Mirror Formula

$$\frac{1}{v}+\frac{1}{u}=\frac{1}{f}$$
$$\frac{1}{v}-\frac{1}{10}=\frac{1}{15}$$
$$\frac{1}{v}=\frac{1}{15}+\frac{1}{10}$$
$$\frac{1}{v}=\frac{2+3}{30}=\frac{5}{30}=\frac{1}{6}$$
$$v=+6\text{ cm}$$

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Interpretation

The image distance is positive, meaning the image forms behind the mirror.

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Final Result

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Final Statement

For an object placed 10 cm in front of a convex mirror of focal length 15 cm, a virtual, erect and smaller image is formed 6 cm behind the mirror.

 

Q13. 

Question

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

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Answer

Magnification +1 means that:

1. The image formed is of the same size as the object

$$m=\frac{\text{height of image}}{\text{height of object}}=+1$$

So, image height = object height.

2. The positive sign indicates the image is erect

Positive magnification → Erect image
Negative magnification → Inverted image

3. The mirror forms a virtual image

A plane mirror always produces a virtual, erect, and same-sized image.

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Final Statement

Magnification +1 means the image is virtual, erect, and equal in size to the object, which is a characteristic of a plane mirror.

Q14. 

Question

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.

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Answer:

Given

• Object height, h = +5.0 cm

• Object distance, u = –20 cm

• Radius of curvature, R = 30 cm

• For spherical mirrors:

$$f=\frac{R}{2}=\frac{30}{2}=15\text{ cm}$$

For a convex mirror, focal length is positive

$$f=+15\text{ cm}$$

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Mirror Formula

$$\frac{1}{v}+\frac{1}{u}=\frac{1}{f}$$
$$\frac{1}{v}-\frac{1}{20}=\frac{1}{15}$$
$$\frac{1}{v}=\frac{1}{15}+\frac{1}{20}$$

$$\frac{1}{v}=\frac{4+3}{60}=\frac{7}{60}$$
$$v=\frac{60}{7}=+8.57\text{ cm (approx)}$$

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Interpretation

• Positive v indicates the image is formed behind the mirror

• Convex mirrors always form virtual, erect and diminished images

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Magnification

$$m=\frac{v}{u}=\frac{8.57}{-20}=-0.4285$$
$$h^{\mathrm{\prime }}=m\times h=-0.4285\times 5=-2.14\text{ cm}$$

Magnitude of height = 2.14 cm
Negative sign means image is upright (virtual & erect).

Q15. 

Question

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 focused image can be obtained? Find the size and the nature of the image.

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Answer:

Given

• Object height: h = +7.0 cm

• Object distance: u = –27 cm (object in front of mirror → negative)

• Focal length for concave mirror: f = –18 cm

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Using Mirror Formula

$$\frac{1}{v}+\frac{1}{u}=\frac{1}{f}$$
$$\frac{1}{v}-\frac{1}{27}=\frac{1}{18}$$
$$\frac{1}{v}=\frac{1}{18}+\frac{1}{27}$$

$$\frac{1}{v}=\frac{3}{54}+\frac{2}{54}=\frac{5}{54}$$
$$v=\frac{54}{5}=10.8\text{ cm}$$

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Position of Image

Screen should be placed at:

v = 10.8 cm in front of the mirror

(Because real images form in front of concave mirrors)

Magnification

$$m=\frac{v}{u}=\frac{10.8}{-27}=-0.4$$
$$h^{\mathrm{\prime }}=m\times h=-0.4\times 7=-2.8\text{ cm}$$

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Q16. Solution

Question

Find the focal length of a lens of power –2.0 D. What type of lens is this?

Answer:

Given

Power of lens,

$$P=-2.0D$$

Formula

$$P=\frac{1}{f}$$

So,

$$f=\frac{1}{P}$$

Substitution

$$f=\frac{1}{-2.0}=-0.5m$$

Convert into cm:

$$f=-0.5\times 100=-50cm$$

Q17. 

Question

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?

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Answer:

Given

Power of lens:

$$P=+1.5D$$

Formula

$$P=\frac{1}{f}$$

Therefore,

$$f=\frac{1}{P}$$

Substitution

$$f=\frac{1}{1.5}=0.67m$$

Convert into cm:

$$0.67\times 100=67cm$$

Final Answer

Explanation

• A positive power always represents a convex (converging) lens

• Convex lenses correct hypermetropia (far-sightedness)

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Page 142 – Questions

Q1. Define the principal focus of a concave mirror.

Answer:
The principal focus of a concave mirror is the point on the principal axis at which all the light rays parallel to the principal axis converge after reflection from the mirror.

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Q2. The radius of curvature of a spherical mirror is 20 cm. What is its focal length?

Answer:
We know: R = 2f
So, f = R/2 = 20/2 = 10 cm

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Q3. Name a mirror that can give an erect and enlarged image of an object.

Answer:
A concave mirror can give an erect and enlarged image when the object is placed between the pole (P) and the focus (F).

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Q4. Why do we prefer a convex mirror as a rear-view mirror in vehicles?

Answer:
Convex mirrors are used as rear-view mirrors because:

• They always form an erect image.

• The image is diminished, so a larger area can be seen.

• They provide a wide field of view, helping the driver see more traffic behind.

Page 145 – Questions

Q1. Find the focal length of a convex mirror whose radius of curvature is 32 cm.

Answer:
Using R = 2f
So, f = R/2 = 32/2 = 16 cm

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Q2. A concave mirror produces three times magnified real image of an object placed at 10 cm in front of it. Where is the image located?

Answer:
Magnification m = –v/u (negative because image is real and inverted)

Given: m = 3, u = –10 cm

$$3=-v\mathrm{/}(-10)\Rightarrow v=30\text{ cm}$$

So, the image is formed 30 cm in front of the mirror.

Page 150 – Questions

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Q1. A ray of light travelling in air enters obliquely into water. Does it bend towards the normal or away from the normal? Why?

Answer:
It bends towards the normal, because light travels slower in water (denser medium) than in air (rarer medium).

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Q2. Light enters from air to glass having refractive index 1.50. What is the speed of light in the glass?

Answer:

$$n=\frac{c}{v}\Rightarrow v=\frac{c}{n}=\frac{3\times {10}^8}{1.50}$$

$$v=2\times {10}^8\text{ m/s}$$

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Q3. Find out, from Table 9.3, the medium having highest optical density. Also find the medium with lowest optical density.

Answer:

• Highest optical density: Diamond (n = 2.42)

• Lowest optical density: Air (n = 1.0003)

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Q4. You are given kerosene, turpentine and water. In which of these does the light travel fastest?

Answer:
Light travels fastest in the medium with lowest refractive index.

Refractive indices:
Water = 1.33, Kerosene = 1.44, Turpentine = 1.47

So, light travels fastest in water.

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Q5. The refractive index of diamond is 2.42. What is the meaning of this statement?

Answer:
It means the speed of light in air is 2.42 times more than the speed of light in diamond, or light slows down by 2.42 times when entering diamond.

Page 158 – Questions

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Q1. Define 1 dioptre of power of a lens.

Answer:
1 dioptre is the power of a lens whose focal length is 1 metre.

$$1D=1m^{-1}$$

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Q2. A convex lens forms a real and inverted image of a needle at 50 cm from it. Where is the needle placed if the image is equal in size to the object? Also find the power of the lens.

Answer:
For a convex lens, an image equal in size to the object is formed when object is at 2F and image is also at 2F.

So, 2F = 50 cm → F = 25 cm

Thus, object is placed 25 cm from lens.

Power:

$$P=1\mathrm{/}f=1\mathrm{/}0.25=4D$$

Power = +4 dioptre

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Q3. Find the power of a concave lens of focal length 2 m.

Answer:

$$P=1\mathrm{/}f=1\mathrm{/}(-2)=\text{–}0.5D$$

Power = –0.5 dioptre (concave lens)