Important Formulae

• Mirror Formula: 1/f = 1/v + 1/u
• Lens Formula: 1/f = 1/v – 1/u
• Magnification (Mirror): m = –v/u
• Magnification (Lens): m = v/u
• Power of Lens: P = 1/f (f in metre)

1. A concave mirror of focal length 15 cm forms an image at 30 cm. Find the object distance and nature of image.
2. An object is placed 25 cm from a concave mirror of focal length 10 cm. Find the image distance.
3. A convex mirror of focal length 20 cm forms an image at 8 cm. Find the object distance and magnification.
4. A concave mirror of radius of curvature 30 cm forms an image 10 cm in front of the mirror. Find the object distance.
5. An object is placed 18 cm from a concave mirror and forms an image at 9 cm. Find the focal length.
6. A convex mirror produces an image of height 3 cm for an object of height 6 cm. If object distance is 24 cm, find the focal length.
7. An object is placed 40 cm from a concave mirror and a real image forms at 10 cm. Find the focal length.
8. Find the position of image formed by a convex mirror of focal length 25 cm when object is at 50 cm.
9. An object of height 5 cm is placed 20 cm in front of a convex mirror of focal length 15 cm. Determine image height.
10. An object is placed 12 cm from a concave mirror of focal length 8 cm. Find image distance and magnification.

11. A convex lens of focal length 20 cm forms image of object placed at 40 cm. Find image distance.
12. An object is placed 30 cm from a convex lens of focal length 10 cm. Locate image and state nature.
13. An image is formed at 30 cm by a convex lens of focal length 15 cm. Find object distance.
14. A concave lens of focal length 12 cm forms image at 8 cm. Find object distance.
15. A convex lens of focal length 25 cm produces magnification 2. If image distance is 50 cm, find object distance.
16. An object of height 4 cm is placed at 15 cm from convex lens of focal length 10 cm. Find image height.
17. A convex lens forms virtual image at 10 cm when object is at 15 cm. Determine focal length.
18. A concave lens of focal length 20 cm produces image at 16 cm. Find object distance.
19. An object of height 8 cm is placed 30 cm in front of convex lens of focal length 15 cm. Find size of image.
20. A convex lens of focal length 30 cm forms real image at 60 cm. Find object distance and magnification.

21. A concave mirror of focal length 18 cm forms image at 6 cm when object is placed at 9 cm. Find magnification.
22. An object is placed 45 cm in front of convex mirror of focal length 15 cm. Find image distance.
23. A convex lens forms real and inverted image at 25 cm when object is at 15 cm. Find focal length.
24. A concave mirror has radius of curvature 40 cm. Object placed at 12 cm. Determine image distance and magnification.
25. An object of height 6 cm is placed at 20 cm from convex lens of focal length 10 cm. Find image distance and size of image.

• Apply New Cartesian Sign Convention.
• Decide the sign of focal length according to mirror or lens type.
• Object distance is taken as per direction of incident light.
• Final answer must include nature of image wherever required.

Important Formulae

• Mirror Formula: 1/f = 1/v + 1/u
• Lens Formula: 1/f = 1/v – 1/u
• Magnification (Mirror): m = –v/u
• Magnification (Lens): m = v/u
• Power of Lens: P = 1/f (f in metre)

1. A concave mirror of focal length –15 cm forms an image at –30 cm. Find the object distance and nature of image.
2. An object is placed at –25 cm from a concave mirror of focal length –10 cm. Find the image distance.
3. A convex mirror of focal length +20 cm forms an image at +8 cm. Find the object distance and magnification.
4. A concave mirror of radius of curvature –30 cm forms an image 10 cm in front of the mirror. Find the object distance.
5. An object is placed at –18 cm from a concave mirror and forms an image at –9 cm. Find the focal length.
6. A convex mirror produces an image of height 3 cm for an object of height 6 cm. If object distance is –24 cm, find the focal length.
7. An object is placed at –40 cm from a concave mirror and a real image forms at –10 cm. Find the focal length.
8. Find the position of image formed by a convex mirror of focal length +25 cm when object is at –50 cm.
9. An object of height 5 cm is placed 20 cm in front of a convex mirror of focal length +15 cm. Determine image height.
10. An object is placed at –12 cm from a concave mirror of focal length –8 cm. Find image distance and magnification.

11. A convex lens of focal length +20 cm forms image of object at –40 cm. Find image distance.
12. An object is placed at –30 cm from a convex lens of focal length +10 cm. Locate image and state nature.
13. An image is formed at +30 cm by a convex lens of focal length +15 cm. Find object distance.
14. A concave lens of focal length –12 cm forms image at –8 cm. Find object distance.
15. A convex lens of focal length +25 cm produces magnification –2. If image distance is +50 cm, find object distance.
16. An object of height 4 cm is placed at –15 cm from convex lens of focal length +10 cm. Find image height.
17. A convex lens forms virtual image at –10 cm when object is at –15 cm. Determine focal length.
18. A concave lens of focal length –20 cm produces image at –16 cm. Find object distance.
19. An object of height 8 cm is placed 30 cm in front of convex lens of focal length +15 cm. Find size of image.
20. A convex lens of focal length +30 cm forms real image at +60 cm. Find object distance and magnification.

21. A concave mirror of focal length –18 cm forms image at –6 cm when object is placed at –9 cm. Find magnification.
22. An object is placed at –45 cm in front of convex mirror of focal length +15 cm. Find image distance.
23. A convex lens forms real and inverted image at +25 cm when object is at –15 cm. Find focal length.
24. A concave mirror has radius of curvature –40 cm. Object placed at –12 cm. Determine image distance and magnification.
25. An object of height 6 cm is placed at –20 cm from convex lens of focal length +10 cm. Find image distance and size of image.

• All distances are measured from Pole (mirror) or Optical Centre (lens).
• Distances measured in direction of incident light are positive.
• Distances measured opposite to incident light are negative.
• Focal length of concave mirror is negative; convex mirror is positive.
• Focal length of convex lens is positive; concave lens is negative.

1. Concave Mirror
Given: \( f = -15\,cm \), \( v = -30\,cm \) Mirror Formula: \[ \frac{1}{f} = \frac{1}{v} + \frac{1}{u} \] \[ \frac{1}{-15} = \frac{1}{-30} + \frac{1}{u} \] \[ \frac{1}{u} = \frac{1}{-15} – \frac{1}{-30} \] LCM = 30 \[ \frac{1}{u} = \frac{-2+1}{30} = \frac{-1}{30} \] \[ u = -30\,cm \] Object at 30 cm in front. Image is real and inverted.

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2. Concave Mirror
Given: \( f=-10\,cm \), \( u=-25\,cm \) \[ \frac{1}{-10} = \frac{1}{v} + \frac{1}{-25} \] \[ \frac{1}{v} = \frac{1}{-10} – \frac{1}{-25} \] LCM = 50 \[ \frac{1}{v} = \frac{-5+2}{50} = \frac{-3}{50} \] \[ v = -16.7\,cm \] Real and inverted.

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3. Convex Mirror
Given: \( f=+20\,cm \), \( v=+8\,cm \) \[ \frac{1}{20} = \frac{1}{8} + \frac{1}{u} \] \[ \frac{1}{u} = \frac{1}{20} – \frac{1}{8} \] LCM = 40 \[ \frac{1}{u} = \frac{2-5}{40} = \frac{-3}{40} \] \[ u = -13.3\,cm \] Magnification: \[ m = -\frac{v}{u} = 0.6 \] Virtual and erect.

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4. Concave Mirror
Radius \( R = -30\,cm \) \[ f = \frac{R}{2} = -15\,cm \] Given \( v = -10\,cm \) \[ \frac{1}{-15} = \frac{1}{-10} + \frac{1}{u} \] \[ \frac{1}{u} = \frac{-2+3}{30} \] \[ u = +30\,cm \]

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5. Concave Mirror
Given \( u=-18\,cm \), \( v=-9\,cm \) \[ \frac{1}{f} = \frac{1}{-9} + \frac{1}{-18} \] \[ \frac{1}{f} = \frac{-2-1}{18} = \frac{-3}{18} \] \[ f = -6\,cm \]

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6. Convex Mirror
Given: \( h_o = 6\,cm \), \( h_i = 3\,cm \), \( u=-24\,cm \) \[ m = \frac{h_i}{h_o} = \frac{3}{6} = 0.5 \] \[ m = -\frac{v}{u} \] \[ 0.5 = -\frac{v}{-24} \] \[ v = +12\,cm \] \[ \frac{1}{f} = \frac{1}{12} + \frac{1}{-24} \] \[ \frac{1}{f} = \frac{2-1}{24} = \frac{1}{24} \] \[ f = +24\,cm \]

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7. Concave Mirror
Given \( u=-40\,cm \), \( v=-10\,cm \) \[ \frac{1}{f} = \frac{1}{-10} + \frac{1}{-40} \] \[ \frac{1}{f} = \frac{-4-1}{40} = \frac{-5}{40} \] \[ f = -8\,cm \]

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8. Convex Mirror
Given \( f=+25\,cm \), \( u=-50\,cm \) \[ \frac{1}{25} = \frac{1}{v} + \frac{1}{-50} \] \[ \frac{1}{v} = \frac{1}{25} + \frac{1}{50} \] \[ \frac{1}{v} = \frac{2+1}{50} = \frac{3}{50} \] \[ v = 16.7\,cm \]

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9. Convex Mirror
\[ \frac{1}{15} = \frac{1}{v} + \frac{1}{-20} \] \[ \frac{1}{v} = \frac{1}{15} + \frac{1}{20} \] \[ \frac{1}{v} = \frac{4+3}{60} = \frac{7}{60} \] \[ v = 8.57\,cm \] \[ m = -\frac{v}{u} = 0.43 \] \[ h_i = 0.43 \times 5 = 2.15\,cm \]

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10. Concave Mirror
\[ \frac{1}{-8} = \frac{1}{v} + \frac{1}{-12} \] \[ \frac{1}{v} = \frac{-3+2}{24} \] \[ v = -24\,cm \] \[ m = -\frac{-24}{-12} = -2 \] Real and inverted.

11. Convex Lens
Given: \( f = +20\,cm \), \( u = -40\,cm \) Lens Formula: \[ \frac{1}{f} = \frac{1}{v} – \frac{1}{u} \] \[ \frac{1}{20} = \frac{1}{v} – \left(\frac{1}{-40}\right) \] \[ \frac{1}{20} = \frac{1}{v} + \frac{1}{40} \] \[ \frac{1}{v} = \frac{1}{20} – \frac{1}{40} \] LCM = 40 \[ \frac{1}{v} = \frac{2 – 1}{40} = \frac{1}{40} \] \[ v = +40\,cm \] Image real and inverted.

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12. Convex Lens
Given: \( f = +10\,cm \), \( u = -30\,cm \) \[ \frac{1}{10} = \frac{1}{v} – \left(\frac{1}{-30}\right) \] \[ \frac{1}{10} = \frac{1}{v} + \frac{1}{30} \] \[ \frac{1}{v} = \frac{1}{10} – \frac{1}{30} \] LCM = 30 \[ \frac{1}{v} = \frac{3 – 1}{30} = \frac{2}{30} \] \[ v = +15\,cm \] Image real and inverted.

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13. Convex Lens
Given: \( f = +15\,cm \), \( v = +30\,cm \) \[ \frac{1}{15} = \frac{1}{30} – \frac{1}{u} \] \[ \frac{1}{u} = \frac{1}{30} – \frac{1}{15} \] LCM = 30 \[ \frac{1}{u} = \frac{1 – 2}{30} = \frac{-1}{30} \] \[ u = -30\,cm \]

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14. Concave Lens
Given: \( f = -12\,cm \), \( v = -8\,cm \) \[ \frac{1}{-12} = \frac{1}{-8} – \frac{1}{u} \] \[ \frac{1}{u} = \frac{1}{-8} – \frac{1}{-12} \] LCM = 24 \[ \frac{1}{u} = \frac{-3 + 2}{24} = \frac{-1}{24} \] \[ u = -24\,cm \]

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15. Convex Lens
Given: \( f = +25\,cm \), \( v = +50\,cm \) \[ \frac{1}{25} = \frac{1}{50} – \frac{1}{u} \] \[ \frac{1}{u} = \frac{1}{50} – \frac{1}{25} \] LCM = 50 \[ \frac{1}{u} = \frac{1 – 2}{50} = \frac{-1}{50} \] \[ u = -50\,cm \]

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16. Convex Lens
Given: \( h_o = 4\,cm \), \( u = -15\,cm \), \( f = +10\,cm \) \[ \frac{1}{10} = \frac{1}{v} – \left(\frac{1}{-15}\right) \] \[ \frac{1}{10} = \frac{1}{v} + \frac{1}{15} \] \[ \frac{1}{v} = \frac{1}{10} – \frac{1}{15} \] LCM = 30 \[ \frac{1}{v} = \frac{3 – 2}{30} = \frac{1}{30} \] \[ v = +30\,cm \] Magnification: \[ m = \frac{v}{u} = \frac{30}{-15} = -2 \] \[ h_i = -2 \times 4 = -8\,cm \] Image height = 8 cm (real and inverted).

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17. Convex Lens
Given: \( u = -15\,cm \), \( v = -10\,cm \) \[ \frac{1}{f} = \frac{1}{-10} – \frac{1}{-15} \] LCM = 30 \[ \frac{1}{f} = \frac{-3 + 2}{30} = \frac{-1}{30} \] \[ f = -30\,cm \]

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18. Concave Lens
Given: \( f = -20\,cm \), \( v = -16\,cm \) \[ \frac{1}{-20} = \frac{1}{-16} – \frac{1}{u} \] LCM = 80 \[ \frac{1}{u} = \frac{-5 + 4}{80} = \frac{-1}{80} \] \[ u = -80\,cm \]

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19. Convex Lens
Given: \( u = -30\,cm \), \( f = +15\,cm \) \[ \frac{1}{15} = \frac{1}{v} – \left(\frac{1}{-30}\right) \] \[ \frac{1}{15} = \frac{1}{v} + \frac{1}{30} \] \[ \frac{1}{v} = \frac{1}{15} – \frac{1}{30} \] LCM = 30 \[ \frac{1}{v} = \frac{2 – 1}{30} = \frac{1}{30} \] \[ v = +30\,cm \] Magnification: \[ m = \frac{v}{u} = \frac{30}{-30} = -1 \] \[ h_i = -1 \times 8 = -8\,cm \]

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20. Convex Lens
Given: \( f = +30\,cm \), \( v = +60\,cm \) \[ \frac{1}{30} = \frac{1}{60} – \frac{1}{u} \] \[ \frac{1}{u} = \frac{1}{60} – \frac{1}{30} \] LCM = 60 \[ \frac{1}{u} = \frac{1 – 2}{60} = \frac{-1}{60} \] \[ u = -60\,cm \] Magnification: \[ m = \frac{v}{u} = \frac{60}{-60} = -1 \]

21. Concave Mirror
Given: \( f = -18\,cm \), \( v = -6\,cm \) Mirror formula: \[ \frac{1}{f} = \frac{1}{v} + \frac{1}{u} \] \[ \frac{1}{-18} = \frac{1}{-6} + \frac{1}{u} \] \[ \frac{1}{u} = \frac{1}{-18} – \frac{1}{-6} \] LCM = 18 \[ \frac{1}{u} = \frac{-1 + 3}{18} = \frac{2}{18} \] \[ u = +9\,cm \] Magnification: \[ m = -\frac{v}{u} = -\frac{-6}{9} = 0.67 \]

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22. Convex Mirror
Given: \( f = +15\,cm \), \( u = -45\,cm \) \[ \frac{1}{15} = \frac{1}{v} + \frac{1}{-45} \] \[ \frac{1}{v} = \frac{1}{15} + \frac{1}{45} \] LCM = 45 \[ \frac{1}{v} = \frac{3 + 1}{45} = \frac{4}{45} \] \[ v = 11.25\,cm \] Image formed behind mirror (virtual and erect).

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23. Convex Lens
Given: \( v = +25\,cm \), \( u = -15\,cm \) Lens formula: \[ \frac{1}{f} = \frac{1}{v} – \frac{1}{u} \] \[ \frac{1}{f} = \frac{1}{25} – \left(\frac{1}{-15}\right) \] \[ \frac{1}{f} = \frac{1}{25} + \frac{1}{15} \] LCM = 75 \[ \frac{1}{f} = \frac{3 + 5}{75} = \frac{8}{75} \] \[ f = 9.4\,cm \]

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24. Concave Mirror
Radius of curvature \( R = -40\,cm \) \[ f = \frac{R}{2} = -20\,cm \] Given \( u = -12\,cm \) \[ \frac{1}{-20} = \frac{1}{v} + \frac{1}{-12} \] \[ \frac{1}{v} = \frac{1}{-20} – \frac{1}{-12} \] LCM = 60 \[ \frac{1}{v} = \frac{-3 + 5}{60} = \frac{2}{60} \] \[ v = 30\,cm \] Magnification: \[ m = -\frac{v}{u} = -\frac{30}{-12} = 2.5 \]

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25. Convex Lens
Given: \( f = +10\,cm \), \( u = -20\,cm \) \[ \frac{1}{10} = \frac{1}{v} – \left(\frac{1}{-20}\right) \] \[ \frac{1}{10} = \frac{1}{v} + \frac{1}{20} \] \[ \frac{1}{v} = \frac{1}{10} – \frac{1}{20} \] LCM = 20 \[ \frac{1}{v} = \frac{2 – 1}{20} = \frac{1}{20} \] \[ v = 20\,cm \] Magnification: \[ m = \frac{v}{u} = \frac{20}{-20} = -1 \] Image height: \[ h_i = -1 \times 6 = -6\,cm \] Image height = 6 cm (real and inverted).

1. What steps were taken to increase agricultural production?
Iron tools such as ploughshares were used to clear forests and cultivate land. Irrigation facilities like wells, canals, and tanks were developed. Farmers began using better seeds and farming techniques to increase production.

2. What is urbanization?
Urbanization is the process by which villages grow into towns and cities due to increase in population, trade, and economic activities.

3. Which period is referred to as the second urbanisation? Which was the first one?
The period around 600 BCE onwards is called the second urbanisation. The first urbanisation took place during the Indus Valley Civilization.

4. Describe the distinctive features of Northern Black Polished Ware.
Northern Black Polished Ware (NBPW) was a fine, glossy pottery with a smooth black surface. It was highly polished and considered a luxury item, mainly used by wealthy people.

5. What evidence points towards India’s trade relations with the Roman world?
Roman gold coins found in South India and references in Roman texts indicate active trade between India and Rome. Indian spices, silk, and precious stones were exported to Rome.

1. Types of towns during the second urbanisation:
Various types of towns developed during this period:

• Administrative towns such as Pataliputra.

• Temple towns such as Mathura.

• Commercial towns such as Ujjain.

• Port towns such as Tamralipti.
These towns grew due to trade, administration, and religious importance.

2. Economic advancement through coins:
Punch-marked coins made of silver and copper were widely used. Later, gold coins were introduced by rulers like the Kushanas and Guptas. The use of coins shows growth in trade and a money-based economy.

3. Define shrenis and their role:
Shrenis were associations or guilds of merchants and craftsmen. They controlled production, fixed prices, and maintained quality. Shrenis helped in the growth of towns by encouraging trade and providing loans.

4. Importance of Mathura:
Mathura was an important centre of trade and religion. It was located on important trade routes and was a major centre for Buddhism, Jainism, and Hinduism.

5. Life in villages and role of Grama Bhojaka:
Most people lived in villages and were engaged in agriculture. The Grama Bhojaka was the village headman who collected taxes on behalf of the king and maintained law and order.

Grihapatis	Krishakas
Large landowners.	Small farmers who worked on land.
Owned slaves and labourers.	Often worked for landowners.

Dasas	Karmakaras
Slaves who had no freedom.	Hired labourers who worked for wages.
Worked without payment.	Received wages for their work.

Vellalar	Uzhavar
Large landowners in Tamil region.	Small farmers who cultivated land.
Owned vast agricultural lands.	Worked on their own or others’ land.

1. The post-Mauryan phase is called the age of second urbanization.
During this period, many new towns developed due to increased trade, use of coins, and growth of craft production.

2. Trade with Rome was considered in favour of India.
India exported spices, silk, and precious stones to Rome and received gold coins in return. This led to an inflow of gold into India.

Given Data

32, 32, 30, 35, 33, 34, 28, 29, 30, 31, 31, 29, 30, 31, 32, 31, 31, 28, 28, 31, 28, 28, 33, 31, 32, 38, 39

Total observations: \( n = 27 \)

Frequency Table

Value (x)	Frequency (f)
28	5
29	2
30	3
31	7
32	4
33	2
34	1
35	1
38	1
39	1

Mode

Highest frequency = 7

$$ \text{Mode} = 31 $$

Median

$$ \text{Median position} = \frac{n+1}{2} $$

$$ = \frac{27+1}{2} = \frac{28}{2} = 14^{th}\ \text{observation} $$

The 14th observation lies in 31.

$$ \text{Median} = 31 $$

$$ \text{Mean} = \frac{\sum fx}{\sum f} $$

x	f	fx
28	5	140
29	2	58
30	3	90
31	7	217
32	4	128
33	2	66
34	1	34
35	1	35
38	1	38
39	1	39

$$ \sum fx = 845 $$

$$ \sum f = 27 $$

$$ \text{Mean} = \frac{845}{27} = 31 + \frac{8}{27} $$

$$ \text{Mean} \approx 31.30 $$

$$ \text{Range} = \text{Maximum value} – \text{Minimum value} $$

$$ = 39 – 28 = 11 $$