12. Electricity

Learning points

Electric current and circuit

Electric Potential and Potential Difference

Circuit Diagram

Ohm's Law

Factors on which the resistance of a conductor depends

Resistance of a system of resistors

Resistors in series

Resistors in parallel

Heating effect of electric current

Practical Applications of heating effect of electric current

Electric Power

Questions Answers

NCERT Textbook.

1. What does an electric circuit mean?

Answer

An electric circuit is pathway in which current can follow. A circuit has electric devices, switching devices, source of energy etcetera. These are connected by conducting wire.

2. Define the unit of current.

Answer

When 1C of charge flows through a conductor in 1 s, it is called 1 ampere(A) current.

I = Q/t

The unit of electric current is ampere (A).

3. Calculate the number of electrons constituting one coulomb of charge.

Answer

We know that one electron possesses 1.6×10–¹⁹ C.

Therefore, 

Number of electron  = Total charge/Charge on 1 electron

or, No. of electron = 1/1.6×10–¹⁹

or, No. of electron = 6.25×10¹⁸ electrons. Ans.

1. Name a device that helps to maintain a potential difference across a conductor.

Answer

A cell, battery, power supply, etc helps to maintain potential difference across a conductor. 

2. What is meant by saying that the potential difference between two points is 1 V?

Answer 

When 1 J of work is required to move a charge of 1 C from one point to another, then it is said that the potential difference between two points is 1 V.

V = W/Q

or, 1 V = 1J/1C

3. How much energy is given to each coulomb of charge passing through a 6 V battery?

Answer 

From question it is clear that 

Work done = ?

Potential difference = 6 V

Charge = 1C.

We know that 

Potential difference = Work done/Charge

Therefore, 

Work done = Potential difference × Charge

or, Work done = 6 V×1C = 6 Joules Ans.

4. Two conducting wires of the same material and of equal lengths and equal diameters are first connected in series and then parallel in a circuit across the same potential difference. The ratio of heat produced in series and parallel combination would be

(a) 1:2 

(b) 2:1

(c) 1:4

(d) 4:1

Answer

According to combination of resistance we know that

In the case of series

R(ES) = R1 + R2

From question it is clear that

R1 = R2 = R

R(S) = 2R

Similarly,

In the case of parallel

1/R(P) = 1/R1 + 1/R2

1/R(P) = 1/R + 1/R

R(P) = R/2

For same potential difference V,

We know that

V = IR

I = V/R

I(S) = V/R(S) = V/2R

Similarly,

I(P) = V/R(P) = 2V/R

H = I^2RT

Here,

I = Current follow through the conductor

R = Resistance

T = Time

Therefore,

H(S) = (V/2R)^2.R.T = V^2T/4R  ------- (i)

Similarly,

H(P) = (2V/R)^2 . R.T = 4V^2T/R ------- (ii)

Now, from (i) and (ii)

Therefore, ratio of heat produced in series and parallel

H(S)/H(P) = 1:4 Ans.

Continues 

Answers of the rest questions.....

11. The Human Eye and the Colourful World

The ability of the eye to focus on both near and distant objects, by adjusting its focal length, is called the accommodation of the eye.

The smallest distance, at which the eye can see objects clearly without strain, is called the point of the eye or the least distance of distinct vision. For a young adult with noumal vision, it is about 25 cm.

The common refractiove defects of vision include myopia, hypermetropia and presbyopia. Myopia (short - sightedness - the image of distant objects is focussed before the retina) is corrected by using a concave lens of suitable power. Hypermetropia (far - sightedness - the image of nearby objects is focussed beyond the retina) is corrected by using a convex lens of suitable poser. The eye loses its power of accommodation at old age.

The splitting of white light into its component colours is called dispersion.

Scattering of light causes the blue colour of sky and the reddening of the Sun at sunrise and sunset.

In this chapter, we shall use these ideas to study some of the optical phenomena in nature. We shall also discuss about rainbow for mation, splitting of white light and blue colour of the sky.

Questions and Answers of NCERT Textbook. 

Learning points

The Human eye.

Power of Accommodation.

Defects of Vision and their correction.

Refraction of light through a prism.

Dispersion of white light by a glass prism.

Atmospheric Refraction.

Scattering of Light

Q.No.1

What is mehant by power of accomodation of the eye ?

Answer

To see the near by object clearly, the ciliary muscles contact making the eye lens thicker. Thus, the focal length of the eye lense decreases and nearby objects become visible to the eye. Similarly, to see the far objects clearly, the ciliary muscles contact making the eye lens thinner. Thus, the focal length of the eye lens increases and far objects become visible to the eyes.

Hence, the human eye lens is able to adjust its focal length to view both nearby and far objects on the retina. This ability is called the power of accommodation of the eyes.

Q.No.2.

A person with a myopic eye cannon see objects beyond 1.2 m distinctly. What should be the type of the corrective lens used to restore power vision?

Answer 

The person is able to see nearby objects clearly, but he or she unable to see objects beyond 1.2 m. This happens because the image of an object beyond 1.2 m is formed in the front of the retina and not at the retina.

To correct this defect of vision, he must use a concave lens. The concave lens will bring the image back to the retina.

Q.No.3.

What is the far point and near point of the human eye with normal vision?

Answer 

For normal eyes, the least distance of distinct vision is 25 cm whereas far distance of distinct vision is infinite. 

Q.No. 4.

A student has difficulty reading the blackboard while sitting in the last row. What could be the defect the child is suffering from? How can it be corrected?

Answer 

Last row it means that distant objects. Student is unable to see distant objects clearly. He is suffering from myopia or far sightedness. This defect can be corrected by using concave lens.

1. The human eye can focus on objects at different distances by adjusting the focal length of the eye lens. This is due to

(a) presbyopia.

(b) accommodation.

(c) near-sightedness.

(d) far-sightedness.

Answer 

This is possible due to the power of accommodation of the eye lens. 

The correct option will be (b)

2. The human eye forms the image of an object at its

(a) cornea. (b) iris. (c) pupil. (d) retina.

Answer 

The human eye forms the image of an object at its retina. 

The correct option will be (d)

3. The least distance of distinct vision for a young adult with normal vision is about

(a) 25 m. (b) 2.5 cm. (c) 25 cm. (d) 2.5 m.

Answer 

The least distance of distinct vision for a young adult with normal vision is about 25 cm.

The correct option will be (c)

4. The change in focal length of an eye lens is caused by the action of the

(a) pupil. (b) retina. (c) ciliary muscles. (d) iris.

Answer 

The change in focal length of an eye lens is caused by the action of the ciliary muscles. 

5. A person needs a lens of power –5.5 dioptres for correcting his distant vision. For correcting his near vision he needs a lens of power +1.5 dioptre. What is the focal length of the lens required for correcting (i) distant vision, and (ii) near vision?

Answer 

From question it is clear that 

P = -5.5 D for distant vision. 

We know that 

P = 1/f

f = 1/P

Now putting the value of P.

f = 1/-5.5 = -0.181 m = -18.1 cm.

P = +1.5 for near vision. 

Similarly, 

f = 1/P 

f = 1/1.5 = 0.667 m = 66.7 cm.

6. The far point of a myopic person is 80 cm in front of the eye. What is the nature and power of the lens required to correct the problem?

Answer

From question it is clear that

Object distance = u = infinity 

Image distance = v = -80 cm

Focal length = f = ?

According to lens formula,  We know that

1/v - 1/u = 1/f

Now, putting the value of u, v and f.

-1/80 - 1/infinity  = 1/f

or, -1/80 = 1/f

f = -80 cm = -0.8 m

Power of lens = P = 1/f 

or, P = -1/0.8

P = -1.25 D. 

Therefore,  A concave lens of power -1.25 is required by the person to correct this defect. 

7. Make a diagram to show how hypermetropia is corrected. The near point of a hypermetropic eye is 1 m. What is the power of the lens required to correct this defect? Assume that the near point of the normal eye is 25 cm.

Answer

A person suffering from hypermetropia can see the distant objects clearly but faces difficulty in seeing nearby by objects clearly. This defect of vision is corrected by using a convex lens. A convex lens of suitable power converges the incoming light in such a way that the image is formed on the retina, as shown in the above figure.

The convex lens actually creates a virtual image of a nearby objects (N' in the figure) at the near point of vision (N) of the person suffering from hypermetropia. 

The given person will be able to clearly see the object kept at 25 cm (least distance of distinct vision), if the image of the the object is formed at his near point, which is given as 1 m.

Object distance = u = -25 cm

Image distance = v = -1 m = -100 cm.

Focal length = f ?

According to lens formula 

We know that 

1/v - 1/u = 1/f

-1/100 - 1/-25 = 1/f

or, -1/100 + 1/25 = 1/f

Therefore,  f = 100/3 cm = 1/3 m.

We know that 

P = 1/f

P = 3 D 

A convex lens of power +3 D is required to correct the defect. 

8. Why is a normal eye not able to see clearly the objects placed closer than 25 cm?

Answer

The least distance of distinct vision of a normal eye is 25 cm. Therefore, a normal not able to see clearly the objects placed closer than 25 cm.

9. What happens to the image distance in the eye when we increase the distance of an object from the eye ?

Answer

Since the size of eyes cannot increase or decrease, the image distance remains constant. When we increase the distance of an object from the eye, the image distance in the eye does not change. The increase in the object distance is compensated by the change in the focal length of the eye lens. The focal length of the eyes changes in such a way that the image is always formed at the retina of the eye.

10. Why do stars twinkle ?

Answer

Stars emit their own light and they twinkle due to the atmospheric refraction of light. Stars are very far away from the earth. Hence, they are considered as point sources of light. When the light coming from stars enters the earth’s atmosphere, it gets refracted at different levels because of the variation in the air density at different levels of theatmosphere. When the star light refracted by the atmosphere comes more towards us, it appears brighter than when it comes less towards us. Therefore, it appears as if the stars are twinkling at night.

11. Explain why the planets do not twinkle? 

Answer 

Planets do not twinkle because they appear larger in size than the stars as they are relatively closer to earth. Planets can be considered as a collection of a large number of point-size sources of light. The different parts of these planets produce either brighter or dimmer effect in such a way that the average of brighter and dimmer effect is zero. Hence, the twinkling effects of the planets are nullified and they do not twinkle.

12.Why does the Sun appear reddish early in the morning? 

Answer

During sunrise, the light rays coming from the Sun have to travel a greater distance in the earth’s atmosphere before reaching our eyes. In this journey, the shorter wavelengths of lights are scattered out and only longer wavelengths are able to reach our eyes. Since blue colour has a shorter wavelength and red colour has a longer wavelength, the red colour is able to reach our eyes after the atmospheric scattering of light. Therefore, the Sun appears reddish early in the morning.

13.Why does the sky appear dark instead of blue to an astronaut? 

Answer

The sky appears dark instead of blue to an astronaut because there is no atmosphere in the outer space that can scatter the sunlight. As the sunlight is not scattered, no scattered light reach the eyes of the astronauts and the sky appears black to them.

3. Metals and Non-Metals

QUESTIONS

1. Give an example of a metal which

(i) is a liquid at room temperature.

Answer

Mercury is an example of metal which is a liquid at room temperature.

(ii) can be easily cut with a knife.

Answer

Lithium, sodium and potassium are the example of metals which can be easily cut with a knife.

(iii) is the best conductor of heat.

Answer

Silver and copper are the metals which are the best conductor of heat.

(iv) is a poor conductor of heat.

Answer

Lead and Mercury are metals which are poor conductor of heat.

2. Explain the meanings of malleable and ductile.

Answer

Malleable is a physical property of metal due to which it is beaten into thin sheet.

Ductile is also a physical property of metal due to which it is drawn into thin wire.

Questions

1. Why is sodium kept immersed in kerosene oil?

Answer

Sodium shows high reactivities towards oxygen. Sodium reacts so vigorously that they catch fire if kept in the open. Hence, to protect it and to prevent accidental fires, sodium is kept immersed in kerosene oil.

2. Write equations for the reactions of

(i) iron with steam

Answer

Iron reacts with steam to form ferric oxide and hydrogen.

Chemical reaction is given below.

3Fe(s) + 4H2O(g) → Fe3O4(s) + 4H2(g)

(ii) calcium and potassium with water

Calcium reacts with water.

Answer

Chemical reaction is given below.

Ca(s) + 2H2O(l) → Ca(OH)2(aq) + H2(g)

Potassium reacts with water.

Chemical reaction is given below.

2K(s) + 2H2O(l) → 2KOH(aq) + H2(g) + heat energy

3. Samples of four metals A, B, C and D were taken and added to the following solution one by one. The results obtained have been tabulated as follows.

Use the Table above to answer the following questions about metals A, B, C and D.

(i) Which is the most reactive metal?

Answer

(ii) What would you observe if B is added to a solution of Copper(II) sulphate?

Answer

(iii) Arrange the metals A, B, C and D in the order of decreasing reactivity.

Answer

4. Which gas is produced when dilute hydrochloric acid is added to a reactive metal? Write the chemical reaction when iron reacts with dilute H2SO4.

Answer

5. What would you observe when zinc is added to a solution of iron(II) sulphate? Write the chemical reaction that takes place.

Answer

14. Sources of Energy

1. What is a good source of energy?

Answer

We can say that a good source of energy would be one

 • which would do a large amount of work per unit volume or mass,

 

• be easily accessible,

 

• be easy to store and transport, and

 

• perhaps most importantly, be economical.

2. What is a good fuel?

Answer

It releases more heat on burning. It produce a little of smoke and is easily available?

3. If you could use any source of energy for heating your food, which one would you use and why?

Answer

I can use LPG gas because it doesn't produce smoke and easily available and it releases more heat on burning.

Electricity

Q.No.1. What does an electric circuit mean ?

Answer

An electric circuit means a continuous and closed path of an electric current.

Q.No. 2. Define the unit of current.

Answer

The electric current is expressed by a unit called ampere (A), named after the French scientist, Andre-Marie Ampere . One ampere is constituted by the flow of one coulomb of charge per second, that is, 1 A = 1 C/1 s.

Q.No. 3. Calculate the number of electrons constituting one coulomb of charge.

Answer

We know that

1.6×10-¹⁹ coulomb charge = 1 Electron.

or, 1 coulomb charge = 1÷(1.6×10-¹⁹) Electrons.

or, 1 coulomb charge = 6×10¹⁸ Electrons.

Q.No.1. Name a device that helps to maintain a potential difference across a conductor.

Answer 

Battery or a combination of cells helps to maintain a potential difference across a conductor. 

2. What is meant by saying that the potential difference between two points is 1 V?

Answer 

One volt is the potential difference between two points in a current carrying conductor when 1 joule of work is done to move a charge of 1 coulomb from one point to the other.

Therefore,

1 volt = 1 joule/1 coulomb

1 volt = 1joule per coulomb. 

1 V = 1 J/C.

3. How much energy is given to each coulomb of charge passing through a 6 V battery?

Answer 

Work done = Potential difference×Charge 

or, Work done = 6×1

or, Work done = 6 joule.

Continue .....

प्रकाश – परावर्तन तथा अपवर्तन

प्रश्न

1. अवतल दर्पण के मुख्य फोकस की परिभाषा लिखिए।

उत्तर

दर्पण पर मुख्य अक्ष के समांतर कुछ किरणें आपतित हो रही हैं। सभी दर्पण की मुख्य अक्ष के एक बिंदु पर मिल रही/प्रतिच्छेदी हैं। यह बिंदु अवतल दर्पण का मुख्य फोकस कहलाता है।

2. एक गोलीय दर्पण की वक्रता त्रिज्या 20  cm है। इसकी फोकस दूरी क्या होगी?

उत्तर

f = R/2

or, f = ±20/2

or, f = ±10 cm.

3. उस दर्पण का नाम बताइए जो बिंब का सीधा तथा आवर्धित प्रतिबिंब बना सके।

उत्तर

अवतल दर्पण

4. हम वाहनों में उत्तल दर्पण को पश्च-दृश्य दर्पण के रूप में वरीयता क्यों देते हैं?

उत्तर

उत्तल दर्पणों को इसलिए भी प्राथमिकता देते हैं, क्योंकि ये सदैव सीधा प्रतिबिंब बनाते हैं यद्यपि वह छोटा होता है। इनका दृष्टि-क्षेत्र भी बहुत अधिक है क्योंकि ये बाहर की ओर वक्रित होते हैं।

1. उस उत्तल दर्पण की फोकस दूरी ज्ञात कीजिए जिसकी वक्रता-त्रिज्या  32  cm है।

उत्तर

f = R÷2

or, f = 32÷2

or, f = 16 cm.

2. कोई अवतल दर्पण आपने सामने  10  cm दूरी पर रखे किसी बिंब का तीन गुणा आवर्धित (बड़ा) वास्तविक प्रतिबिंब बनाता है। प्रतिबिंब दर्पण से कितनी दूरी पर है।

उत्तर

दर्पण के अवर्धन सूत्र से

प्रतिबिंब की ऊंचाई/बिंब की ऊंचाई = प्रतिबिंब दूरी/बिंब दूरी

or, प्रतिबिंब दूरी = -3×10

or, प्रतिबिंब दूरी = -30 cm

1. वायु में गमन करती प्रकाश की एक किरण  जल  में तिरछी प्रवेश करती है। क्या प्रकाश किरण अभिलंब की ओर झुकेगी अथवा अभिलंब से दूर हटेगी? बताइए क्यों?

उत्तर

प्रकाश की किरण अभिलंब की ओर झुकेगी क्योंकि प्रकाश की किरण विरल से सघन माध्यम में प्रवेश कर रही है।

2. प्रकाश वायु से  1.50 अपवर्तनांक की काँच की प्लेट में प्रवेश करता है। काँच में प्रकाश की चाल कितनी है? निर्वात में प्रकाश की चाल  3  ×  108  m/s है।

उत्तर

कांच का अपवर्तनंक = हवा में प्रकाश की चाल/कांच में प्रकाश की चाल

or, कांच में प्रकाश की चाल = 3×10⁸/1.5

or, प्रकाश में कांच की चाल = 2×10⁸ m/s

3. सारणी 10.3 से  अधिकतम प्रकाशिक घनत्व के माध्यम को ज्ञात कीजिए। न्यूनतम प्रकाशिक  घनत्व के माध्यम को भी ज्ञात कीजिए।

4. आपको किरोसिन,  तारपीन का तेल तथा जल दिए गए हैं। इनमें से  किसमें प्रकाश सबसे अधिक तीव्र गति से चलता है? सारणी  10.3 में दिए गए आँकड़ों का उपयोग कीजिए।

उत्तर

जल का अपवर्तनांक सबसे कम है अतः जल में प्रकाश सबसे तीव्र गति में चलता है।

5. हीरे  का अपवर्तनांक  2.42 है। इस कथन का क्या अभिप्राय है?

इसका अर्थ है कि वायु में प्रकाश का वेग तथा हीरे में प्रकाश के वेग का अनुपात 2.42 है।

10. Light Reflection and Refraction

QUESTIONS

1. Define the principal focus of a concave mirror.

Answer:

 A number of rays parallel to the principal axis are falling on a concave mirror. They are all meeting/intersecting at a point on the principal axis of the mirror. This point is called the principal focus of the concave mirror.

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

Answer:

Focal length = Radius of curvature ÷ 2

or, Focal length = ±20÷2

or, Focal length = ±10 cm.

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

Answer:

Concave Mirror.

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

Answer:

Convex mirrors are preferred because they always give an erect, though diminished, image. Also, they have a wider field of view as they are curved outwards. Thus, convex mirrors enable the driver to view much larger area than would be possible with a plane mirror.

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

Answer:

Focal length = Radius of curvature = Radius of curvature÷2

or, Focal length = 32÷2

or, Focal length = 16 cm.

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

Answer:

Height of image÷Height of object = -(Image distance÷Object distance)

or, 3÷1 = -(Image distance÷10)

or, Image distance = -(3x10)

or, Image distance = 30 cm.

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

Answer:

The ray of light bends towards the normal.

Water is denser with respect to air.

2. Light enters from air to glass having refractive index 1.50. What is the speed of light in the glass? The speed of light in vacuum is 3 × 108 m s–1.

Answer:

Refractive index of glass with respect to air or vacuum = Speed of light in vacuum ÷ speed of light in glass.

or, 1.5 = 3×10⁸÷speed of light in glass.

or, Speed of light in glass = 3×10⁸÷1.5.

or, Speed of light in glass = 2×10⁸ m/s Ans.

3. Find out, from Table 10.3, the medium having highest optical density. Also find the medium with lowest optical density.

Answer:

Diamond having highest optical density whereas air with lowest optical density.

4. You are given kerosene, turpentine and water. In which of these does the light travel fastest? Use the information given in Table 10.3.

Answer:

Water having lowest refractive index out of these three. So, the light travels fastest in the water.

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

Answer:

This means that the ratio of the speed of light in air and the speed of light in diamond is equal to 2.42.

1. Define 1 dioptre of power of a lens.

Answer:

The power of a lens is the reciprocal of its focal length. The unit of power of lens is dioptre when focal length is in metre. 1 dioptre means the focal length is 1 metre.

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

Answer:

From question it is clear that

v = 50 cm.

Height of the image = Height of the object

So, v÷u = 1

or, u = -50 cm.

As per image formation property

Convex lens forms real and inverted image of eual size at 2f.

So, 2f = 50 cm.

or, f = 50÷2

or, f = 25 cm.

or, f = 25÷100 m.

or, Power of lens = 1÷f

or, Power of lens = 100÷25

or, Power of lens = 4 dioptre.

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

Answer:

From question it is clear that

f = -2 m.

Power of lens = 1÷f

Power of lens = 1÷(-2)

or, Power of lens = -0.5 dioptre.

EXERCISES

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

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

Answer

Clay cannot be used to make a lens.

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?

(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

The correct option will be (d)

3. 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 of the lens and its principal focus.

Answer

The correct option will be (b)

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

(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.

Answer

The correct option will be (a)

5. 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

The correct option will be (d)

6. 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

The correct option will be (b)

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

The range of distance should be less than 15 cm.

The nature of image will be virtual.

The image is larger than object.

The ray diagram is given below

8. 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.

Answer

(a)

Concave mirror is used for headlights of car because it converges the ray of light in its focal length.

(b)

Convex mirror is used for side/rear-view mirror of a vehicle because it has large view range and erract and virtual image is formed by it.

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

This lens will produce a complete image when object is placed between optical centre and main focus of it.

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

From question it is clear that

u = -25 cm.

Length of object = 5 cm

f = 10 cm.

From lens formula we know that

1/v = 1/f + 1/u

1/v = 1/10 + 1/-25

or, 1/v = (5 - 2)÷50

or, 1/v = 3/50

or, v = 50/3 cm.

Length of image = Length of object×v÷u

or, Length of image = 5×50÷(25×3)

or, Length of image = 10/3 cm.

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

From question it is clear that

f = -15 cm

v = -10 cm

u = ?

1/u = 1/v - 1/f

or, 1/u = -1/10 + 1/15

or, 1/u = (-3+ 2)/30

or, 1/u = -1/30

u = -30 cm.

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

From question it is clear that

u = -10 cm

f = 15 cm

v = ?

From mirror formula

1/v = 1/f - 1/u

or, 1/v = 1/10 + 1/15

or, 1/v = (3 + 2)/30

or, 1/v = 5/30

or, 1/v = 1/6

v = 6 cm.

Nature of image will be virtual.

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

Answer

Plane mirror makes virtual image of equal height.

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

From question it is clear that

u = -20 cm

f = 30/2 = 15 cm

v = ?

From mirror formula

1/v = 1/f - 1/u

or, 1/v = 1/15 + 1/20

or, 1/v = 7/60

or, v = 60/7 cm.

Image will be virtual.

Size of image = Size of object × v/u

or, Size of image = 5×60÷20×7

or, Size of image = 15/7 cm.

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

Size of object = 7 cm

From mirror formula

1/v = 1/f - 1/u

or, 1/v = -1/18 + 1/27

or, 1/v = (-3+2)/54

or, 1/v = -1/54

or, v = -54 cm

Nature of image will be real.

Size of image = Size of object×image distance÷object distance

or, Size of image = 7×54÷27

or, Size of image = 14 cm.

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

Answer

Focal length = 1/Power of lens

or, f = 1/-2

or, f = -0.5 m

This is type of a concave 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

f = 1/P

or, f = 1/1.5

or, 0.66 m

The prescribed lens is converging.