Chemical Equations

Introduction:

Changes:

Changes are classified into two types, they are Physical change and Chemical change.

Physical Change:

• Physical change is the change in which there is no formation of a new substance.
• Physical changes are Temporary changes and they can be reversed.

Example:1. Melting of wax

               2. Ice melting and water vapour formation. 

               Eg: On heating, Ice -----> water ------> water vapour

Chemical Change:

• A chemical change is a process in which a new substance is formed with new properties that are totally different from the actual substance.
• It is an irreversible and permanent change.

Example: Burning of crackers, the process of digestion, ripening of fruits, etc.

Observations regarding Chemical Change:

Following are some of the observations from which it can be concluded that a chemical change or reaction has taken place:

1. Change in colour
2. Change in physical state
3. Liberation of gas
4. Formation of precipitate
5. Change in temperature

Activity - 1: Reaction of Quick Lime (Calcium Oxide) with water (Change in Temperature)

Aim: To demonstrate the reaction of quick lime (Calcium Oxide) with water.

Materials Required: Beaker, Litmus paper, quick lime, water.

Procedure:

1.    Take a beaker and add 1gm of Calicium Oxide (Quick Lime) into it.

2.    Then add 10ml of water to the same water

3.    Now touch the beaker with your finger, it observes to be hot.

4.    The hotness of the beaker is due to the reaction of Calcium Oxide(quick lime) with water,         heat energy is released. Hence it is an exothermic reaction.

        CaO_(s) + H₂O_(aq) → Ca(OH)_2(aq) + Heat

5.    Calcium Oxide dissolves in water and produces a colourless solution. 

6.    A red litmus paper turns to a blue colour when it is dipped in water. This clearly indicates that the solution is basic in nature.

Result: In this activity, the reaction is exothermic and the solution formed is basic in nature.

Activity 1 Based Questions:

Q1: What do you notice?

A:    It is noticed that the beaker is hot when we touch it and the reason behind that is the action of calcium oxide (quick lime) with water, heat is emitted.

Q2: What is the nature of the solution?

A:    As red litmus paper turns out to be blue colour when dipped in the solution, it indicates that the solution is basic in nature.

Activity - 2: Formation of Precipitate

Aim: To show the formation of precipitation in some chemical reactions.

Materials Required: Two beakers, sodium sulphate (Na₂SO₄) solution, barium chloride (BaCl₂) solution and water.

Procedure:

    1. Take a beaker and add 100ml of water into it. Then dissolve a small amount of sodium sulphate (Na₂SO₄) into it.

   2. Take another beaker and add 100ml of water into it. Then dissolve a small amount of barium chloride (BaCl₂) into it.

     3. On observing the two beakers, both the solutions present in the beakers are colourless.

     4. Now add Na₂SO₄ solution to BaCl₂ solution.

     5. Then, the formation of a white coloured precipitate (BaSO₄) is formed.

         Na₂SO_4(aq)+BaCl_2(aq) → BaSO_4(s)↓+ 2NaCl_(aq)

Result: From this activity, it is concluded that some chemical reactions are characterized by the formation of a precipitate. This reaction is a double displacement reaction. 

Activity 2 Based Questions:

Q1:    What are the colours of the above solutions?

A:       Both the solutions are colourless.

Q2:    Can you name the solutions obtained by adding Na₂SO₄ solution to BaCl₂ solution?

A:       Barium Sulphate(BaSO₄)and Sodium Chloride(NaCl).

Q3:    Do you observe any change in mixing these solutions?

A:       Yes, a white coloured precipitate of Barium Sulphate is formed.

Activity - 3: Formation Hydrogen Gas

Aim: To show the formation of hydrogen gas by the action of dilute HCl on zinc.

Materials Required: Conical flask, dilute HCl, zinc granules, matchbox.

Procedure:

    1. Take some zinc granules in a conical flask and add 5ml of dilute HCl into it.

    2.  It can be observed that the bubbles formed when zinc granules were added to dilute HCl.

    3. Now place the burning match-stick near the mouth of the conical flask. The match-stick turns off with a pop-up sound which indicates the evolution of hydrogen gas.

    4. As heat is evolved during the reaction, the conical flask is turned out to be hot.

        Zn_(s) + 2HCl_(aq) → ZnCl_2(aq) + H_2(g)↑

Result: From this activity, we can conclude that the reaction is a single displacement reaction because zinc displaces hydrogen to produce zinc chloride and hydrogen gas.

Activity 3 Based Questions:

Q1: What changes do you notice?

A:    Hydrogen gas formation.

Q2: What happens to the burning match stick?

A:   Burning match-stick puts off with a pop-up sound when hydrogen gas is liberated.

Q3: Is there any change in temperature?

A:    Yes, the temperature increases.

Chemical Equation:

Def: A chemical equation is the symbolic representation of reactants and products involved in the reaction.

• Reactants: The substances which are involved in the chemical reaction are known as Reactants.
• Products: The new substance(s) which forms after a chemical reaction, known as Products.
• If the number of reactants and products are more than one in a chemical reaction, then they are indicated by the '+' sign between them.
• In any chemical equation, reactants are written on the left-hand side and products are written on the right-hand side.

Representation / Writing a chemical equation:

• The chemical equations can be represented or written in two forms namely Word Equation and Symbolic/formula equation. 
• The symbolic equation is easier to write than the words equation as word representation is much lengthy and time is taken.
• A compound is represented by using its chemical formulae that include symbols of the elements and it also uses subscript which indicates the number of atoms present in the compound.

         Example: Calcium oxide reacts with water and produces calcium hydroxide. The chemical equation for the mentioned reaction can be represented in the following forms:

• Word Equation: calcium oxide + water -------> calcium hydroxide

• Symbolic / formula equation: CaO + H₂O → Ca(OH)₂

Think and Discuss Question:

Q1: You have brushed the wall with an aqueous suspension of Ca(OH)₂. After two days the wall turned to white colour.

    (i) What are the steps involved in the whitewashing of the wall?

    (ii) Write the balanced chemical reactions using the appropriate symbols.

A: (i) Aqueous suspension of slaked lime (Ca(OH)₂)is not white. After applying on the wall, it reacts with carbon dioxide (CO₂) gas present in the air and forms calcium carbonate (CaCO3), water (H₂O). The resultant water gets evaporated and calcium carbonate which is quite white in colour gets stuck to the wall.

    (ii) Balanced chemical reaction:

         Ca(OH)_2(aq) + CO_2(g) = CaCO_3(s) + H₂O_(l)

        

Balancing Chemical Equations:

A chemical equation needs to be balanced to follow the law of conservation of mass. According to the law of conservation of mass, when a chemical reaction occurs, the mass of the products formed should be equal to the mass of reactants consumed. This means atoms are neither be created nor destroyed during the chemical reaction.

Unbalanced/Skeleton Chemical Equation:

A chemical equation in which the number of atoms of each element on the reactant side is not equal to the number of atoms of the same element on the product side is known as an unbalanced chemical equation.

Example: Na₂SO₄ + BaCl₂ → BaSO₄ + NaCl

In the above example, the number of atoms of Na and Cl is not equal on both sides.

Balanced Chemical Equation:

A chemical equation in which the number of atoms of each element on the reactant side is equal to the number of atoms of the same element on the product side is known as a balanced chemical equation.

Example: 2Mg + O₂ → 2MgO

Explanation of Steps involved in Balancing Chemical Equation through Example:

Hydrogen reacts with Oxygen and forms Water:

Step 1: Write an unbalanced chemical equation by using the chemical formula of all substances.

H₂ + O₂ → H₂O

Step 2: Compare the atoms of each element on both sides.

There exist 2 'O' atoms on LHS but there is only one 'O' atom on RHS. To balance it, add a coefficient of 2 to the reactant (H₂) and to the product water (H₂O). Thus equation can be rewritten as:

2H₂ + O₂ → 2H₂O

Step 3: Divide the coefficients of all the substances with a suitable number to get the lowest ratio of coefficients for reactants and products. But in the above equation, there is no common factor. Hence there is no requirement for the division of coefficients.

Step 4: Verification of balanced chemical equation.

Count the number of atoms of each element on both sides and verify it.

Therefore, 2H₂ + O₂ → 2H₂O is a balanced chemical equation.

Making Chemical Equation More Informative:

The chemical equation can be more informative if it is expressed with the following characteristics:

• Physical state
• Heat changes
• Gas evolution
• Formation of precipitate

Interpreting a Balanced Chemical Equation:

    A chemical equation provides information about the involvement of the number of molecules of reactants and products in a chemical reaction. From this equation, the relative masses of reactants and products can also be obtained, since the molecular masses are expressed in Unified Masses (U).

    The chemical equation also gives information about the molar ratios of reactants and products, if they are expressed in grams. If gases are involved, we can equate their masses to their volumes and calculate the volumes or those gases liberated at a given condition of temperature and pressure using the molar mass and molar volume relationship. We can also get the number of molecules and atoms of different compounds by using molar mass and Avagadro's number.

    From the equation, we can get the below-mentioned factors:

• mass - mass relationship
• mass - volume relationship
• volume - volume relationship
• mass - volume - number of molecules relationship etc.

Example Problems:

Q1: Al_(s) + Fe₂O₃ → Al₂O₃ + Fe

       (Atomic masses of Al=27U, Fe=56U, and O=16U)

        2Al_(s) + Fe₂O₃ → Al₂O₃ + 2Fe

        (2*27)U + (2*56+3*16)U → (2*27+3*16)U + (2*56)U

                    (or)

        54U + 160U → 102U + 112U

                    (or)

        54g +160g → 102g + 112g 

Suppose that you are asked to calculate the amount of aluminium required to get 1120kg of iron by the above reaction.

Sol: According to the balanced chemical equation,

        54g of Al required to get 112g of Fe

        

        i.e., 54g → 112g

        The amount of Al required to get 1120kg of Fe to be 'x' g

        i.e., 'x'g → 1120kg

        here 1120kg = 1120000g

        ⇒ 54g → 112g

            x g → 1120000g

        ⇒ x = 54 * 1120000 / 112

            x = 540000g

            x = 540kg

            Therefore the amount of aluminium required to get 1120kg of iron is 540kg.

Q2: Calculate the volume, mass and number of molecules of hydrogen liberated when 230g of sodium reacts with excess water at STP.

        (Atomic masses of Na=23U, O=16U, and H=1U)

        2Na_(s) + 2H₂O_(l) → 2NaOH_(aq) + H_2(g)↑ is a balanced equation

        (2*23)U + 2(2*1+1*16)U → 2(23+16+1)U+ (2*1)U

        46U + 36U → 80U + 2U     

                        (or)

        46g + 36g → 80g + 2g

Sol: According to the balanced chemical equation,

Mass Calculation:

46g of sodium → 2g of hydrogen

230g of sodium → ?

= 230*2 / 46

= 10g

Therefore, hydrogen gas liberated from the reaction of 230g of sodium with excess water is 10g

Volume Calculation:

At STP, 2g of hydrogen occupies 22.4 litres

10g of hydrogen occupies ?

= 10*22.4 / 2

= 112 litres

Therefore, 10g of hydrogen occupies 112 litres at STP

The number of Molecules Calculation:

2g of hydrogen contains 6.623*10²³  number of molecules

10g of hydrogen contains?

= 10*6.623*10²³ / 2

= 3.01*10²⁴ molecules.

Therefore, 10g of hydrogen contains 3.01*10²⁴ molecules.

Q3. Calculate the volume and number of molecules of CO₂ liberated at STP. If 50g of CaCO₃ is treated with dilute HCl which contains 7.3g of dissolved HCl gas.

CaCO_3(s) + 2HCl_(aq) → CaCl_2(aq) + H₂O_(l) + CO_2(g) is a balanced equation.

According to metric-stoichio equation, 100g of CaCO₃ reacts with 73g of HCl to liberate 44g of CO₂.

Sol: In the given problem, 50g of CaCO₃ is treated with 7.3g of HCl.

But, according to the metric-stoichio equation, 100g of CaCO₃ reacts with 73g of HCl.

As 50g of CaCO₃ is half of the 100g then it should be react with 36.5g of HCl, which is half of 73g, but it react with 7.3g of HCl only.

Hence the product CO₂ obtained depends upon the amount of HCl, which is in the least amount but not on the amount of CaCO₃ which is in excess amount.

Here, the reactant HCl which is in least amount is called limiting reagent, because it limits the amount of product to be formed. 

Mass Calculation:

73g of HCl → 44g of CO₂

7.3g of HCl → ?

= 7.3*44 / 7.3

= 4.4g

Therefore, 7.3g of HCl produce 4.4g of CO₂

Volume Calculation:

44g of CO₂ occupies 22.4 litres at STP

4.4g of CO₂ occupies ?

= 4.4*22.4 / 44

= 2.24 litres.

Therefore, 4.4g CO₂ of occupies 2.24 litres.

The number of Molecules Calculation:

44g of CO₂ contains 6.623*10²³ molecules

4.4g of CO₂ contains ?

= 4.4*6.623*10²³ / 44

= 6.623*10²² mol.

Therefore, 4.4g of CO₂ occupies 6.623*10²² molecules.

      

         

        

        

            

Want to be a successful student?

Developing the Qualities of a Successful Student

Successful students know how to focus on their studies when it matters while also taking breaks when they need them. They can manage their time wisely, stick with meaningful study schedules, and make the foremost of their time within the classroom. In the process, successful students also skill to possess an honest time and love gaining knowledge the maximum amount as they enjoy getting stellar grades.

1. Make your studies a priority

Successful students know how to succeed because they’ve made their studies their top priority. Though it’s important to make time for friends, family, extracurricular activities, and even some solo downtime, you ought to never neglect the time you would like to spend studying. If you've got a crucial exam arising and don’t feel prepared, then you ought to probably skip the large party two days before it. If you’re really behind on your French, then you may need to skip that new episode of Criminal Minds for the time being. This doesn’t mean that you simply can never do the items you would like to try to do, but that you simply should recognize when studying should be at the highest of your list.

That said, you'll ’t ignore everything within the world with great care you can study. If a lover or loved one has a crisis, you can’t ditch him or her just to review, either.

 2. Be punctual

Develop the habit of judging time and learn how to arrive where you need to be on time. In fact, you ought to decide to be a touch early anywhere you go so you've got time to urge situated, focused, and prepared to find out once you get there. Whether you've got to require a test or have a study date with your Classmate, it’s important to get on time if you would like to be a successful student.

3. Work honestly

This means you ought to do your own work, avoid copying, and avoid cheating at the least cost. Cheating won’t get you anywhere, and what could seem sort of a shortcut at some point can actually get you into tons of trouble subsequent. It’s never worthwhile to cheat a test, and you’re much better off not doing well on an exam you’re not prepared for than getting caught cheating. Even if you’re not caught, cheating causes you to think it’s okay to require shortcuts when it involves life and studying, and it can cause some bad habits down the line.

Don’t fall into peer pressure, either. In some schools, cheating is taken into account as the norm, and it looks like numerous kids do it that you simply might also take part. This kind of group thinking is extremely dangerous and may keep you from reaching your full potential.

4. Don’t compare yourself to anyone else

Successful students succeed on their own terms. They don’t care what their brother, neighbor, or lab partner does in class because they know that within the end, all that matters is their own success. If you get too caught up in what people do, then you’re sure to be disappointed in yourself or to become so competitive that your mind gets poisoned. Learn to brush the others aside and to specialize in doing the simplest that you simply can do.

Don't compare yourself with anyone. With this, you are insulting yourself.

5. Work on making incremental progress

If you would like to be a successful student, then you shouldn’t aim to go from a “C” to an “A” average. Instead, you ought to work on getting to a “C+” then a “B-,” then on, so your progress is manageable then you don’t get disappointed. Successful students know that it’s hard to enhance by leaps and bounds and specialize in the small print rather than jumping ahead to the ultimate product. If you want to be a successful student, then you've got to be okay with improving little by little.

6.  Get excited about the material

Successful students aren’t just machines who work to urge “A”s regardless of what. They actually care and have an interest in the material that they study, and their passion for knowledge helps them improve. Of course, you can’t get excited about every little thing you’re learning, from photosynthesis to linear equations, but you'll attempt to find something that you simply care about in every class. This will keep you focused and will make it more fun for you to learn.

If you actually find something you’re hooked in to in school, then you ought to do some outside reading to urge even more excited about the subject. For example, if you liked reading The Sun Also Rises in school, try reading A moveable fest or a number of Hemingway’s other works on your own, too.

7. Pay attention.

If you would like to be a successful student, then listening in school is completely crucial to your success. While you don’t need to love every single subject that comes your way, you ought to be motivated enough to concentrate on your teachers, to avoid texting your friends, and to be focused enough to really hear what your teacher is telling you, and to be able to pick up on the most important aspects of each lesson.

In order to pay attention, it’s important to keep your eye on the teacher.

If you’re confused about something, you'll quickly invite clarification. If the lesson goes on and you are feeling yourself getting more and more lost, it’ll be hard to concentrate.

Reflection of Light at Curved Surfaces Part-5 (Problem Solving)

Problems on Concave Mirror:

Example Problem: 

• An object 4cm in size, is placed at 25cm in front of a concave mirror of a focal length of 15cm. At what distance from the mirror would a screen be placed in order to obtain a sharp image? Find the nature and size of the image. 

Ans:  Given Data (according to sign convention):

           Object height (ho) = +4cm

          Object distance (u) = - 25cm

          Focal length (f) = - 15cm

          Image distance (v) = ?

          Image height (hi) = ?

          Substitute the above values in the mirror formula,

          Mirror Formula:  1/f = 1/u + 1/v

                                    

                                      1/(-15) = 1/v + 1/(-25)

                                      1/(-15) = 1/v - 1/25

                                          1/v = -1/15 + 1/25

                            

                                       1/v =  -5+3 / 75

                                       1/v = -2/75

                                          v = -75/2

                                           

                                          v = -37.5cm

            Hence screen should be placed at a distance of 37.5cm in front of the mirror. The image is real.

            Magnification m = hi / ho = -v / u

by substituting the above values, we get  hi / 4 = -(-37.5) / (-25)

                                                                       hi = -(37.5 × 4) / 25

                                                                               hi = -150 / 25

                                                                       hi = -6cm

So the image is inverted and enlarged.

WORKOUT PROBLEMS:

1.    Find the distance of the image when an object is placed on the principal axis at a distance of 10cm in front of a concave mirror whose radius of curvature is 8cm.

Sol: Given Data (according to sign convention):

        Image distance (v) =?

        Object distance (u) = -10cm

        Radius of Curvature (R) = -8cm     (since concave mirror)

        we know that R = 2f => f = R/2 => f = -8/2 => f = -4cm

        Applying Mirror Formula: 1/f = 1/v + 1/u

                                                 1/(-4) = 1/v + 1/(-10)

                                                 -1/4 = 1/v - 1/10

                                                  1/v = -1/4 +1/10

                                                  1/v = -5+2/20 => 1/v = -3/20

                                                     v = -20/3 

                                    

                                                     v = -6.6cm

Hence the image is at a distance of 6.6cm in front of the mirror. Therefore the image is real.

                                            

2.    An object is placed 20cm from a concave mirror with a radius of curvature of 60cm. Find the position of the image and its magnification.

Sol: Given Data (according to sign convention):

        Object distance (u) = -20cm

        Radius of Curvature (R) = -60cm     (since concave mirror)

        Image distance (v) =?

        we know that R = 2f => f = R/2 => f = -60/2 => f = -30cm

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

                        

                                            1/(-30) = 1/v + 1/(-20)

                                              -1/30 = 1/v - 1/20

                                                  1/v = -1/30 + 1/20

                                                  1/v = -2+3/60 => 1/v = 1/60

                                                     v = 60cm

Position: Hence the image is placed at a distance of 15cm behind the mirror. Therefore the image is virtual.

Magnification (m) = -v/u 

                             = -60/(-20) = 3 => m = 3

Characteristics of image: The image is virtual, erect, and enlarged.

3.    The focal length of a concave mirror is 30cm. Find the position of the object in front of the mirror so that the image is 3 times the size of the object.

Sol: Given Data (according to sign convention):

       Focal length (f) = -30cm    (since concave mirror)

       Magnification (m) = 3

Case: 1 - If the image is real and inverted:

        m = -v/u = -3

         v = 3u

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

                                     1/(-30) = 1/3u + 1/u

                                        -1/30 = 4/3u

                                           -3u = 120

                                              u = -40cm

                              Therefore v = 3(-40) = -120cm 

Hence the object is placed at a distance of 40cm in front of the concave mirror.

Case: 2 - If the image is virtual and erect:

            m = -v/u = 3

             v = -3u

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

                                        1/(-30) = 1/(-3u) + 1/u

                                           -1/30 = -1/3u +1/u

                                           -1/30 = -1+3/3u 

                                           -1/30 = 2/3u => -3u = 60

                                                 u = -20cm

                                 Therefore v = -3(-20) = +60cm

4.    An object, 4cm in size, is placed 25cm in front of a concave mirror of focal length 15cm. At what distance from the mirror, should a screen be placed in order to obtain a sharp image? Find the nature and size of the image.

Sol: Given Data (according to sign convention):

        Height of the object (ho) = 4cm

        Object distance (u) = -25cm

        Focal length (f) = -15cm     (since concave mirror)

        Image distance (v) =?

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

                                    1/(-15) = 1/v + 1/(-25)

                                       -1/15 = 1/v - 1/25

                                           1/v = -1/15 + 1/25

                                            1/v = -5+3/75 => 1/v = -2/75 => v = -75/2

                                               v = -37.5cm

Position: Hence the screen should be placed at 37.5cm from the mirror. The image is real.

 

Magnification (m) = hi / ho = -v/u

                            hi = -v × ho / u

                            hi = -(-37.5) × 4 / (-25)

                            hi = +150 / (-25)

                   Size: hi = - 6cm

Characteristics of image: The image is real, inverted, and enlarged.

5.    Focal length of a concave mirror is 20cm. Where should an object be placed in front of the concave mirror so as to obtain an image which is real, inverted as of the same size as that of the object?

Sol: Given Data (according to sign convention):

        Focal length (f) = -20cm

        Object distance (u) =?

To obtain an image that is real, inverted, and of the same size as that of the object, the object should be placed at the centre of curvature (C).

Also, we observed that in a concave mirror, the centre of curvature is at a distance twice the focal length,

Therefore 2f = 2 × (-20)

                 2f = -40cm     (since concave mirror)

Hence object should be placed at a distance of 40cm from the concave mirror.

6.    An object of height 5cm is placed at a 30cm distance on the principal axis in front of a concave mirror of focal length 20cm. Find the image distance and size of the image.

Ans:  

            

            Given Data (according to sign convention):

            Object height (ho) = 5cm

            Object distance (u) = -30cm

            Focal length (f) = -20cm     (since concave mirror)

            Image distance (v) =?

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

                                      1/(-20) = 1/v + 1/(-30)

                                         -1/20 = 1/v -1/30

                                            1/v = -1/20 + 1/30

            

                                            1/v = -3+2 / 60 => 1/v = -1/60

                                               v = -60cm

Hence the image is placed at a distance of 60cm from the mirror. Therefore the image is real.

Magnification (m) = hi/ho = -v/u

                                  hi/5 = -(-60)/(-30)

        

                                  hi/5 = -2

                                     hi = -10cm

Characteristics of image: The image is real, inverted and enlarged.

7.    An object of 6cm is placed at a distance of 30cm in front of a concave mirror of focal length of 10cm. At what distance from the mirror, will the image be formed? What are the characteristics of the image?

Sol: Given Data (according to sign convention):

        Object height (ho) = 6cm 

        Object distance (u) = -30cm 

        Focal length (f) = -10cm  (since concave mirror)

        Image distance (v) =?

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

                                    1/(-10) = 1/v + 1/(-30)

                                       -1/10 = 1/v -1/30

 

                                          1/v = -1/10 + 1/30

                                          1/v = -3+1 / 30 => 1/v = -2/30 => 1/v = -1/15

                                             v = -15cm

Hence the image is formed at a distance of 15cm from the concave mirror. Therefore the image is real.

Magnification (m) = hi/ho = -v/u

                                hi/6 = -(-15) / (-30)

                                hi/6 = -1/2

                                

                          Size: hi = -3cm

Characteristics of image: The image is real, inverted and diminished.

8.    A spherical mirror produces an image of magnification -1 on a screen placed at a distance of 50cm from the mirror.

    (a)    Write the type of mirror.

    (b)    FInd the distance of the image from the object.

    (c)    What is the focal length of the mirror.

    (d)    Draw the ray diagram to show the image formation

Sol: (a) Type of mirror: It is a concave mirror as its magnification is negative and the image formed is real.

        (b) Distance of the object from the mirror: 

                Magnification, m = -v/u =-1

                Object distance, u = -50cm

                Substitute 'u' value in magnification formula, we get

                -v/(-50) = -1

                          v = -50cm

Hence the distance of the object from the mirror is 50cm

        (c) Focal Length of the mirror:

                By using mirror formula, 1/f = 1/v + 1/u

                substituting 'u' and 'v' values in the above formula

                                                        1/f = 1/(-50) + 1/(-50)

                                                        1/f = -1/50 - 1/50

                    

                                                        1/f = -2/50 => 1/f = -1/25

                                                           f = -25cm

        (d) Ray Diagram:

9.    Mahira wants to project the image of a candle flame on a screen 60cm in front of a mirror by keeping the flame at a distance of 15cm from its pole.

        (a) Write the type of mirror she should use.

        (b) Find the linear magnification of the image produced.

        (c) What is the distance between the object and its image.

        (d) Find its focal length and radius of curvature.

Ans: (a) Type of mirror: It is a concave mirror as it forms a real image that presents on the same side of the mirror.

        (b) Magnification of the image produced:

            Given object distance, u = -15cm and Image distance, v = -60cm

            Magnification, m = -v/u

                                    m = -(-60) / (-15)

                                    m = -4cm (-ve sign indicates that image formed is real and inverted)

        (c) Distance between object and its image: 

                 v-u = 60-15 (as we taking the difference between distances, so no need to use sign convention)

                    v-u = 45cm

        (d) Finding Focal Length and Radius of Curvature:

                Given object distance, u = -15cm and Image distance, v = -60cm

            By using the mirror formula, 1/f = 1/v + 1/u

                                                          1/f = 1/(-60) + 1/(-15)

                                                           1/f = -1/60 - 1/15

                    

                                                           1/f = -1-4 / 60

                                                           1/f = -5/60

                                                            1/f = -1/12

                                                               f = -12cm

            we know that the Radius of curvature, R = 2f 

                                                                    R = 2(-12)

                                                                    R = -24cm

10.    Find the distance at which an object to be placed for getting a real, inverted and enlarged image at 45cm using a concave mirror of focal length 20cm.

Sol:    Given Data (according to sign convention):

            Image distance, v = -45cm

            focal length, f = -20cm    (since concave mirror)

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

                                        1/(-20) = 1/(-45) + 1/u

                                           -1/20 = -1/45 + 1/u

                                              1/u = -9+4 / 180

                                               1/u = -5/180 => u = -180/5

                                                   u = -36cm

Problems on Convex Mirror:

WORKOUT PROBLEMS:

1.    A 4.5cm candle is placed 12cm away from a convex mirror of focal length 15cm. Give the location of the image and magnification. Describe what happens to the image as the needle is moved farther from the mirror.

Sol:    Given Data (according to sign convention):

            Object height (ho) = 4.5cm

            Object distance (u) = -12cm

            Focal length (f) = +15cm   (since convex mirror)

            Image distance (v) =?

            Magnification (m) =?

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

                                         1/(15) = 1/v + 1/(-12)

                                           1/15 = 1/v - 1/12

                                             1/v = 1/15 + 1/12

                                             1/v = 4+5 / 60

                                             1/v = 9/60

                                                v = 6.6cm

Position: The image is placed at a distance of 6.6cm behind the mirror.

        Magnification (m) = hi/ho = -v/u

                                m = -v/u = -6.6/(-12) = 0.5

                                m = 0.6

If we move the object farther from the mirror, the image will be diminished

        Also m = hi/ho

              0.6 = hi/4.5

                hi = 0.6 × 4.5

                hi = 2.7cm

Characteristics of the image: The image is virtual, erect and diminished.

2.    A convex mirror with a radius of curvature of 3m is used as a rearview mirror for a vehicle. If a bus is located at 5m from this mirror, find the position, nature and size of the image.

Sol:    Given Data (according to sign convention):

            Radius of curvature (R) = +3m    (since convex mirror)

            Object distance (u) = -5m

            Image distance (v) =?

            Magnification (m) =?

            we know that, R = 2f 

                                    f = R/2 = 3/2

                                    f = +1.5m    (since convex mirror)

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

                                                1/1.5 = 1/v + 1/(-5)

                                                1/1.5 = 1/v - 1/5

                                                   1/v = 1/1.5 +1/5

                                                   1/v = 10+3 / 15

                                                   1/v = 13/15

                                                      v = 15/13

                                                      v = +1.15m 

Position: The image is formed at a distance of 1.15m inside the mirror. As v is positive, it indicates that the image is virtual.

Size:    Magnification (m) = hi/ho = -v/u

                                    m = -v/u

                                    m = -1.15/(-5)

                                    m = 0.23

As the magnification of the image is positive and less than zero, the image is erect and smaller than the object (diminished).

Characteristics of the image: The image is virtual, erect and diminished.

3.    A 2.5cm candle is placed 12cm away from a convex mirror of focal length 30cm. Find the position of the image and its magnification.

Sol:    Given Data (according to sign convention):

            Object height (ho) = 2.5cm

            Object distance (u) = -12cm

            Focal length (f) = +30cm  (since convex mirror)

            Image distance (v) =?

            Magnification (m) =?

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

                                         1/(30) = 1/v + 1/(-12)

                                            1/30 = 1/v - 1/12

                                              1/v = 1/30 + 1/12

                                              1/v = 2+5 / 60

    

                                              1/v = 7/60

                                                 v = 60/7

                                                 v = +8.6cm

Position: The image is formed at a distance of 8.6cm inside the mirror. As v is positive, it indicates that the image is virtual.

            Magnification (m) = hi/ho = -v/u

                                    m = -v/u

                                    m = -8.6/(-12)

                                    m = 0.7

As the magnification of the image is positive and less than zero, the image is erect and smaller than the object (diminished).

Characteristics of the image: The image is virtual, erect and diminished.