Showing posts with label Class 10 Science (NCERT Solution). Show all posts
Showing posts with label Class 10 Science (NCERT Solution). Show all posts

CLASS 10 Science CHAPTER 9 Heredity And Evolution (NCERT Solution)

CLASS 10 Science                                                                                                    CHAPTER 9                                                                                                             
Heredity  And Evolution (NCERT Solution)

Question 1: If a trait A exists in 10% of a population of an asexually reproducing species and a trait B exists in 60% of the same population, which trait is likely to have arisen earlier?
Answer: In asexual reproduction, the reproducing cells produce a copy of their DNA through
some chemical reactions. However, this copying of DNA is not accurate and therefore, the newly
formed DNA has some variations.


It can be easily observed in the above figure that in asexual reproduction, very few variations are
allowed. Therefore, if a trait is present in only 10% of the population, it is more likely that the trait
has arisen recently. Hence, it can be concluded that trait B that exists in 60% of the same population has arisen earlier than trait A.

Question 2: How does the creation of variations in a species promote survival?
Answer: Sometimes for a species, the environmental conditions change so drastically that their
survival becomes difficult. For example, if the temperature of water increases suddenly, most of
the bacteria living in that water would die. Only few variants resistant to heat would be able to
survive. If these variants were not there, then the entire species of bacteria would have been
destroyed. Thus, these variants help in the survival of the species. However, not all variations are useful. Therefore, these are not necessarily beneficial for the individual organisms.

Question 3: How do Mendel’s experiments show that traits may be dominant or recessive?
Answer: Mendel selected true breeding tall (TT) and dwarf (tt) pea plants. Then, he crossed these
two plants. The seeds formed after fertilization were grown and these plants that were formed
represent the first filial or F1 generation. All the F1 plants obtained were tall.


Then, Mendel self-pollinated the F1 plants and observed that all plants obtained in the
F2 generation were not tall. Instead, one-fourth of the F2 plants were short.


From this experiment, Mendel concluded that the F1 tall plants were not true breeding. They were
carrying traits of both short height and tall height. They appeared tall only because the tall trait is
dominant over the dwarf trait.

Question 4: How do Mendel’s experiments show that traits are inherited independently?
Answer: Mendel crossed pea plants having round green seeds (RRyy) with pea plants having
wrinkled yellow seeds (rrYY).


.
Since the F1 plants are formed after crossing pea plants having green round seeds and pea plants
having yellow wrinkled seeds, F1 generation will have both these characters in them. However, as
we know that yellow seed colour and round seeds are dominant characters, therefore, the F1 plants
will have yellow round seeds.
Then this F1 progeny was self-pollinated and the F2 progeny was found to have yellow round
seeds, green round seeds, yellow wrinkled seeds, and green wrinkled seeds in the ratio of 9:3:3:1.

In the above cross, more than two factors are involved, and these are independently inherited.

Question 4: A man with blood group A marries a woman with blood group O and their daughter
has blood group O. Is this information enough to tell you which of the traits − blood group A or O
− is dominant? Why or why not?
Answer: No. This information is not sufficient to determine which of the traits − blood group A
or O − is dominant. This is because we do not know about the blood group of all the progeny.
Blood group A can be genotypically AA or AO. Hence, the information is incomplete to draw any
such conclusion.

Question 5: How is the sex of the child determined in human beings?
Answer: In human beings, the females have two X chromosomes and the males have one X and
one Y chromosome. Therefore, the females are XX and the males are XY.
The gametes, as we know, receive half of the chromosomes. The male gametes have 22 autosomes
and either X or Y sex chromosome.
Type of male gametes: 22+X OR 22+ Y.
However, since the females have XX sex chromosomes, their gametes can only have X sex
chromosome.
Type of female gamete: 22+ X


Sex determination in humans
Thus, the mother provides only X chromosomes. The sex of the baby is determined by the type of
male gamete (X or Y) that fuses with the X chromosome of the female.

Question 6: What are the different ways in which individuals with a particular trait may increase
in a population?
Answer: Individuals with a particular trait may increase in a population as a result of the following:
(i) Natural selection: When that trait offers some survival advantage.
(ii) Genetic drift: When some genes governing that trait become common in a population.
(iii) When that trait gets acquired during the individual’s lifetime.

Question 7: Why are traits acquired during the life-time of an individual not inherited?
Answer: This happens because an acquired trait involves change in non-reproductive tissues
(somatic cells) which cannot be passed on to germ cells or the progeny. Therefore, these traits
cannot be inherited.

Question 8: Why are the small numbers of surviving tigers a cause of worry from the point of view of genetics?
Answer: Small numbers of tigers means that fewer variations in terms of genes are available. This means that when these tigers reproduce, there are less chances of producing progeny with some useful
variations. Hence, it is a cause of worry from the point of view of genetics.

Question 9: What factors could lead to the rise of a new species?
Answer: Natural selection, genetic drift and acquisition of traits during the life time of an
individual can give rise to new species. ill geographical isolation be a major factor in the
speciation of a self-pollinating plant species? Why or why not? Geographical isolation can prevent
the transfer of pollens among different plants. However, since the plants are self-pollinating,
which means that the pollens are transferred from the anther of one flower to the stigma of the
same flower or of another flower of the same plant, geographical isolation cannot prevent
speciation in this case.

Question 10: Will geographical isolation be a major factor in the speciation of an organism that
reproduces asexually? Why or why not?
Answer: Geographical isolation prevents gene flow between populations of a species whereas asexual reproduction generally involves only one individual. In an asexually reproducing organism, variations can occur only when the copying of DNA is not accurate. Therefore, geographical isolation cannot prevent the formation of new species in an asexually reproducing organism.

Question 11: Give an example of characteristics being used to determine how close two species are in evolutionary terms.
Answer: The presence of feathers in dinosaurs and birds indicates that they are evolutionarily related. Dinosaurs had feathers not for flying but instead these feathers provided insulation to these warm-blooded animals. However, the feathers in birds are used for flight. This proves that reptiles and birds are closely related and that the evolution of wings started in reptiles.

Question 12: Can the wing of a butterfly and the wing of a bat be considered homologous organs?
Why or why not?
Answer: The wing of a butterfly and the wing of a bat are similar in function. They help the butterfly and the bat in flying. Since they perform similar function, they are analogous organs and not homologous.

Question 13: What are fossils? What do they tell us about the process of evolution?
Answer: Fossils are the remains of organisms that once existed on earth. They represent the
ancestors of plants and animals that are alive today. They provide evidences of evolution by
revealing the characteristics of the past organism and the changes that have occurred in these
organisms to give rise to the present organisms.

Question 14: Why are human beings who look so different from each other in terms of size, colour and looks said to belong to the same species?
Answer: A species is a group of organisms that are capable of interbreeding to produce a fertile
offspring. Skin colour, looks, and size are all variety of features present in human beings. These
features are generally environmentally controlled. Various human races are formed based on these
features. However, there is no biological basis to this concept of races. Therefore, all human beings are a single species as humans of different colour, size, and looks are capable of reproduction and can produce a fertile offspring.

Question 15: In evolutionary terms, can we say which among bacteria, spiders, fish and
chimpanzees have a ‘better’ body design? Why or why not?
Answer: Evolution cannot always be equated with progress or better body designs. Evolution simply creates more complex body designs. However, this does not mean that the simple body designs are inefficient. In fact, bacteria having a simple body design are still the most cosmopolitan organisms found on earth. They can survive hot springs, deep sea, and even freezing environment. Therefore, bacteria, spiders, fish, and chimpanzees are all different branches of evolution.

Question 16: A Mendelian experiment consisted of breeding tall pea plants bearing violet flowers
with short pea plants bearing white flowers. The progeny all bore violet flowers, but almost half of them were short. This suggests that the genetic make-up of the tall parent can be depicted as
(a) TTWW
(b) TTww
(c) TtWW
(d) TtWw
Answer: (c) The genetic make-up of the tall parent can be depicted as TtWW Since all the progeny bore violet flowers, it means that the tall plant having violet flowers has WW genotype for violet flower colour. Since the progeny is both tall and short, the parent plant was not a pure tall plant. Its genotype
must be Tt. Therefore, the cross involved in the given question is
TtWw × ttww
         ↓
TtWw − ttww
Therefore, half the progeny is tall, but all of them have violet flowers.
Question 17: An example of homologous organs is
(a) our arm and a dog’s fore-leg.
(b) our teeth and an elephant’s tusks.
(c) potato and runners of grass.
(d) all of the above.
Answer: (b)An example of homologous organs is our teeth and an elephant’s tusks.

Question 18: In evolutionary terms, we have more in common with
(a) a Chinese school-boy.
(b) a chimpanzee.
(c) a spider.
(d) a bacterium.
Answer: (a) In evolutionary terms, we have more in common with a Chinese school boy.

Question 19: A study found that children with light- coloured eyes are likely to have parents with
light- coloured eyes. On this basis, can we say anything about whether the light eye colour trait is
dominant or recessive? Why or why not?
Answer: Let us assume that children with light- coloured eyes can either have LL or Ll or ll genotype. If the children have LL genotype, then their parents will also be of LL genotype.
LL × LL
     ↓
LL
If the children with light-coloured eyes have ll genotype, then their parents will also have ll
genotype.
ll×ll
 ↓
ll
Therefore, it cannot be concluded whether light eye colour is dominant or recessive.

Question 20: How are the areas of study − evolution and classification − interlinked?
Answer: Classification involves grouping of organism into a formal system based on similarities
in internal and external structure or evolutionary history.
Two species are more closely related if they have more characteristics in common. And if two species are more closely related, then it means they have a more recent ancestor.
For example, in a family, a brother and sister are closely related and they have a recent common
ancestor i.e., their parents. A brother and his cousin are also related but less than the sister and her
brother. This is because the brother and his cousin have a common ancestor i.e., their grandparents
in the second generation whereas the parents were from the first generation.
With subsequent generations, the variations make organisms more different than their ancestors.

Question 21: Explain the terms analogous and homologous organs with examples.
Answer: Homologous organs are similar in origin (or are embryologically similar) but perform
different functions. For example, the forelimbs of humans and the wings of birds look different
externally but their skeletal structure is similar. It means that their origin is similar (as wings in
birds are modifications of forearm) but functions are different - the wings help in flight whereas
human forearm helps in various activities.

Analogous organs, on the other hand, have different origin but perform similar functions. For example, the wings of a bird and a bat are similar in function but this similarity does not mean that these animals are more closely related. If we carefully look at these structures, then we will find that the wings of a bat are just the folds of skin that are stretched between its fingers whereas the wings of birds are present all along the arm. Therefore, these organs are analogous organs.

Question 22: Outline a project which aims to find the dominant coat colour in dogs.
Answer: Dogs have a variety of genes that govern coat colour. There are at least eleven identified
gene series (A, B, C, D, E, F, G, M, P, S, T) that influence coat colour in dog.
A dog inherits one gene from each of its parents. The dominant gene gets expressed in the
phenotype. For example, in the B series, a dog can be genetically black or brown.
Let us assume that one parent is homozygous black (BB), while the other parent is homozygous
brown (bb)
bb BB
B B
bBbBb
bBbBb
In this case, all the offsprings will be heterozygous (Bb). Since black (B) is dominant, all the offsprings will be black. However, they will have both B and b alleles.
If such heterozygous pups are crossed, they will produce 25% homozygous black (BB), 50 %
heterozygous black (Bb), and 25% homozygous brown (bb) offsprings.
B b
BBBBb
bBbBb

Question 23: Explain the importance of fossils in deciding evolutionary relationships.
Answer: Fossils are the remains of the organism that once existed on earth. They represent the
ancestors of the plants and animals that are alive today. They provide evidences of evolution by
revealing the characteristics of the past organisms and the changes that have occurred in these
organisms to give rise to the present organisms. Let us explain the importance of fossils in
deciding evolutionary history with the help of the following example.
Around 100 million years ago, some invertebrates died and were buried in the soil in that area.
More sediment accumulated on top of it turning it into sedimentary rock.
At the same place, millions of years later, some dinosaurs died and their bodies were buried on top
of the sedimentary rock. The mud containing dinosaurs also turned into a rock.
Then, millions of years later, some horse-like creatures died in that area and got fossilized in rocks
above the dinosaur fossils.
Some time later, due to soil erosion or floods in that area, the rocks containing horse-like fossils
are exposed.
If that area is excavated deeper, then the dinosaur and invertebrates fossils can also be found.
Thus, by digging that area, scientists can easily predict that horse-like animals evolved later than
the dinosaurs and the invertebrates.
Thus, the above example suggests that the fossils found closer to the surface of the earth are more
recent ones than the fossils present in deeper layers.www.ncrtsolutions.in


Question 24: What evidence do we have for the origin of life from inanimate matter?
Answer: A British scientist, J.B.S. Haldane, suggested that life originated from simple inorganic
molecules. He believed that when the earth was formed, it was a hot gaseous mass containing
elements such as nitrogen, oxygen, carbon, hydrogen, etc. These elements combined to form molecules like water (H2O), carbon dioxide (CO2), methane (CH4), ammonia (NH3), etc.
After the formation of water, slowly the earth surface cooled and the inorganic molecules interacted with one another in water to form simple organic molecules such as sugars, fatty acids, amino acids, etc. The energy for these reactions was provided by solar radiations, lightning, volcanic eruptions, etc.
This was proved by the experiment of Stanley L. Miller and Harold C. Urey in 1953. They took a mixture of water (H2O), methane (CH4), ammonia (NH3), and hydrogen gas (H2) in a chamber and sparks were passed through this mixture using two electrodes. After one week, 15 % of the carbon from methane was converted into amino acids, sugars, etc. These organic molecules are polymerized and assembled to form protein molecules that gave rise to life on earth.
Miller and Urey experiment


Question 25: Explain how sexual reproduction gives rise to more viable variations than asexual
reproduction. How does this affect the evolution of those organisms that reproduce sexually?
Answer: In sexual reproduction, two individuals having different variations combine their DNA
to give rise to a new individual. Therefore, sexual reproduction allows more variations, whereas in
asexual reproduction, chance variations can only occur when the copying of DNA is not accurate.
Additionally, asexual reproduction allows very less variations because if there are more variations,
then the resultant DNA will not be able to survive inside the inherited cellular apparatus.
However, in sexual reproduction, more variations are allowed and the resultant DNA is also able
to survive, thus making the variations viable.
Variation and Evolution: Variants help the species to survive in all the conditions. Environmental
conditions such as heat, light, pests, and food availability can change suddenly at only one place.
At that time, only those variants resistant to these conditions would be able to survive. This will
slowly lead to the evolution of a better adapted species. Thus, variation helps in the evolution of
sexually reproducing organisms.

Question 26: How is the equal genetic contribution of male and female parents ensured in the
progeny?
Answer: In human beings, every somatic cell of the body contains 23 pairs of chromosomes. Out
of these 23 pairs, the first 22 pairs are known as autosomes and the remaining one pair is known
as sex chromosomes represented as X and Y.
Females have two X chromosomes and males have one X and one Y chromosome. The gamete receives half of the chromosomes. Therefore, the male gametes have 22 autosomes and either X or Y chromosome.
The female gamete, on the other hand, has 22 autosomes and X chromosome.
During reproduction, the male and female gametes fuse and thus the progeny receives 22 autosomes and one X or Y chromosome from male parent and 22 autosomes and one X chromosome from the female parent.


Question 27: Only variations that confer an advantage to an individual organism will survive in a
population. Do you agree with this statement? Why or why not?
Answer: In species, variations that offer survival advantages are naturally selected. Individuals
adjust to their environments with the help of these selected variations and consequently these
variations are passed on to their progeny. Evolution of organisms occurs as a result of this natural
selection. However, there can be some other variations, which do not offer any survival advantage and arise only accidentally. Such variations in small populations can change the frequency of some genes
even if they are not important for survival. This accidental change in the frequency of genes in small populations is referred to as genetic drift. Thus, genetic drift provides diversity (variations) without any survival advantage.

( From www.ncrtsolutions.in )

CLASS 10 Science CHAPTER 10 Light- Reflection And Refraction (NCERT Solution)

CLASS 10 Science 
CHAPTER 10
Light- Reflection And Refraction (NCERT Solution)

Question 1: Define the principal focus of a concave mirror.
Answer: Light rays that are parallel to the principal axis of a concave mirror converge at a
specific point on its principal axis after reflecting from the mirror. This point is known as the
principal focus of the concave mirror.

Question 2: The radius of curvature of a spherical mirror is 20 cm. What is its focal length?
Answer: Radius of curvature, R = 20 cm
Radius of curvature of a spherical mirror = 2 × Focal length ( f )
R = 2f
f = R / 2     = 20 /2  = 10
Hence, the focal length of the given spherical mirror is 10 cm.

Question 3: Name the mirror that can give an erect and enlarged image of an object.
Answer: When an object is placed between the pole and the principal focus of a concave mirror,
the image formed is virtual, erect, and enlarged.

Question 4: Why do we prefer a convex mirror as a rear-view mirror in vehicles?
Answer: Convex mirrors give a virtual, erect, and diminished image of the objects placed in front
of them. They are preferred as a rear-view mirror in vehicles because they give a wider field of
view, which allows the driver to see most of the traffic behind him.

Question 5: Find the focal length of a convex mirror whose radius of curvature is 32 cm.
Answer: Radius of curvature, R = 32 cm
Radius of curvature = 2 × Focal length ( f )
f = R/ 2   = 32/2 = 16 cm
Hence, the focal length of the given convex mirror is 16 cm.

Question 6: A concave mirror produces three times magnified (enlarged) real image of object
placed at 10 cm in front of it. Where is the image located?
Answer: Magnification produced by a spherical mirror is given by the relation,
Let the height of the object, ho = h
Then, height of the image, hi = −3h (Image formed is real)
Object distance, u = −10 cm
v = 3 × (−10) = −30 cm
Here, the negative sign indicates that an inverted image is formed at a distance of 30 cm in front of the given concave mirror.

Question 7: 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 light ray bends towards the normal.
When a ray of light travels from an optically rarer medium to an optically denser medium, it gets
bent towards the normal. Since water is optically denser than air, a ray of light travelling from air
into the water will bend towards the normal.

Question 8: 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 a medium nm is given by,
Speed of light in vacuum, c = 3 × 108 m s−1
Refractive index of glass, ng = 1.50
Speed of light in the glass,

Question 9: Find out, from Table, the medium having highest optical density. Also find the
medium with lowest optical density.
Material medium         Refractive index   Material medium        Refractive medium index
Air                                    1.0003                  Canada Balsam            1.53
Ice                                    1.31                               -                             -
Water                               1.33                       Rock salt                      1.54
Alcohol                           1.36                                -                              -
Kerosene                         1.44                     Carbon disulphide          1.63
Fused                                                              Dense                  
quartz                             1.46                       flint glass                       1.65                           
Turpentine oil                1.47                        Ruby                              1.71
Benzene                         1.50                        Crown  glass                   1.52        
Sapphire                        1.77                         Diamond                       2.42
Answer: Highest optical density = Diamond
Lowest optical density = Air
Optical density of a medium is directly related with the refractive index of that medium. A medium which has the highest refractive index will have the highest optical density and vice- versa.
It can be observed from table 10.3 that diamond and air respectively have the highest and lowest  refractive index. Therefore, diamond has the highest optical density and air has the lowest optical density.

Question 10: You are given kerosene, turpentine and water. In which of these does the light travel
fastest? Use the information given in Table.
Material medium         Refractive index                             Material medium         Refractive index
Air                               1.0003                                             Canada Balsam           1.53
Ice                                1.3                                                              -                           -
Water                            1.33                                                Rock salt                      1.54
Alcohol                        1.36                                                            -                          -
Kerosene                      1.44                                              Carbon disulphide           1.63
Fused quartz                 1.46                                                 Dense
flint glass                      1.65                                                 Turpentine oil               1.47
Ruby                             1.71                                               Benzene
Crown                           1.50                                                  Sapphire                        1.77
glass                                        1.52                                                            Diamond 2.42
Answer: Speed of light in a medium is given by the relation for refractive index (nm). The relation is given as 


It can be inferred from the relation that light will travel the slowest in the material which has the highest refractive index and travel the fastest in the material which has the lowest refractive index .It can be observed from table 10.3 that the refractive indices of kerosene, turpentine, and water are
1.44, 1.47, and 1.33 respectively. Therefore, light travels the fastest in water.

Question 11: The refractive index of diamond is 2.42. What is the meaning of this statement?
Answer: Refractive index of a medium nm is related to the speed of light in that medium v by the
relation: 


.
Where, c is the speed of light in vacuum/air
The refractive index of diamond is 2.42. This suggests that the speed of light in diamond will
reduce by a factor 2.42 compared to its speed in air.

Question 12: Define 1 dioptre of power of a lens.
Answer: Power of lens is defined as the reciprocal of its focal length. If P is the power of a lens of
focal length F in metres, then
    P = 1/f ( in meters )
The S.I. unit of power of a lens is Dioptre. It is denoted by D.
1 dioptre is defined as the power of a lens of focal length 1 metre.
1 D = 1 m−1

Question 13: 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: When an object is placed at the centre of curvature, 2F1, of a convex lens, its image is
formed at the centre of curvature, 2F2, on the other side of the lens. The image formed is inverted
and of the same size as the object, as shown in the given figure.


It is given that the image of the needle is formed at a distance of 50 cm from the convex lens.
Hence, the needle is placed in front of the lens at a distance of 50 cm.
Object distance, u = −50 cm
Image distance, v = 50 cm
Focal length = f
According to the lens formula,
Hence, the power of the given lens is +4 D.

Question 14: Find the power of a concave lens of focal length 2 m.
Answer: Focal length of concave lens, f = 2 m
Power of a lens, P = 1/ f (in meters)  = 1/ (-50)  =  -0.5 D
Here, negative sign arises due to the divergent nature of concave lens.
Hence, the power of the given concave lens is −0.5 D.

Question 15: Which one of the following materials cannot be used to make a lens?
(a) Water
(b) Glass
(c) Plastic
(d) Clay
Answer: (d) A lens allows light to pass through it. Since clay does not show such property, it cannot be used to make a lens.

Question 16: 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: 

Question 17: Where should an object be placed in front of a convex lens to get a real image of the
size of the object?www.ncrtsolutions.in
(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: (b) When an object is placed at the centre of curvature in front of a convex lens, its image is formed at the centre of curvature on the other side of the lens. The image formed is real, inverted, and of the same size as the object.

Question 18: 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: (a) By convention, the focal length of a concave mirror and a concave lens are taken as
negative. Hence, both the spherical mirror and the thin spherical lens are concave in nature.

Question 19: No matter how far you stand from a mirror, your image appears erect. The mirror is
likely to be
(a) plane
(b) concave
(c) convex
(d) either plane or convex
Answer: (d) A convex mirror always gives a virtual and erect image of smaller size of the object
placed in front of it. Similarly, a plane mirror will always give a virtual and erect image of same
size as that of the object placed in front of it. Therefore, the given mirror could be either plane or
convex.

Question 20: 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 gives a magnified image of an object when it is placed between the radius of curvature and focal length. Also, magnification is more for convex lenses having shorter focal length. Therefore, for reading small letters, a convex lens of focal length 5 cm should be used.

Question 21: 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: Range of object distance = 0 cm to15 cm
A concave mirror gives an erect image when an object is placed between its pole (P) and the principal focus (F). Hence, to obtain an erect image of an object from a concave mirror of focal length 15 cm, the
object must be placed anywhere between the pole and the focus. The image formed will be virtual,
erect, and magnified in nature, as shown in the given figure.


Question 19: 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 (b) Convex (c) Concave
(a) Concave mirror is used in the headlights of a car. This is because concave mirrors can produce
powerful parallel beam of light when the light source is placed at their principal focus.
(b) Convex mirror is used in side/rear view mirror of a vehicle. Convex mirrors give a virtual, erect, and diminished image of the objects placed in front of it. Because of this, they have a wide field of view. It enables the driver to see most of the traffic behind him/her.
(c) Concave mirrors are convergent mirrors. That is why they are used to construct solar furnaces.
Concave mirrors converge the light incident on them at a single point known as principal focus. Hence, they can be used to produce a large amount of heat at that point.

Question 20: 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: The convex lens will form complete image of an object, even if its one half is covered
with black paper. It can be understood by the following two cases.
Case I When the upper half of the lens is covered
In this case, a ray of light coming from the object will be refracted by the lower half of the lens.
These rays meet at the other side of the lens to form the image of the given object, as shown in the
following figure.


Case II When the lower half of the lens is covered
In this case, a ray of light coming from the object is refracted by the upper half of the lens. These
rays meet at the other side of the lens to form the image of the given object, as shown in the
following figure.



Question 21: 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: Object distance, u = −25 cm
Object height, ho = 5 cm
Focal length, f = +10 cm
According to the lens formula,


The negative value of image height indicates that the image formed is inverted. The position, size, and nature of image are shown in the following ray diagram.



Question 22: 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: Focal length of concave lens (OF1), f = −15 cm
Image distance, v = −10 cm
According to the lens formula,


The negative value of u indicates that the object is placed 30 cm in front of the lens. This is shown
in the following ray diagram.



Question 23: 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: Focal length of convex mirror, f = +15 cm
Object distance, u = −10 cm
According to the mirror formula,

The magnification produced by a plane mirror is +1. It shows that the image formed by the plane
mirror is of the same size as that of the object. The positive sign shows that the image formed is
virtual and erect.

Question 24: 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: Object distance, u = −20 cm
Object height, h = 5
Radius of curvature, R = 30 cm
Radius of curvature = 2 × Focal length
R = 2 f
f = 15 cm
According to the mirror formula,

The positive value of image height indicates that the image formed is erect. 
Therefore, the image formed is virtual, erect, and smaller in size.

Question 25: 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.
Answer: Object distance, u = −27 cm
Object height, h = 7 cm
Focal length, f = −18 cm
According to the mirror formula,

The negative value of image height indicates that the image formed is inverted.

Question 26: Find the focal length of a lens of power −2.0 D. What type of lens is this?
Answer: 

A concave lens has a negative focal length. Hence, it is a concave lens.

Question 27: 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: 

A convex lens has a positive focal length. Hence, it is a convex lens or a converging lens.

( From www.ncrtsolutions.in )

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