PSEB Class 10 Biology Chapter 7 Control And Coordination Question and Answers

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Control And Coordination Questions

Question 1. What is the difference between reflex action and walking?
Answer:

Animals Nervous System Difference Between Reflex Action And Walking

Question 2. What happens at the synapse between two neurons?
Answer:

The synapse between two neurons is a junctional complex that permits an axon end of a neuron to pass its electrical or chemical signal to the dendrite end of another neuron.

  • There is a fluid-filled narrow space between the two neurons. It is called synaptic cleft. The presynaptic axon end of the neuron is bulb-like.
  • It contains many synaptic vesicles having neurotransmitter chemicals. The postsynaptic dendrite end of the next neuron has a trough-like depression. Its membrane has receptors for the neurotransmitter.
  • As the impulse reaches the presynaptic bulb, the synaptic vesicles move to the cleft side and burst open.
  • The released neurotransmitter sensitizes the receptors that in turn create the electrical impulse.
  • Synapse allows a unidirectional flow of information from the axon end of one neuron to the dendrite end of the next neuron.

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PSEB Class 10 Biology Chapter 7 Control And Coordination

Question 3. Which part of the brain maintains posture and equilibrium of the body?
Answer: Cerebellum.

Question 4. How do we detect the smell of an agarbatti (incense stick)?
Answer: Burning of agarbatti releases numerous molecules of incense. They enter the nose and come in contact with particular sensory hair of olfactory cells.

An impulse is generated. It passes into the olfactory bulb which relays it to the temporal lobes of the cerebrum for interpretation.

Question 5. What is the role of the brain in reflex action?
Answer: Like the spinal cord, the brain can also play a direct role in certain reflex actions called cerebral reflexes, for example., closing of eyes or flashing of light.

In such cases, brain acts as a relay station in which the impulse of sensory neurons is transferred to motor neurons through an interneuron.

Question 6. What is reflex action? Explain the mechanism of reflex action with a suitable example.
Answer:

Reflex action is a nerve-mediated automatic and spontaneous response to a certain stimulus without consulting the will of the individual.

  • The stimulus for reflex action is picked up by its receptor. The receptor is connected to a sensory neuron. An impulse generated in it is passed to the central nervous system.
  • The central nervous system acts as a relay or modulator. With the help of an interneuron, it transfers the impulse to a motor neuron. The motor neuron takes the impulse to the effector organ for response.

Question 7. Name the three major regions of the brain. Which part of the brain maintains posture and equilibrium of the body?
Answer:

Three regions of the brain. Fore brain, midbrain, and hindbrain. Maintenance of posture and equilibrium. Cerebellum.

Question 8.

  1. Distinguish between voluntary and involuntary actions of our body.
  2. Choose involuntary actions among the following: reading, beating of heart, salivation in mouth on viewing tasty food, talking.

Answer:

  1. Voluntary action is under the control of the will and is performed with the help of striated muscles. Involuntary action is performed without instructions from will with the help of smooth muscles.
  2. The beating of the heart and salivation in the mouth on viewing a tasty food are involuntary actions.

Question 9.

  1. Name two main regions of our central nervous system,
  2. Which one of them plays a major role in sending commands to muscles to act without Involving the thinking process,
  3. Name the phenomenon involved.

Answer:

PSEB Class 10 Biology Chapter 7 Question And Answers

  1. Brain and spinal cord.
  2. The spinal cord has a major role in sending instructions to muscles without involving the thinking process.
  3. Reflex action.

Question 10. Write the function of the following :

  1. Sensory neuron
  2. Cranium
  3. Vertebral column
  4. Motor neuron.

Answer:

  1. Sensory Neuron. Conveying impulse or sensation from receptor to central nervous system.
  2. Cranium. Covering and protection of the brain.
  3. Vertebral Column. Covering and protection of the spinal cord.
  4. Motor Neuron. Conveying impulse for action from the central nervous system to the effector organ.

Question 11. Identify the part of a neuron

  1. Where information is acquired
  2. Through which information travels
  3. Where the impulse must be converted into a chemical signal?

Answer:

  1. Dendrite
  2. Axon
  3. Synapse.

Question 12.

  1. What is a reflex arc?
  2. Why have reflex arcs evolved in animals?

Answer:

  1. A reflex arc is nerve nerve-based pathway performed by an impulse from the receptor of stimulus to the effector organ through the central nervous system without consulting the will of the individual.
  2. Reflex arcs evolved in animals even before the evolution of intelligence as a survival mechanism because they can provide immediate responses to harmful stimuli.

Question 13. Stale the sequence of events through a reflex arc which occurs when a bright light is focussed on your eyes.
Answer:

  • Bright light is sensed by receptors present over the eyes. The sensation is picked up by sensory neurons and taken to the superior quadrigeminal.
  • Intemeurons present in the superior corpora quadrigemina transfer the impulse to motor neurons. The motor neurons close the eyelids as well as narrow the pupil.

Question 14.

  1. Name one gustatory receptor and one olfactory receptor present in human beings.
  2. Write a and b in the given flow chart of neurons through which information travels as an impulse.

Dendrite a → b → End point of a neuron.

Answer:

  1. Gustatory Receptor. Taste buds over the tongue.
    1. Olfactory Receptor. Olfactory epithelium in nasal chambers.
  2. Dendrite → Cell body → Axon → End point of a neuron.

Question 15. What is a nerve impulse? State the direction followed by a nerve impulse while traveling in the body of an organism.
Answer:

A nerve impulse is a progressive electrical wave or signal that develops in response to a stimulus and travels along a nerve fiber to reach another nerve, muscle, or gland.

Direction Of Nerve Impulse

Dendrite → cell body → Axon → Axon end → Dendrite/gland/muscle.

Control And Coordination PSEB Class 10 Notes

Question 16. List in tabular form two differences between reflex action and walking.
Answer:

Animals Nervous System Difference Between Reflex Action And Walking.

Question 17. Which part of the nervous system controls reflex arcs? Trace the sequence of events that occur when we touch a hot object. Mention the part of the neuron that requires information and the form in which information travels.
Answer:

Reflex arcs are controlled by the central nervous system (many in the spinal cord) without analyzing them. CNS functions as a coordinator or relay station for transferring sensory information to motor information.

Events. A hot object is the stimulus. As it is touched, the stimulus is picked up by skin-based receptors. From receptors, the information is taken up as an impulse by sensory neurons.

  • Sensory neurons take the impulse to the spinal cord. In the spinal cord, the impulse is transferred to motor neurons through interneurons.
  • The motor neurons take the impulse to the muscles of the hand which pull away the arm. Information travels in neurons in the form of electrochemical impulses.

Question 18. What is the structural and functional unit of the nervous system? Name any two components. Which structure in a neuron helps to conduct a nerve impulse

  1. Towards the cell body
  2. Away from the cell body.

Answer:

Structural and Functional Unit. Neuron.

Components of a Neuron. Dendrites, cell body, and axon.

  1. Dendrite
  2. Axon

Question 19.

  1. Which part of the brain controls Involuntary actions?
  2. Write the function of any two regions of It.

Answer:

Hind brain is a part of the brain that controls involuntary functions. It has three parts—pons, cerebellum, and medulla oblongata.

  1. Medulla Oblongata. It regulates blood pressure, heart rate, and breathing. Several reflex actions have centers in the medulla oblongata, viz., swallowing, vomiting, sneezing, coughing, salivation, and peristalsis.
  2. Cerebellum. It involuntarily co-ordinates muscular activities of the body. The cerebellum maintains equilibrium during different activities or postures of the body.

Question 20.

  1. If the cerebellum is not functioning properly, state the activities of our body that are affected.
  2. How do muscle cells move?

Answer:

PSEB Class 10 Biology Important Questions

  1. Impairment of the cerebellum shall result in non-coordination of body activities and defects in maintaining balance or posture of the body.
  2. Muscle cells move in response to excitation provided by a neuromuscular junction. The motor impulse passes to the bulb end of the motor end plate of an axon.
    • The bulb is in near contact with the sole plate of the muscle cell. As the nerve impulse reaches the bulb, it activates the synaptic vesicles.
    • The vesicles move to the membrane and burst open to release molecules of neurotransmitters.
    • They activate receptors over the end plate and cause activation of muscle interior releasing calcium and causing contraction of muscle cells.

Question 21. Name the part of the brain that controls

  1. Voluntary actions
  2. Involuntary actions.

How is the brain protected from injury and shock?

Answer:

  1. Voluntary Actions. Cerebrum.
  2. Involuntary Actions. Hindbrain.

Protection of the Brain.

  1. Covering of brain by cranium,
  2. Holding of the brain by meninges,
  3. Presence of shock-resistant fluid medium both inside and outside the brain.

Question 22.

  1. Name the system which facilitates communication between the central nervous system and other parts of the body.
  2. Mention two types of nerves it consists of along with their organs of origin.

Answer:

  1. Peripheral Nervous System (PNS) facilitates communication between the central nervous system (CNS) and other parts of the body.
  2. The peripheral nervous system consists of two types of nerves
    1. Cranial from the brain, 12 pairs
    2. Spinal from spinal cord, 31 pairs.

Question 23. Trace the sequence of events that occur when a bright light is focused on your eyes.
Answer:

It is an example of a cerebral reflex.

Receptor cells of eyes → Sensory neurons → CNS (Brain) → Motor neurons → Eye muscles → Pupil contracts and eyelids close.

Question 24. List in tabular form three distinguishing features between the cerebrum and cerebellum.
Answer:

Animals Nervous System Distinguishing Features Between Cerebrum And Cerebellum

Question 25.

  1. Name a part of the human brain that controls
    1. Voluntary actions and
    2. Involuntary actions.
  2. Write the function of the peripheral nervous system. Name the components of this system stating their origin.

Answer:

PSEB Class 10 Science Chapter 7 Solutions

  1. Voluntary actions. Forebrain/cerebrum.
    1. Involuntary Actions. Hind brain/medulla oblongata.
  2. Peripheral Nervous System (PNS) connects different parts of the body with the central nervous system. It consists of nerves. Depending upon their origin, nerves arc of two types, cranial (from the brain) and spinal (from the spinal cord).

Question 26. Name parts in the diagram. What is the term given to the sequence of events occurring in the diagram?

Animals Nervous System Reflex Arc.

Answer:

  1. —Receptor (skin)
  2. —Sensory neuron
  3. —Spinal cord
  4. —Relay neuron
  5. —Motor neuron.

Term. Reflex arc.

Question 27. A cheetah on seeing a prey, moves towards it at a very high speed. What causes the movement of his muscles? How does the chemistry of cellular components of muscles change during the event?
Answer:

  • The sight of prey generates a nerve impulse that reaches the muscles. A neurotransmitter is released. It causes the activation of muscles through the release of calcium.
  • Calcium brings about a shortening of muscles through the sliding of myosin fibrils over actin fibrils. Shortening and relaxing of muscles allow the cheetah to move at high speed.

PSEB Class 10 Biology Chapter 9 Heredity Question And Answers

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Heredity Short Question And Answers

In each question, two statements are given- one labelled Assertion and the other labelled Reason. Select the correct answer to these questions from the codes (A), (B), (C) and (D) as given below :

  • Both assertion and reason are true and reason is a correct explanation of the assertion.
  • Both assertion and reason are true but reason is not the correct explanation of the assertion.
  • Assertion is true but reason is false.
  • The assertion is false but the reason is true.

Question 1. Assertion (A). The sex of a child in human beings will be determined by the type of chromosome he or she inherits from the father.

Reason (R). A child who inherits an ‘X’ chromosome from his father would be a girl (XX) while a child who inherits a ‘Y’ chromosome from the father would be a boy (XY).
Answer:  (1)

The sex of the children is determined by what they inherit from their father and not the mother. Justify.
Answer:

Women are homogametic. They produce one type of ova 22 + X. Men are heterogametic. They have two types of sperms, androsperms (22 + Y) and gynosperms (22 + X). During fertilization, it is the sperm type which determines the sex of the child—male (22 + X and 22 + Y or 44 + XY) or female (22 + X and 22 + X or 44 +XX).

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Question 2. State the importance of chromosomal differences between sperms and eggs of humans.
Answer:

The chromosomal difference between sperm and eggs of humans is basic to the determination of the sex of the child. Eggs are of one type carrying the genetic complement of 22 + X. Sperms are of two types, androsperms with 22 + Y chromosomes and gynosperms with 22 + X chromosomes. The two types of sperm determine the sex of the child. Justify.

Question 3. Name the organism Mendel used for his experiments. Explain F1 and F2 progeny obtained by Mendel when he bred tall and short varieties of the organisms he experimented with.
Answer:

The experimental organism used by Mendel was the Garden Pea (Pisum sativum). Mendel crossed a pure tall

Pea plant with a pure dwarf Pea plant. All the plants of the Fj generation were tall. On self-breeding plants of the F1 generation, Mendel obtained an F2 progeny of both tall and dwarf in the ratio of 3: 1. The presence of dwarf plants in the F2 generation shows that the factor of dwarfness was present in F1 plants but it did not express its effect due to its recessive nature. The factor for tallness is dominant.

PSEB Class 10 Biology Chapter 9 Heredity

Question 4. The genotype of a plant bearing purple flowers is PP and one with white flowers is pp. When these are crossed

What colour of flowers would you find in F1 progeny?

Give the percentage of white flowers if F1 plants are self-pollinated.

In what ratio would you find PP and Pp in F2 progeny? Draw a flowchart in support of your answer.

Answer:

Purple (b) 25 % (c) 1: 2

Heredity F2 progeny

1 Pure: 2 hybrid purple: 1 pure white

Question 5. Why are some Pea plants tall and others short in nature? Explain with reference the role of genes in controlling characteristics.
Answer:

Genes are segments of DNA. They function in controlling body structure and its working through the synthesis of mRNAs. wRNAs take part in the formation of proteins. The proteins help in the synthesis of enzymes and other biochemicals.

The character of height is controlled by growth hormones. If the gene-controlling height is dominant, it helps in the synthesis of more hormones so that the plant becomes tall. If the gene is recessive, a less efficient protein and a smaller quantity of growth hormones would be produced. As a result, the plant remains dwarf.

Question 6. Explain how Mendel’s experiments show that traits are inherited independently.
Answer:

Mendel performed a dihybrid cross between round yellow-seeded and wrinkled green-seeded Pea plants. In F1 generation he obtained only round yellow seeded plants. Both traits are dominant and are expressed in hybrids.

He raised the F2 generation after the self-breeding of F1 plants. There were four types of F2 plants—round yellow (9/16), round green (3/16), wrinkled yellow (3/16) and wrinkled green (1/16). The combinations of round green and wrinkled yellow-seeded plants were new. They could arise only if the traits of the two characters could assort independently.

Question 7. One plant with dominant tall height bearing purple flowers is crossed with another plant having recessive dwarf and white flowers.

  1. What does the plant’s F1 generation look like?
  2. What type of cross is it?
  3. What is the phenotypic ratio in F2 generation and name the type of plants obtained according to phenotypic ratio?

Answer:

  1. Plants of generation are all tall and purple-flowered.
  2. Dihybrid cross as it involves two different characters.
  3. 9: 3 : 3: 1. Tall and purple flowered (9/16) tall and white flowered (3/16) dwarf and purple flowered (3/16) and dwarf and white flowered (1/16).

Question 8.

1. Name the sex chromosomes present in human males and human females.

2. With the help of a flowchart determine genetically in human beings the sex of the spring if the sperm carrying X chromosome fertilizes the egg.
Answer:

A. Sex chromosomes.

1. Human Male-XY

2. Human female-XX.

B. 22 + X (egg) and 22 + X (sperm) will produce a female baby (44 + XX).

PSEB Class 10 Biology Chapter 9 Question And Answers

Heredity Female Baby

Question 9. Guinea pigs having black colour when crossed with guinea pigs having the same colour produced 100 offspring out of which 75 were black and 25 were white. Now find out :

  1. What is the possible genotype of guinea pigs?
  2. Which trait is dominant and which is recessive?
  3. What is this cross called and what is the ratio of F2 progeny obtained from this cross?

Answer:

  1. Genotype of Guinea Pigs. Bb, Bb.
  2. Black colour is dominant (B) while white colour (b) is recessive.
  3. Monohybrid cross. F2 progeny. 1 pure black (BB): 2 hybrid black (Bb): one white (bb).

Question 10.

  1. Name an animal which can change sex. What does it indicate?
  2. Name the trait which can express itself in the next generation.

Answer:

  1. Ophryotrocha(an annelid). It is male in the young state and female later. Sex is here determined non-genetically.
  2. Dominant trait.

Heredity PSEB Class 10 Notes

Question 11.

  1. Mendel selected the Pea plant for his experiments. List two reasons.
  2. State the meaning of recessive and dominant genes.

Answer:

  • Mendel selected the Pea plant for his experiments because It can be cross-pollinated manually and self-pollinated automatically. It is annual with a short life span and a high yield of seeds.
  • Recessive Gene: It is a gene or factor that cannot express its effect in the presence of its contrasting factor.
    • Dominant Gene: It is a gene or factor which expresses its effect both in homozygous and heterozygous states.

Question 12. In a cross between plants with purple flowers and plants with white flowers, the offspring of the F1 generation had all purple flowers. When F1 generation individuals were self-bred, the F2 generation gave rise to 100 individuals, 75 of which had purple flowers. Make a cross and answer the following.

  1. What is the genotype of F2 individuals?
  2. What is the ratio of purple to white-flowered plants in F2 generation?

Answer:

Heredity Genotypes

1PP (pure purple): 2 Pp (hybrid purple): pp (pure white)

  1. Genotypes of F2 Progeny. 1 PP: 2 Pp: 1 pp.
  2. Purple to white flowered ratio. 75: 25 or 3: 1.

Question 13. A husband has 46 chromosomes. His wife has 46 chromosomes. Then why does not their offspring have 46 pairs of chromosomes which is obtained by fusion of male and female gametes? Support your answer with an illustration.
Answer:

Gametogenesis is not a mitotic division. It is a meiotic division. Therefore, gametes do not possess the same chromosome number as the parents. Instead, they have half the number of chromosomes. As the mother has 46 chromosomes, the ovum will have 23 chromosomes.

Similarly, as the father has 46 chromosomes his sperm will have 23 chromosomes. The fusion of a sperm (23 chromosomes) and an ovum (23 chromosomes) restores the 46 chromosome number in the offspring. There is no chance of having 46 pairs of chromosomes.

PSEB Class 10 Biology Important Questions On Heredity

Heredity pairs Of chromosomes

Question 14. “It is a matter of chance whether a couple will give birth to a boy or a girl”. Justify the statement and support your answer with a neat illustration.
Answer:

The wife produces only one type of ova, that is, 22 + X. The Husband has two kinds of sperm, 22 + X (gymnosperms) and 22 + Y (angiosperms) formed in equal numbers. It is a pure chance which type of sperm comes in contact with the ovum. If it is endosperm, a male child is born. If it is a gymnosperm, a female child is born.

Heredity Neat Illustration

Question 15. A round-seeded plant (RR) Is crossed with a wrinkled-seeded plant (RR). Find out the phenotype and genotype of the F1 and F2 generations.
Answer:

Heredity Phenotype generation

Phenotype of F1 generation = Round Seeded ; Genotype of F, generation = Rr

The phenotype of F2 generation = 3 round: 1 wrinkled

Genotype of F2 generation = 1 pure round (RR): 2 hybrid round (Rr): 1 wrinkled (rr).

Question 16. Mendel studied the inheritance pattern of traits in a Pea plant. According to this study, he obtained a 9 : 3 : 3: 1 ratio of certain traits in the progeny of the F2 generation. Based on it answer these questions.

  1. What traits did he study? How do they represent themselves?
  2. What was the trait of F1 progeny?
  3. Which rule does this inheritance pattern suggest?

Answer:

  1. Mendel crossed round yellow seeded (RRYY) plant with wrinkled green seeded (rryy) plant.
  2. F1 plants were all round and yellow-seeded.
  3. In the F2 generation, four types of plants were formed—round yellow seeded (9/16), round green seeded (3/16), wrinkled yellow seeded (3/16) and wrinkled green seeded (1/16).

The occurrence of round green and wrinkled yellow-seeded plants was a new combination of traits other than the parental types.

This is possible if the traits of the two characters separate or assort independently of each other. The phenomenon is called the law of independent assortment.

PSEB Class 10 Science Chapter 9 Solutions

Question 17. A cross was obtained between a purebred tall Pea plant and a purebred dwarf pea plant and F1 progeny. Later the F1 progeny was selfed to obtain F2 progeny. Answer the following questions :

  1. What is the phenotype of F1 progeny and why?
  2. Give the phenotypic ratio for F2 progeny.
  3. Why is the F2 progeny different from the F1 progeny?

Answer:

  1. F1 progeny is all tall as the trait of tallness is dominant over the trait of dwarfism.
  2. 3 tall: 1 dwarf.
  3. Segregation of the trait of tallness and trait of dwarfism during gamete formation by F1 progeny followed by random fusion of the two types of gametes during the formation of F2 progeny, i.e., T x T (TT), T x t (Tt), t x t (tt).

Question 18. How can you say that in Mendel’s monohybrid cross, the F2 dominants are not the same?
Answer:

Phenotypically F2 dominants appear similar but genotypically they are of two types, pure and hybrid. This can be proved by self-breeding the dominants. One-third of the dominants produce a progeny of only dominants. They are pure for the dominant trait. Two-thirds of the dominants behave as hybrids. They give a progeny of both dominant and recessive traits in the ratio of 3: 1.

Question 19. A cross was made between pure breeding Pea plants one with round and green seeds and the other with wrinkled and yellow seeds.

  1. Write the genotype of Ft progeny. Give reasons for your answer.
  2. Write the different types of F2 progeny obtained along with ratios when F1 progeny was selfed.

Answer:

  1. Across between RRyy and rrYY plants yields an Fj generation ofRrYy(round and yellow seeded plants) even though none of the parents have this set of traits. It is due to the non-expression of recessive alleles for green colour and wrinkled seed in the presence of dominant alleles for yellow colour and round seed.
  2. F2 Progeny. Round yellow (9/16), round green (3/16), wrinkled yellow (3/16) and wrinkled green (1/16).

Question 20. In an experiment, Mendel obtained 1014 plants, out of which 787 had round seeds and 227 had wrinkled seeds in F2 generation.

  1. What is the approximate ratio obtained in F2 generation?
  2. Under which law do you find this ratio?
  3. Why is this law so-called?

Answer:

  1. 787: 227 roughly 3:1.
  2. Law of segregation.
  3. As per the law of segregation, the two alleles of a character separate during gametogenesis and pass to different gametes.
  4. A gamete comes to have only a single allele of a character.

PSEB Class 10 Physics Solutions For Chapter 1 Light-Reflection And Refraction

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PSEB Class 10 Physics Solutions For Chapter 1 Light-Reflection And Refraction

Light

Light is the form of energy that enables us to see. An object reflects light that falls on it. We are able to see the object when this reflected light is received by our eyes.

Properties Of Light

  1. Light travels in a straight line. Due to this, light casts a shadow of the object placed in its path.
  2. Light is an electromagnetic wave. It does not require any medium to travel.
  3. The speed of light is maximum in a vacuum. It is 3×108 ms-1.
  4. Light has a dual nature, i.e., particle as well as wave nature.
  5. When light falls on an object, one or more of reflection, refraction, or absorption may happen.

Types Of Materials Based On Absorption Of Light

Three types of materials based on absorption of light are Transparent, Translucent, and Opaque.

PSEB Class 10 Physics Chapter 1 Light-Reflection And Refraction

PSEB Solutions For Class 10 Physics Chapter 1 Types Of Materials Based On Absorption Of Light

Reflection Of Light

The phenomenon of bouncing back of light into the same medium, when it strikes a polished or smooth surface, is called reflection.

Incident Ray = Light ray which falls on the polished or smooth surface

Reflected ray = Light ray that gets bounced back from the polished or smooth surface.

Laws Of Reflection

Applicable to all types of reflecting surfaces, including spherical surfaces

  1. Angle of Incidence (∠i) = Angle of Reflection (∠r)
  2. Incident Ray, Normal to the angle of incidence, and Reflected Ray all lie in the same plane.

PSEB Solutions For Class 10 Physics Chapter 1 Laws Of Reflection

Image– It is the point at which two rays actually meet or appear to meet. An image can be of two types: Real or Virtual.

PSEB Solutions For Class 10 Physics Chapter 1 Image

Characteristics Of the Image Formed By a Plane Mirror

  1. The image formed is Virtual- formed behind the mirror.
  2. Size of image = Size of Object
  3. Distance between mirror and object = Distance between mirror and image
  4. Laterally inverted- means the left side of the object appears right and the right side of the object appears left.

PSEB Solutions For Class 10 Physics Chapter 1 Image Formed By Plane Mirror

Spherical Mirrors

Mirrors whose reflecting surface forms part of the sphere are called spherical mirrors. They are of two types – Concave mirrors and Convex mirrors, as shown below:

PSEB Solutions For Class 10 Physics Chapter 1 Spherical Mirrors

PSEB Solutions For Class 10 Physics Chapter 1 Convex And Concave Mirror

Some Important Terms Used In Context To Spherical Mirrors

PSEB Solutions For Class 10 Physics Chapter 1 Context To Spherical Mirrors

Pole (P) – Centre of the spherical mirror.

Centre of Curvature (C) – The centre of the hollow glass sphere of which the spherical mirror was a part.

Principal Axis – Line joining the pole (P) and centre of curvature (C).

Radius of Curvature (R) – Distance between pole (P) and centre of curvature (C).

Principal Focus (F) – Point on the principal axis where all rays actually meet or appear to meet.

Focal Length (f) – Distance between pole (P) and principal focus (F).

Aperture (MN) – Effective diameter of the spherical mirror.

Relationship Between Focal Length (f) And Radius Of Curvature (R)

Focal Length (f) = Radius of Curvature (R)/2

Rules For Making Ray Diagrams For Spehrical Mirrors

PSEB Solutions For Class 10 Physics Chapter 1 Rules For Making Ray Diagram For Spherical Mirrors

Simple Trick To Remember Image Formation By Concave Mirror

Step 1: Write position of object starting from infinity till between F and P.

Step 2: Write numbers 5, 4, 3, 2, 1 to all positions of object, and numbers 1, 2, 3, 4, 5 to all positions of image. (Note that numbers given to positions of objects are 5-1 and those given to image positions are 1-5. This is important point to be remembered).

Step 3: Comparing the numbers will give us the size of image, For Examle Since1 is very less than 5, size of image will be highly diminished, 2 is less than 4, therefore image will be diminished, and so on (as given in table below).

Note:

  • For object position numbers 5, 4, 3, 2 and 1, image will be real and inverted.
  • Last position of object is an exception- in this case image will be virtual and erect.

image

Simple Trick To Remember Image Formation By Convex Mirror

Important Note: Image formed by convex mirror is always virtual, erect and small in size.

PSEB Solutions For Class 10 Physics Chapter 1 Image Formation By Concave Mirror

PSEB Solutions For Class 10 Physics Chapter 1 Image Formation By Convex Mirror

Uses Of Concave Mirrors

1. Used in torches, search-lights and vehicles headlights to get powerful parallel beams of light as shown in diagram:

PSEB Solutions For Class 10 Physics Chapter 1 Parallel Beams Of Light

2. Used as shaving mirrors to see a larger image of the face, and also used by dentists to see large images of the teeth of patients. For this, the object has to be placed between pole and focus, as shown in figure:

PSEB Class 10 Physics Solutions Chapter 1

PSEB Solutions For Class 10 Physics Chapter 1 Object Has To Be Placed Between Pole And Focus

3. Large concave mirrors are used to concentrate sunlight to produce heat in solar furnaces.

4. Used as satellite dish TV antennas as shown in figure:

PSEB Solutions For Class 10 Physics Chapter 1 Satellite Dish TV Antennas

Uses Of Convex Mirrors

1. Used as rear-view (wing) mirrors in vehicles because:

  • They always give an erect, though diminished, image.
  • They have a wider field of view as they are curved outwards, as shown in figure.

PSEB Solutions For Class 10 Physics Chapter 1 Wider View by Convex Mirror

2. Used at blind turns and points of merging traffic to give view of traffic from both sides.

3. Used in shops as security mirror.

New Cartesian Sign Conventions – For Reflection By Spherical Mirrors

In the New Cartesian Sign Convention,

  • Pole (P) of the mirror is taken as the origin.
  • Principal axis of the mirror is taken as the x-axis of the coordinate system.
  • The New Cartesian Sign Convention are summarized in the figure provided below (For details, please refer points 1-5 given after the figure):

PSEB Solutions For Class 10 Physics Chapter 1 The New Cartesian Sign Convention

  1. Object is always placed to the left of the mirror, i.e, light from the object falls on the mirror from the left-hand side.
  2. All distances parallel to the principal axis are measured from the pole of the mirror.
  3. All the distances measured to the right of the origin (along + x-axis) are taken as positive while those measured to the left of the origin (along- x-axis) are taken as negative.
  4. Distances measured perpendicular to and above the principal axis (along + y-axis) are taken as positive.
  5. Distances measured perpendicular to and below the principal axis (along -y-axis) are taken as negative.

Conclusions From New Cartesian Sign Conventions For Spherical Mirrors

Distance of Object from Mirror (u) = Always Negative

Focal length of Concave Mirror (f) = Negative

Focal length of Convex Mirror (f) = Positive

Mirror Formula

It gives the relation between distance of object, distance of image and focal length of the mirror. According to mirror formula

∴ \(\frac{1}{v}+\frac{1}{u}=\frac{1}{f}\)

Where, u = distance of object from pole, v = distance of image from pole, and f = focal length of mirror.

Mirror formula is valid for plane mirrors as well as spherical mirrors for all positions of the object.

Important Note: For solving numerical problems, New Cartesian Sign Conventions must be considered while substituting values for u. v, f (or R. if focal length is calculated using relation f = R/2) in the mirror formula.

Magnification Of Spherical Mirrors

Magnification (m) is defined as the ratio of height of image to the height of object.

∴ \(m=\frac{\text { Height of image }}{\text { Height of object }}=\frac{h_i}{h_o}=-\frac{v}{u}\)

Simple Trick To Find The Type Or Size Of Image From Value Of Magnification

PSEB Solutions For Class 10 Physics Chapter 1 Size Of Image From Value Of Magnification

Important Note:

  • Magnification of convex mirror is always negative.
  • Magnification of concave mirror can be positive or negative depending on the position of the object.

Refraction Of Light

The phenomenon of bending of light when it enters obliquely from one transparent medium into another is called refraction of light. Refraction occurs due to change in speed of light in different media.

Light-Reflection And Refraction PSEB Class 10 Notes

Bending Of Light In Refraction

1. If a ray of light passes from a material in which it travels fast (rarer medium) into a material in which it travels slower (denser medium), then the light ray will bend towards the normal, i.e,

PSEB Solutions For Class 10 Physics Chapter 1 Light In Reflection Fast To Slow

2. If a ray of light passes from a material in which it travels slow (denser medium) into a material in which it travels faster (rarer medium), then the light ray will bend away from the normal.

PSEB Solutions For Class 10 Physics Chapter 1 Light In Reflection Slow To Fast

Simple Trick To Remember The Direction Of Bending Of Light In Refraction

Direction of bending of light in refraction can be easily remembered using the mnemonic “FAST SOFA”

PSEB Solutions For Class 10 Physics Chapter 1 FAST SOFA

3. Extent of bending of incident ray depends upon difference in refractive index between the two medium through which the ray is passing.

4. Emergent Ray is parallel to Incident Ray because the extent of bending of ray of light at the opposite parallel faces of the second medium (glass in this case) is equal and opposite as shown:

PSEB Solutions For Class 10 Physics Chapter 1 Emergent Ray Is Parallel To Incident Ray

5. Perpendicular distance between the incident ray and the emergent ray is called lateral displacement or lateral shift (d). It depends upon:

  • Thickness of the glass slab
  • Refractive index of the glass slab

PSEB Solutions For Class 10 Physics Chapter 1 Lateral Displacement Or Lateral Shift

Refractive Index (n)

It is defined as the ratio of velocity of light in a given pair of media, i.e.,

∴ \(n=\frac{\text { Velocity of light in medium } 1}{\text { Velocity of light in medium } 2}\)

Points To Remember About Refractive Index

  1. Refractive index has no units because it is a ratio.
  2. Refractive index does not depend on the physical density of the medium, For Example kerosene having higher refractive index, is optically denser than water, although its mass density is less than water.
  3. Refractive index of water, nwater= 1.33, and refractive index of air, nair = 1. This means that the ratio of the speed of light in air and the speed of light in water is equal to 1.33.
  4. Refractive index of diamond is highest till date. Its value is 2.42. This means speed of light in diamond is 1/ 2.42 times less than the speed of light in air or vacuum.
  5. n21 means refractive index of 2nd medium with respect to 1st medium, whereas n21 means refractive index of 1st medium with respect to 2nd medium, i.e.,

∴ \(n_{21}=\frac{V_1}{V_2} \quad \text { whereas } \quad n_{12}=\frac{V_2}{V_1}\)

Absolute Refractive Index

It is the ratio of the speed of light in vacuum to the speed of light in the given medium, i.e.,

∴ \(n=\frac{c}{v}\)

where c = 3 x 108 ms-1

Important Note: Absolute refractive index can never be less than 1 because the speed of light in any medium is always less than that in a vacuum.

Optically Rarer And Optically Denser Medium

Optically Rarer Medium – Medium with a lower value of refractive index

Optically Denser Medium – Medium with a higher value of refractive index

For Example – Out of Air and Glass, glass is denser medium while air is rarer medium.

PSEB Class 10 Physics Important Questions Chapter 1

Laws Of Refraction

The incident ray, the refracted ray and the normal to the interface of two transparent media at the point of incidence, all lie in the same plane.

The ratio of sine of angle of incidence to the sine of angle of refraction is a constant, for the light of a given colour and for the given pair of media. This law is also known as Snell’s law of refraction. Thus, according to Snell’s law,

∴ \(\frac{\sin i}{\sin r}=\text { constant }\)

The constant value is called the refractive index of the 2nd medium with respect to the 1st medium.

Snell’s law is true for 0 < i < 90°, where “i” is the angle of incidence.

Some Common Examples Of Refraction

  1. The bottom of swimming pool appears higher due to refraction of light on entering water through air.
  2. A pencil partially immersed in water appears to be bent at the interface of air and water.
  3. When a thick glass slab is placed over some printed matter, the letters appear raised.
  4. Lemons placed in a glass tumbler appear bigger.

Spherical Lens – A transparent material bound by two surfaces, of which one or both surfaces are spherical, forms a lens.

Four Types Of Spherical Lens

PSEB Solutions For Class 10 Physics Chapter 1 Four Types Of Spherical Lens

Difference Between Concave Lens And Convex Lens

PSEB Solutions For Class 10 Physics Chapter 1 Difference Between Concave And Convex Lenses

Some Important Terms About Lens

Since both convex and concave lens have two spherical surfaces, it has two centres of curvature and two principal focus.

PSEB Solutions For Class 10 Physics Chapter 1 Two Spherical Surfaces

Centre of Curvature (C)– It is the centre of the sphere of which the lens forms a part. Because a lens has two spherical surfaces, it has two centres of curvature (C1 and C2 as shown in above figures).

Radius of Curvature (R)– It is the radius of the sphere of which the spherical surface of lens forms a part. R1 and R2 are radii of curvature in above figures.

Principal Axis (C1C2 – It is the line passing through the two centres of curvature (C1 and C2) of the lens.

Optical Centre (O) – If a ray of light is incident on a lens such that after refraction through the lens the emergent ray is parallel to the incident ray, then the point at which the refracted ray intersects, the principal axis is called the optical centre of the lens.

Note: Optical centre of lens coincides with the geometric centre of the lens when the radii of curvature of the two surfaces are equal.

Aperture– The diameter of the circular boundary of the lens is called its aperture.

Principal Focus (F)– It is the point through which a parallel light after refraction passes through in the case of a convex lens and appears to pass through in the case of a concave lens. A lens has two focal points, because a lens has two refracting surfaces.

Focal Length (f)– It is the distance of the principal focus from the optical centre of a lens.

1. First Principal Focus (F1) and First Focal Length (f) – It is a fixed point on the principal axis such that rays starting from this point (in convex lens) or appearing to go towards this point (concave lens), after refraction through the lens, become parallel to the principal axis.

PSEB Solutions For Class 10 Physics Chapter 1 First Principal Focus (F1) And First Focal Length (f)

2. Second principal focus (F2) and second focal length (f’)– It is a fixed point on the principal axis such that the light ravs incident parallel to the principal axis, after refraction through the lens, either converge to this point (in convex lens) or appear to diverge from this point (in concave lens).

PSEB Solutions For Class 10 Physics Chapter 1 Second Principal Focus (F2) And Second Focal Length (f')

Important Note: If the medium on both sides of a lens is same, then first and second focal lengths are equal, i.e, f = f.

Rules For Image Formation By Lens

PSEB Solutions For Class 10 Physics Chapter 1 Rules For Image Formation By Lens

Simple Trick To Remember Image Formation By Convex Lens

Step 1: Write position of object starting from infinity till between F and P.

Step 2: Write numbers 1-6 to all positions of object, and numbers 5-1 to all positions of image as shown in figure.

Step 3: Comparing the numbers will give us the size of image, For Example. Since 1 is very less than 5, size of image will be highly diminished, 2 is less than 4, therefore image will be diminished, and so on (as given in table below).

Note:

  • For object position numbers 5-1, image will be real and inverted.
  • Last position of object is an exception- in this case image will be virtual and erect.

This trick will become clear from illustration given below:

PSEB Solutions For Class 10 Physics Chapter 1 Image Formation By Convex Lens

Simple Trick To Remember Image Formation By Concave Lens

Important Note: Image formed by Concave Lens is always Virtual. Erect and Small in size.

PSEB Solutions For Class 10 Physics Chapter 1 Image Formation By Concave Lens

Lens Formula – It gives the relation between distance of object (u), distance of image (v) and focal length (f) for a lens. According to lens formula,

∴ \(\frac{1}{v}-\frac{1}{u}=\frac{1}{f}\)

Note: Lens formula is valid in all situations for all types of spherical lens.

Magnification Of Lens – It is defined as the ratio of the height of an image (hi) to the height of an object (h0). It is also written as ratio of distance of image from optical centre and distance of object from optical centre, i.e,

∴ \(m=\frac{h_i}{h_o}=\frac{V}{u}\)

Simple Trick To Find The Type or Size Of Image From Value Of Magnification

PSEB Solutions For Class 10 Physics Chapter 1 Types Of Image From Value Of Magnification

Power Of Lens (P)– Power (P) of a lens gives the degree of convergence or divergence of light rays achieved by a lens. Power of lens is the reciprocal of its focal length.

∴ \(\text { Power }=\frac{1}{\text { focal length (in meter) }} \mathrm{P}=\frac{1}{f}\)

SI unit of power of a lens is ‘dioptre’ (D). 1 dioptre is the power of a lens having focal length of lm, i.e., ID = lm-1.

PSEB Class 10 Science Chapter 1 Solutions

Points To Remember About Power Of Lens

  1. Power of a Convex Lens = Always positive
  2. Power of Concave Lens = Always Negative.
  3. Opticians use of the term power of lens instead of their focal lengths.
  4. In many optical instruments, number of lenses is combined to increase the magnification and sharpness of the image. In such instruments, the net power (P) of the lenses is given by the algebraic sum of the individual powers PI, P2, P3, … i.e, P = PI + P2 + P3 + …..

Chapter 1 Light-Reflection And Refraction Reason – Assertion Questions And Answers

The following questions consist of two statements- Assertion (A) and Reason (R). Answer these questionsby  selecting the appropriate option given below:

  1. Both A and R are true and R is the correct explanation of A.
  2. Both A and R are true but R is not the correct explanation of A.
  3. A is true but R is false.
  4. A is false but R is true.

Question 1. Assertion (A): Large concave mirrors are used to concentrate sunlight to produce heat in solar cookers.

Reason (R): Concave mirror converges the light rays falling on it to a point.

Answer: 1. Both A and R are true and R is the correct explanation of A.

Question 2. Assertion (A): The focal length of the mirror is f and the distance of the object from the focus is u. The magnification of the mirror is f/u.

Reason (R): Magnification = + Size of image/ Size of object

Answer: 3. A is true but R is false.

Question 3. Assertion (A): For observing traffic at back, the driver mirror is a convex mirror.

Reason (R): A convex mirror has a much larger field of view than a plane mirror.

Answer: 1. Both A and R are true and R is the correct explanation of A.

Question 4. Assertion (A): Higher is the refractive index of a medium or denser the medium, lesser if the velocity of light in that medium.

Reason (R): Refractive index is inversely proportional to velocity.

Answer: 1. Both A and R are true and R is the correct explanation of A.

Question 5. Assertion: The image formed by a concave mirror is certainly real if the object is virtual.

Reason: The image formed by a concave mirror is certainly virtual if the object is real.

Answer: 3. A is true but R is false.

Question 6. Assertion: The bending of a stick appears to take place by different amounts in different liquids.

Reason: Light is refracted in different media by different amount.

Answer: 1. Both A and R are true and R is the correct explanation of A.

Question 7. Assertion: A stick partly immersed in water appears to be bent.

Reason: Refraction of light when it passes from water into air.

Answer: 1. Both A and R are true and R is the correct explanation of A.

Question 8. Assertion: Refractive index of glass with respect to air is different for red light and violet light.

Reason: Refractive index of a pair of media depends on wavelength of light used.

Answer: 1. Both A and R are true and R is the correct explanation of A.

Question 9. Assertion: When objects are observed through hot air, they appear to be moving slightly.

Reason: The hotter air is optically denser and the colder air is optically rarer.

Answer: 3. A is true but R is false.

Chapter 1 Light-Reflection And Refraction Case Or Source Based Questions And Answers

Question 1. Nalini, a young student, was trying to demonstrate some properties of light in her Science project work. She kept X inside the box (as shown in the figure) and with the help of a laser pointer made light rays pass through the holes on one side of the box. She had a small butterpaper screen to see the spots of light being cast as they emerged.

PSEB Solutions For Class 10 Physics Chapter 1 A Small butter-paper screen To See The Spots Of Light Being Castas They Emerged

1) What could be the ‘X’ that she placed inside the box to make the rays behave as shown?

  1. A converging lens
  2. A parallel-sided glass block
  3. A plane mirror
  4. A triangular prism

Answer: 2. A parallel-sided glass block

2) She measured the angles of incidence for both the rays on the left side of the box to be 48.60. She knew the refractive index of the material ‘X’ inside the box was 1.5. What will be the approximate value of angle of refraction?

  1. 450
  2. 400
  3. 300
  4. 600 (use the value: sin 48.60 = 0.75)

Answer: 3. 300

3) Her friend noted the following observations from this demonstration:

  1. Glass is optically rarer than air.
  2. Air and glass allow light to pass through them with the same velocity.
  3. Air is optically rarer than glass.
  4. Speed of light through a denser medium is faster than that of a rarer medium.

The ratio: sin of angle of incidence in the first medium to the ratio of sin of angle of refraction in the second medium, gives the refractive index of the second material with respect to the first one.

Which one of the combination of the above statements given below is correct.

2, 4 and 5 are correct.

3 and 4 are correct.

1, 4 and 5 are correct.

3 and 5 are correct.

Answer: 4. 3 and 5 are correct.

4) If the object inside the box was made of a material with a refractive index less than 1.5 then the

  1. Lateral shift of the rays would have been less.
  2. Lateral shift of the rays would have been more.
  3. Lateral shift of the rays would remain the same as before.
  4. There is not enough information to comment on any of the above statements.

Answer: 1. Lateral shift of the rays would have been less.

Question 2. A concave mirror has focal length 10 cm. Answer the questions using the following table:

PSEB Solutions For Class 10 Physics Chapter 1 A Concave Mirror Has Focal Length 10 cm

1) If object size is 2 cm, what would be the size of the image in case A?

  1. 2 cm
  2. 10 cm
  3. 4 cm
  4. Less than 2 cm

Answer: 1. 2 cm

2) Nature of the image when u = -15 cm

  1. Virtual and inverted
  2. Real and erect
  3. Virtual and erect
  4. Real and inverted

Answer: 4. Real and inverted

3) Nature of the image in case (D)

  1. Virtual and inverted
  2. Real and erect
  3. Virtual and erect
  4. Real and inverted

Answer: 3. Virtual and erect

4) Position of the image in case (C)

  1. At centre of curvature
  2. At focus
  3. At infinity
  4. Between focus and centre of curvature

Answer: 2. At focus

5) What is the distance at which the centre of curvature located?

  1. 5 cm
  2. 10 cm
  3. 20 cm
  4. 15 cm

Answer: 3. 20 cm

Question 3. An object is placed at the following distances from a convex lens of focal length 15 cm:

  1. 35 cm
  2. 30 cm
  3. 20 cm
  4. 10 cm.

Which position of the object will produce:

1) A magnified real image?

  1. 35 cm
  2. 30 cm
  3. 20 cm
  4. 10 cm

Answer: 3. 20 cm

Here, f = 15cm and 2f = 30cm

Because a magnified real image is formed when the object is placed between f and 2f.

2) A magnified virtual image?

  1. 35 cm
  2. 30 cm
  3. 20 cm
  4. 10 cm

Answer: 4. 10 cm

Because a magnified virtual image is formed when the object is placed between f and the lens.

3) A diminished real image?

  1. 35 cm
  2. 30 cm
  3. 20 cm
  4. 10 cm

Answer: 1. 35 cm

Because a diminished real image is formed when the object is placed beyond 2f.

4) An image of same size as the object?

  1. 35 cm
  2. 30 cm
  3. 20 cm
  4. 10 cm

Answer: 2. 30 cm

Because an image of same size as the object is formed when the object is placed at 2f.

5) At what distance will the image be formed when an object is placed at 30 cm for this converging lens?

  1. 10 cm
  2. 20 cm
  3. 25 cm
  4. 30 cm

Answer: 4. 30 cm

When an object is placed at 2F of converging lens the image will be formed at 2F’ at right side of lens and image formed is real, inverted and of same size.

Question 4. The figure given below illustrates the ray diagram for the formation of image by a concave mirror. The position of the is beyond the centre of the curvature of the concave mirror. On basis of the given figure answer the questions given below.

PSEB Solutions For Class 10 Physics Chapter 1 The Formation Of Image By A Concave Mirror

1) If focal length of the concave mirror is 10 cm, the image formed will be at a distance _______.

  1. Between 10 cm and 15 cm
  2. Between 10 cm and 20 cm
  3. Beyond 20 cm
  4. At 20 cm

Answer: 2. Between 10 cm and 20 cm

Explanation: The focal length of the mirror is 10 cm. This mean, the radius of curvature is 20 cm. From figure, it is clear that the image is formed between the focus and centre of curvature. Thus, the image is formed between 10 cm and 20 cm.

2) In case of concave mirror, the image distance from the pole of the mirror is

  1. Always positive
  2. Always negative
  3. Negative or positive depending upon the position of the object
  4. None of these

Answer: 3. Negative or positive depending upon the position of the object.

3) If the size of the object in the given figure is 5 cm and the magnification produced is – 0.5. The size of the image is

  1. -2.5 cm
  2. -0.1 cm
  3. 2.5 cm
  4. 0.1 cm

Answer: 1. -2.5 cm

= h2/h1; h2 = (- 0.5 x 5)/ 10; h2 =- 2.5 cm

4) A negative sign in the magnification value indicate that the image is

  1. Real and inverted
  2. Real and erect
  3. Virtual and erect
  4. Virtual and inverted

Answer: 1. Real and inverted

5) If the value of magnification is greater than 1 then it indicates that the image formed is

  1. Diminished
  2. The same size as that of the object
  3. Enlarged
  4. Value of magnification cannot specify whether the image is diminished or magnified.

Answer: 3. Enlarged

Question 5. When a spherical mirror is held towards the sun and its sharp image is formed on a piece of carbon paper for some time, a hole is burnt in the carbon paper. Answer the following questions in reference to the above activity.

1) What is the nature of spherical mirror?

  1. Convex mirror
  2. Concave mirror
  3. Plane mirror
  4. Plano convex mirror

Answer: 2. Concave mirror

2) Why is a hole burnt in the carbon paper?

  1. Sun rays are dispersed by the spherical mirror
  2. The Sun’s heat rays are concentrated at the point of sun’s image
  3. Sun rays get diverged after reflection from spherical mirror
  4. Due to atmospheric refraction

Answer: 2. The Sun’s heat rays are concentrated at the point of sun’s image

3) At which point of the spherical mirror the carbon paper is placed?

  1. Between pole and focus
  2. Between centre of curvature and focus
  3. Anywhere between infinity and focus
  4. At focus

Answer: 4. At focus

4) What name is given to the distance between spherical mirror and carbon paper?

  1. Image distance
  2. Object distance
  3. Focal length
  4. Principal axis

Answer: 3. Focal length

5) For the above-mentioned spherical mirror, the image formed by it when the object is placed at its centre of curvature is

  1. Twice the size of the object
  2. Greater than size of object
  3. Equal to the size of the object
  4. Less than the size of the object

Answer: 3. Equal to the size of the object