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CBSE Class 10 Science – Lens Formula Solution

Given:

Object height, h₁ = 5 cm
Object distance, u = −30 cm
Focal length of lens, f = +20 cm

Formula Used:

Lens formula:
1/v − 1/u = 1/f

Solution:

Substituting the given values in lens formula:

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

1/v = 1/20 − 1/30

Taking LCM (60):

1/v = (3 − 2) / 60
1/v = 1/60

∴ v = +60 cm

Calculation of Image Size:

Magnification, m = v / u
m = 60 / (−30) = −2

Image height, h₂ = m × h₁
h₂ = (−2) × 5 = −10 cm

Position of image = +60 cm

Size of image = 10 cm (Inverted)

Given:

Height of bus, h₁ = 3 m
Object distance, u = −6.0 m
Focal length of convex mirror, f = +1.5 m

Formula Used:

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

Solution:

Substituting the given values:

1/v + 1/(−6) = 1/1.5

1/v − 1/6 = 2/3

1/v = 2/3 + 1/6

Taking LCM (6):

1/v = (4 + 1)/6
1/v = 5/6

∴ v = 6/5 = +1.2 m

Calculation of Image Size:

Magnification, m = −v/u

m = −(1.2)/(−6) = 0.2

Image height, h₂ = m × h₁
h₂ = 0.2 × 3 = 0.6 m

Position of image: +1.2 m (behind the mirror)

Size of image: 0.6 m (erect and diminished)

Given:

Focal length of concave mirror, f = −18 cm
Magnification, m = +2

Formula Used:

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

Magnification:
m = −v/u

Solution:

From magnification formula:

+2 = −v/u

∴ v = −2u

Substituting v = −2u in mirror formula:

1/(−2u) + 1/u = 1/(−18)

−1/(2u) + 1/u = −1/18

(−1 + 2)/(2u) = −1/18

1/(2u) = −1/18

∴ u = −9 cm

Final Answer:
The object should be placed at a distance of 9 cm in front of the mirror, i.e., between the pole and the focus of the concave mirror, to obtain a virtual and magnified image.

Given:

Height of object, h₁ = 5 cm
Object distance, u = −30 cm
Focal length of converging lens, f = +20 cm

Formula Used:

Lens formula:
1/v − 1/u = 1/f

Magnification:
m = v/u

Solution:

Substituting the given values in lens formula:

1/v − 1/(−30) = 1/20

1/v + 1/30 = 1/20

1/v = 1/20 − 1/30

Taking LCM (60):

1/v = (3 − 2)/60
1/v = 1/60

∴ v = +60 cm

Calculation of Image Size:

Magnification, m = v/u

m = 60/(−30) = −2

Image height, h₂ = m × h₁
h₂ = (−2) × 5 = −10 cm

Position of image: +60 cm from the lens (on the other side)

Size of image: 10 cm (inverted)

Given:

Object distance, u = −20 cm
Image distance, v = −10 cm

Formula Used:

Lens formula:
1/v − 1/u = 1/f

Power of lens:
P = 1/f (in metre)

Solution:

Substituting the given values in lens formula:

1/(−10) − 1/(−20) = 1/f

−1/10 + 1/20 = 1/f

Taking LCM (20):

(−2 + 1)/20 = 1/f

−1/20 = 1/f

∴ f = −20 cm

Calculation of Power of Lens:

Focal length in metre, f = −0.20 m

P = 1/(−0.20) = −5 D

Final Answer:
(i) Focal length of the lens = −20 cm
(ii) Power of the lens = −5 dioptre

Given:

Height of object, h₁ = 2.0 cm
Height of image, h₂ = 8.0 cm
Object distance, u = −6.0 cm

Formula Used:

Magnification:
m = h₂ / h₁ = v / u

Lens formula:
1/v − 1/u = 1/f

Solution:

(i) Nature of image:

Since the object and image are on the same side of the convex lens, the image is virtual and erect.

(ii) Position of image:

Magnification,

m = h₂ / h₁ = 8 / 2 = 4

m = v / u

4 = v / (−6)

∴ v = −24 cm

(iii) Focal length of the lens:

Substituting values in lens formula:

1/(−24) − 1/(−6) = 1/f

−1/24 + 1/6 = 1/f

Taking LCM (24):

(−1 + 4)/24 = 1/f

3/24 = 1/f

∴ f = +8 cm

Final Answer:
(i) Nature of image: Virtual, erect and magnified
(ii) Position of image: 24 cm from the lens on the same side as the object
(iii) Focal length of the lens: +8 cm

(a) Focal length of the convex lens:

From the table, when the object distance is −30 cm, the image distance is +30 cm.

This shows that the object is placed at 2F and the image is formed at 2F.

Therefore, the focal length of the convex lens is 15 cm.

(b) Incorrect set of observation:

The 8^th observation (u = −12 cm, v = +120 cm) is not correct.

Reason: When the object is placed within the focal length of a convex lens, the image formed is virtual and erect and cannot be obtained on a screen.

Since the image is shown as formed on the screen, this observation is incorrect.

(c) Ray diagram (description):

Consider the correct observation where the object is placed at −30 cm:

Draw a ray diagram showing (Do Your Self):

The image formed at 2F on the other side of the lens, real and inverted

An object placed at 2F in front of a convex lens

One ray parallel to the principal axis passing through the focus after refraction

Another ray passing through the optical centre without deviation

(a) Defect of vision:

The person is suffering from hypermetropia (long-sightedness).

(b) Causes of this defect:

Two causes of hypermetropia are:

• Increase in focal length of the eye lens due to weak ciliary muscles.
• Eyeball being shorter than normal.

(c) Focal length of the lens used:

Power of lens, P = +2.0 D

Relation between power and focal length:

P = 1/f (where f is in metre)

f = 1/P = 1/2.0 = 0.5 m

Final Answer:
The focal length of the lens used is +0.5 m (or 50 cm).

(a) Defect of vision:

The person is suffering from myopia (short-sightedness).

(b) Causes of this defect:

Two main causes of myopia are:

• Elongation of the eyeball.
• Excessive curvature of the eye lens leading to increased converging power.

(c) Corrective lens used:

A concave (diverging) lens is used to correct myopia.

(d) Ray diagram (description):

To draw the ray diagram:

• Show parallel rays coming from a distant object entering the concave lens.
• The concave lens diverges the rays.
• The diverged rays appear to come from the far point of the myopic eye.
• The eye lens then focuses these rays on the retina.

Thus, the concave lens helps the image to form on the retina instead of in front of it.

(a) Defect of vision:

The person is suffering from hypermetropia (long-sightedness).

(b) Causes of this defect:

Two causes responsible for hypermetropia are:

• The eyeball is shorter than normal.
• The focal length of the eye lens increases due to weak ciliary muscles.

(c) Correction using eye-glasses:

Hypermetropia is corrected by using a convex lens.

The convex lens converges the rays of light coming from a nearby object so that the image is formed on the retina.

(d) Ray diagram (description):

To draw the labelled ray diagram:

• Show an object placed at a normal reading distance.
• Draw rays diverging from the object and passing through a convex lens placed in front of the eye.
• The convex lens converges the rays before they enter the eye.
• The eye lens then focuses the rays on the retina.

Type of glasses: The glasses used are convergent (convex) in nature.

(a) Type of lens and focal length:

Power of lens, P = −0.25 D

Since the power is negative, the lens is a concave (diverging) lens.

Relation between power and focal length:

P = 1/f (f in metre)

f = 1/(−0.25) = −4 m

Thus, the focal length of the lens is −4 m.

(b) Eye defect corrected by this lens:

A concave lens is used to correct myopia (short-sightedness).

(c) Nature and size of the image formed:

When an object is placed between F and 2F of a concave lens:

The image is formed between the optical centre and the focus on the same side of the lens.

The image formed is virtual.

The image is erect.

The image is diminished in size.

(a) Dispersion of white light by a glass prism:

To draw the diagram:

• Draw a triangular glass prism.
• Show a beam of white light incident on one face of the prism.
• After refraction through the prism, show the white light splitting into seven colours (VIBGYOR).
• Label the colours in order: Violet, Indigo, Blue, Green, Yellow, Orange and Red.
• Show violet deviated the most and red deviated the least.

(b) Spectrum:

The band of seven colours obtained on dispersion of white light is called a spectrum.

Reason for formation of spectrum:

The spectrum is formed because different colours of white light have different wavelengths and hence they travel with different speeds in glass. As a result, they are refracted by different amounts on passing through the prism.

Rainbow:

A rainbow is a natural spectrum of seven colours (VIBGYOR) formed in the sky due to the dispersion, refraction and total internal reflection of sunlight by water droplets present in the atmosphere.

Formation of a rainbow:

A rainbow is formed due to the following processes taking place inside a water droplet:

• Refraction and dispersion of sunlight when it enters the water droplet.
• Total internal reflection of light inside the droplet.
• Refraction again when light comes out of the droplet.

Labelled ray diagram (description):

To draw the labelled diagram:

Label the emergent rays showing red on the top and violet at the bottom.

Draw a circular water droplet.

Show a ray of white sunlight incident on the droplet.

Show refraction and dispersion into different colours at the first surface.

Show total internal reflection inside the droplet.

Show refraction again at the second surface.

(a) Meaning of 1 volt potential difference:

The potential difference between two points is said to be 1 volt if 1 joule of work is done in moving a charge of 1 coulomb from one point to the other.

Mathematically,

V = W / Q
1 volt = 1 joule / 1 coulomb

(b) Circuit symbols and their functions:

(i) Ammeter

• It is used to measure the electric current flowing in a circuit.

(ii) Rheostat (Variable resistor)

It is used to change or control the current in an electric circuit by varying resistance.

Potential Difference:

The potential difference between two points in an electric circuit is defined as the work done in moving a unit charge from one point to the other.

S.I. Unit:

The S.I. unit of potential difference is volt (V).

Definition of 1 volt:

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

Expression in terms of work and charge:

V = W / Q

where,

Q = charge (coulomb)

V = potential difference (volt)

W = work done (joule)

(a) One volt potential difference:

The potential difference between two points in an electric field is said to be one volt if one joule of work is done in moving a charge of one coulomb from one point to the other.

(b) Electric circuit and calculation of current:

Schematic diagram (description):

The circuit consists of:

• A cell of emf 1.5 V
• A 5 Ω resistor
• A 10 Ω resistor
• A plug key

All components are connected in series.

Calculation:

Given:
Voltage of the cell, V = 1.5 V
Resistance R₁ = 5 Ω
Resistance R₂ = 10 Ω

Total resistance, R = R₁ + R₂ = 5 + 10 = 15 Ω

Using Ohm’s law:

V = IR

I = V / R = 1.5 / 15 = 0.1 A

Final Answer:
The current drawn from the cell when the key is closed is 0.1 ampere.

(a) Relationship between resistance and resistivity:

The resistance (R) of a cylindrical conductor of length L and area of cross-section A is given by:

R = ρL / A

where ρ (rho) is the resistivity of the material of the conductor.

S.I. unit of resistivity:

From the relation:

ρ = RA / L

S.I. unit of resistance, R = ohm (Ω)
S.I. unit of area, A = m²
S.I. unit of length, L = m

Therefore, S.I. unit of resistivity:

ρ = Ω × m² / m = Ω m

Thus, the S.I. unit of resistivity is ohm metre (Ω m).

(b) Reason for using alloys in electrical heating devices:

Alloys are used in electrical heating devices because:

They have a high melting point and do not oxidise easily at high temperatures.

They have high resistivity, so they produce more heat.

Given:

Resistance of wire, R = 7 Ω
Radius of wire, r = 0.01 cm = 1 × 10⁻⁴ m
Resistivity of material, ρ = 44 × 10⁻⁶ Ω m

Formula Used:

Resistance of a wire:

R = ρL / A

Area of cross-section,

A = πr²

Solution:

A = π (1 × 10⁻⁴)²
A = π × 10⁻⁸ m²

Substituting values in the formula:

7 = (44 × 10⁻⁶ × L) / (π × 10⁻⁸)

L = (7 × π × 10⁻⁸) / (44 × 10⁻⁶)

L = (7π / 44) × 10⁻²

Taking π ≈ 3.14:

L ≈ (7 × 3.14 / 44) × 10⁻²
L ≈ 0.5 × 10⁻² m

L ≈ 5 × 10⁻³ m

Final Answer:
The length of the wire is 5 × 10⁻³ m (or 0.005 m).

(i) Classification of materials:

On the basis of given resistivity values:

• Material A (10¹⁷ Ω m) → Insulator
• Material B (44 × 10⁻⁶ Ω m) → Alloy
• Material C (1.62 × 10⁻⁸ Ω m) → Conductor

(ii) Example and use of each material:

Conductor (Material C):

• Example: Copper
• Use: Used for making connecting wires in electric appliances like electric iron or stove.

Alloy (Material B):

• Example: Nichrome
• Use: Used as the heating element in electric iron or electric stove.

Insulator (Material A):

Use: Used to insulate electrical components to prevent leakage of current and electric shock.

Example: Mica / Plastic

Given:

Length of wire, L = 1 m
Resistance, R = 35 Ω
Diameter, d = 0.2 mm = 2 × 10⁻⁴ m
Radius, r = 1 × 10⁻⁴ m
π = 22/7

Formula Used:

Resistance of a wire:

R = ρL / A

Area of cross-section:

A = πr²

Solution:

A = (22/7) × (1 × 10⁻⁴)²
A = (22/7) × 10⁻⁸ m²

From R = ρL / A,

ρ = RA / L

ρ = 35 × (22/7 × 10⁻⁸) / 1

ρ = 110 × 10⁻⁸ Ω m

ρ = 1.1 × 10⁻⁶ Ω m

Effect of doubling length and diameter:

Resistivity depends only on the nature of the material and temperature.

Even if the length and diameter of the wire are both doubled, the resistivity remains unchanged.

Final Answer:
Resistivity of the wire = 1.1 × 10⁻⁶ Ω m.
Resistivity does not change on doubling the length and diameter of the wire.

(i) Resistors connected in series:

Arrangement (description):

The three resistors of 2 Ω, 3 Ω and 6 Ω are connected end to end so that the same current flows through each resistor.

Calculation of equivalent resistance:

For series combination:

R_s = R₁ + R₂ + R₃

R_s = 2 + 3 + 6 = 11 Ω

Equivalent resistance in series = 11 Ω

(ii) Resistors connected in parallel:

Arrangement (description):

The three resistors of 2 Ω, 3 Ω and 6 Ω are connected between the same two points so that the potential difference across each resistor is the same.

Calculation of equivalent resistance:

For parallel combination:

1/R_p = 1/R₁ + 1/R₂ + 1/R₃

1/R_p = 1/2 + 1/3 + 1/6

Taking LCM (6):

1/R_p = (3 + 2 + 1)/6 = 6/6 = 1

R_p = 1 Ω

Equivalent resistance in parallel = 1 Ω

Given:

Two resistors of 10 Ω and 15 Ω are connected in parallel.
Two resistors of 60 Ω and 40 Ω are connected in parallel.
These two parallel combinations are connected in series.

Calculation:

Equivalent resistance of 10 Ω and 15 Ω in parallel:

1/R₁ = 1/10 + 1/15

1/R₁ = (3 + 2)/30 = 5/30

R₁ = 6 Ω

Equivalent resistance of 60 Ω and 40 Ω in parallel:

1/R₂ = 1/60 + 1/40

1/R₂ = (2 + 3)/120 = 5/120

R₂ = 24 Ω

Total resistance of the circuit (series combination):

R = R₁ + R₂ = 6 + 24 = 30 Ω

Total resistance of the circuit = 30 Ω

(a) Magnetic field lines through points 1, 2 and 3:

On the given diagram of the current carrying circular loop:

• Draw a magnetic field line passing through point 1 near P, curved around the wire.
• Draw a magnetic field line passing through point 2 at the centre of the loop; it should be nearly straight and perpendicular to the plane of the loop.
• Draw a magnetic field line passing through point 3 near Q, similar in shape to that at point 1 but on the opposite side.

(b) Factors affecting the intensity of magnetic field at the centre of the loop:

• The magnitude of current flowing through the loop.
• The radius (or number of turns) of the circular loop.

(c) Rule used to determine the direction of magnetic field:

The direction of magnetic field produced due to a current carrying straight conductor is determined using the Right-Hand Thumb Rule.

(a) Magnetic field pattern due to a straight current carrying conductor:

Draw a horizontal white board and show a straight conductor passing perpendicular through its centre.

• Mark the conductor at the centre with a ⊗ sign to indicate that the current is flowing vertically downwards.
• Draw several concentric circular magnetic field lines around the conductor on the board.
• Mark arrows on the circular field lines to show their direction.
• Since the current is vertically downward, the magnetic field lines should be shown in the clockwise direction.

Thus, mark:

• (i) Direction of current → vertically downward (⊗)
• (ii) Direction of magnetic field lines → clockwise circles around the conductor

(b) Right Hand Thumb Rule:

The Right Hand Thumb Rule states that if a straight current carrying conductor is held in the right hand such that the thumb points in the direction of current, then the curled fingers give the direction of the magnetic field lines around the conductor.

Verification using the rule:

When the thumb of the right hand points downward (direction of current), the curled fingers move in a clockwise direction.

Hence, the directions of the magnetic field lines marked in the diagram are in accordance with the right hand thumb rule.

(a) Factors affecting the magnitude of force:

The magnitude of force acting on a current carrying conductor placed in a uniform magnetic field depends on:

• The strength of the magnetic field.
• The magnitude of current flowing through the conductor.
• The length of the conductor within the magnetic field.

(b) Condition for maximum force:

The magnitude of force on the conductor is maximum when the conductor is placed perpendicular to the direction of the magnetic field.

(c) Rule to determine the direction of force:

The direction of force acting on the conductor is determined using Fleming’s Left Hand Rule.

One application:

Fleming’s Left Hand Rule is used in the working of an electric motor to determine the direction of motion of the conductor.

(a) Rule to determine the force on a current carrying conductor:

The rule used to determine the direction of force on a current carrying conductor placed in a uniform magnetic field is Fleming’s Left Hand Rule.

Statement:
If the forefinger, thumb and middle finger of the left hand are stretched mutually perpendicular to each other, such that the forefinger points in the direction of the magnetic field and the middle finger points in the direction of current, then the thumb gives the direction of force (motion) acting on the conductor.

(b) Force on a moving positive charge in a magnetic field:

(i) Case of maximum force:

The force experienced by the charge is maximum in case III.

Reason:
The magnetic force is maximum when the velocity of the charge is perpendicular to the direction of the magnetic field.

(ii) Case of minimum force:

The force experienced by the charge is minimum (zero) in case I.

Reason:
When the charge moves parallel to the magnetic field, no magnetic force acts on it.

(a) Importance of the earth wire:

The earth wire is very important in domestic electric appliances because it provides a safe path for leakage current.

If due to insulation failure the live wire comes in contact with the metallic body of the appliance, the earth wire allows the current to flow to the ground, keeping the body of the appliance at zero potential.

This prevents electric shock to the user and also helps in the proper functioning of the fuse.

(b) Precautions to avoid overloading:

Use proper rating fuses and MCBs in the electric circuit.

Do not connect too many electrical appliances to a single socket at the same time.