SAQ for Light Reflection and Refraction Class 10 Science NCERT

Answer

Here, h₁ = +4 cm, f = -10 cm, u = -15cm, v = ?, h₂ = ?

1/f = 1/v + 1/u

⇒ 1/v = 1/f – 1/u

⇒ 1/v = 1/-10 cm - 1/-15 cm

∴ v = -30 cm

h₂/h₁ = - v/u

∴ h₂ = - v/u × h₁

= -(-30 cm/-15 cm) × 4 cm

 = 8 cm

Detailed Answer:

Using, the mirror formula,

1/f = 1/v + 1/u

⇒ 1/v + 1/-15 = 1/-10

⇒ 1/v = 1/-10 + 1/15

⇒ 1/v = (-15 + 10)/150

= -5/150

⇒ v = -30 cm

Thus, to obtain a sharp image of the object the screen should be placed in front of the mirror at a distance of 30 cm from the mirror.

m = -v/u = h_i/h_o

⇒ m = - (-30/-15) = -2

⇒ -2 = h₁/4

⇒ h_i = - 8 cm

Hence, the height of image will be 8 cm.

Answer

Given, u = - 18 cm, f = -12 cm, v = ?

1/f = 1/v + 1/u

⇒ 1/v + 1/-18 = 1/-12

⇒ 1/v = - 1/12 + 1/18

⇒ 1/v = (-3 + 2)/36

= -1/36

⇒ v = -36 cm

Magnification, m = -v/u

Also, m = h_i/h_o

m = - (-36/-18) = - 2

⇒ -2 = h_i/3

⇒ h_i = - 6 cm

Hence, the height of image will be 6 cm.

Answer

The nature of the mirror is concave since the image formed is real.

Given, u = -30 cm, v = -60 cm, h = -2.4 cm,

Using mirror formula,

1/f = 1/v + 1/u

⇒ 1/f = -1/60 + (1/-30)

= - 1/60 – 1/30

= -3/60

= - 1/20

Therefore, f = -20 cm

Magnification, m = v/u = -(Height of image)/(Height of object)

v/u = -(-60/-30) = h’/2.4

⇒ h’ = 60/-30 × 2.4 = -4.8 cm

The required height of the image is -4.8 cm

The image formed by the mirror is inverted.

Answer

Convex mirror

Convex mirrors can be used as rear view mirror in vehicles because they are used always give erect image with large field of view.

Answer

Object position: At C (Centre of curvature)

Object distance = 40 cm,

Position of the image- at infinity,

Reason: Focal length of the mirror = 20 cm,

If the object is moved 20 cm towards the mirror then its new position would be at the focus of the mirror.

Answer

(i) Concave mirror

(ii) u = -20 cm, v = -80 cm and m = ?

m = -v/u = -(-80 cm)/(-20 cm) = -4

(iii) v – u = -60 cm

(iv)

Answer

M = v/u

- 1 = -(-30)/u

u = -30 cm 

Using mirror formula: u = - 30 cm, v = -30 cm

1/f = 1/v + 1/u

⇒ 1/f = 1/-30 + 1/-30

⇒ f = -15 cm

Now, u’ = - 10 cm, f = -15 cm

Using mirror formula;

1/v’ = 1/f – 1/u’

⇒ v’ = +30 cm

Nature of the image is virtual and erect.

Answer

H = + 5 cm; u = - 30 cm; f = + 15 cm; v = ?

1/f = 1/v + 1/u

∴ 1/v = 1/f – 1/u = 1/(+15) - 1/(-30)

= 1/15 + 1/30 

= (2 + 1)/30

= 3/30

= 1/10

∴ v = + 10 cm

h₂/h₁ = -v/u

⇒ h₂ = -(+10)/(-30) × (+5)

= +5/3

= +1.67 cm

Nature = virtual, erect

Answer

Answer

(i) He should use a concave mirror, as it forms a real image on the same side of the mirror.

(ii) Object distance, u = -15 cm

Image distance, v = - 60 cm

Magnification, m = -v/u = -(-60)/(-15) = -4 cm

The minus sign in magnification shows that the image formed is real and inverted.

(iii) The image is formed at a distance of 45 cm from the object.

(iv)

In this case, the image is formed beyond the centre of curvature. This image is real, inverted and enlarged.

Answer

The type of a mirror is convex mirror.

Convex mirror can be used as rear – view mirrors in automobiles because it gives a wider field of view as the mirror is curved outward. It produces erect and diminished image of the traffic behind the driver of the vehicle.

Answer

(i) He should use a concave mirror, as it forms a real image on the same side of the mirror.

(ii) Object distance, u = -12 cm

Image distance, v = -48 cm

Magnification, m = -v/u = (-48)/(-12) = -4 cm

The minus sign in magnification shows that the image formed is real and inverted.

(iii) The image is formed at a distance of 36 cm from the object.

(iv)

In this case, the image is formed beyond the centre of curvature. This image is real, inverted and enlarged.

Answer

(i) m = - v/u; u = -10 cm

- 3 = -v/-10

⇒ v = + 3 × -10 = -30 cm

⇒ -1/30 – 1/10 = 1/f

⇒ f = -7.5 cm

(ii)

Answer

If θ is a submultiple of 180°, then the number of images formed,

n = 360°/θ – 1

If θ is not a submultiple of 180, then the number of images formed is the integer next higher than = 360°/θ – 1

Answer

(i) (a) Shaving mirror – Concave mirror

(b) Rear view mirror – Convex mirror.

(c) Reflector in search – lights – Concave mirror

(ii) (a) Real image can be obtained on screen but virtual image cannot be obtained.

(b) Reflected/Refracted rays actually met where real image is formed while for virtual they only appear to meet.

(c) Real image is always inverted while virtual image is always erect.

Answer

(i) As magnification is negative, the image formed is real.

Hence, it is a concave mirror.

(ii) m = - v/u = - 1

∴ u = v = - 50 cm

Distance of the image from the object;

v – u = - 50 – (- 50) = 0 cm

(iii) By using mirror formula:

1/f = 1/v + 1/u

= 1/(-50) + 1/(-50) = -1/25

∴ f = -25 cm

(iv)

Answer

(i) Object should be placed at < 40 cm (less than 40 cm) in front of mirror, i.e., between focus and pole, as concave mirror forms a virtual, erect and magnified image when object is placed between focus and pole.

(ii) Image will be bigger than the object.

(iii)

Answer

Type of mirror used in:

(i) Headlights of a car: Concave mirror

Concave mirror is used because light from the bulb placed at the focus of it gets reflected and produces a powerful parallel beam of light to illuminate the road.

(ii) Rear view mirror of vehicles: Convex mirror

Convex mirror is used because it always produces a virtual and erect image whose size is smaller than the object. Therefore, it enables the driver to see wider field of view of traffic behind the vehicle in a small mirror.

(iii) Solar furnace: Concave mirror

Concave mirror has the property to converge the sunlight coming from sun along with heat radiation at its focus. As a result, temperature at its focus increases and the substance placed at the focal point gets heated to a high temperature.

Answer

(i) A convex mirror always forms an erect, virtual and diminished image of an object placed anywhere in front of it. Thus, convex mirrors enable the driver to view much larger traffic behind him that would not be possible with a plane mirror.

(ii) A concave mirror is used as a shaving or make-up mirror because it forms an erect and enlarged image of the face when it is held closer to the face.

Answer

Answer

(a) The relative extent to which the image of an object is magnified with respect to object size. It is the ratio size of the image to the size of the object.

(b) v = -40 cm, u = -20 cm, f = ?

1/f = 1/v + 1/u

⇒ 1/f = 1/-40 + 1/-20

⇒ f = -40/3 cm

Answer

We will look our face in each mirror, turn by turn.

(i) If the image formed is of same size as our face but laterally inverted for all positions, then it is a plane mirror.

(ii) If the image formed is erect and enlarged initially but gets inverted as the fact is moved away, then it is a concave mirror.

(iii) If the image formed is erect and smaller in size for all positions, then it is a convex mirror.

Answer

(i) Such a ray falls normally on the mirror. So its angle of incidence is 0°.

(ii) The concave mirror with smaller aperture forms the sharper image because it is free from spherical aberration.

(iii) No change. The focal length of a convex mirror does not depend on the nature of the medium.

Answer

(i) There are two laws of refraction :

(a) The ratio of the sine of the angle of incidence to the sine of the angle of refraction is constant. This is known as Snell’s law. Mathematically, it can be expressed as:

s_in_i/s_in_r = n₁₂

Here, n₁₂ is the relative index of medium 1 with respect to medium 2.

(b) The incidence ray, the refracted ray and the normal to the interface of two media at the point of incidence lie on the same plane.

Given,

c = 3 × 10⁸ m/s 

v = 1.4 × 10⁸ m/s

(ii) Absolute refractive index = (Speed of light in vacuum/Speed of light in medium)

= (3 × 10⁸ m/s)/(1.4 × 10⁸ m/s)

= 2.14

Answer

Given,

Speed of light in vacuum = 3 × 10⁸ m/s

Refractive index of the medium = (Speed of light in vacuum)/(Speed of light in medium)

⇒ = 1.5 = (3 × 10⁸)/v

⇒ v = (3 × 10⁸)/1.5

= 2 × 10⁸ m/s

Hence, the speed of light in the medium of refractive index 1.5 × 2 × 10⁸ m/s.

Answer

(i) Ratio of height of the image to height of the object in magnification.

m (mirror) = h’/h = -v/u

where,

m = magnification

h = height of object

h’ = height of image

v = Distance of image

u = Distance of object

(ii) It means image formed is two times the size of object and the image formed is inverted, formed in front of the lens. 

Answer

(i) the ability of a lens to converge/diverge a beam of light rays is expressed in terms of its power (P). It is inverse of focal length, f (in metre).

Hence, power of a lens is given by the relation P = 1/f(in meters)

F = 1/P = + ½ = + 0.5 m

(ii) Convex lens

Answer

Answer

If image formed by a lens is always erect & diminished for all value of ‘u’ therefore, the lens is concave lens.

Object At any finite distance

Object at infinity

We know that power of a concave lens is negative

∴ P = -10D

P = 1/f

f = 1/P = -1/10

= -0.1 m or – 10 cm

Answer

1^st law: The incident ray, refracted ray and normal to the interface at the point of incidence lie in the same plane.

2^nd Law: The sine of angle of incidence bears a constant ratio with sine of angle of refraction for a given pair of media. Or, s_in_i/s_in_r = constant

Absolute refractive index of a medium, = {speed of light in air (c) or vacuum/speed of light in medium (v)}

Answer

Power of lens = Ability to converge/diverge light rays passing through it/reciprocal of the focal length in metres/ 1/f (in meters),

SI unit of power is Dioptre

Power of 1^st lens P₂ = 100/f = 100/-20 cm = -5D

Nature : Diverging lens/Concave lens.

Power of 2^nd lens P₂ = 100/f = 100/-20 cm = -5D

Nature : Diverging lens/Concave lens

Answer

Answer

u = - 12 cm, f = + 8 cm, v = ?

1/v – 1/u = 1/f

⇒ 1/v – 1/(-12) = 1/8

⇒ v = +24 cm

Image is real and inverted.

Answer

m = -2

v = -30 cm

u = ?

m = -v/u

⇒ 2 = (-30)/u

⇒ u = -30/2

⇒ u = -15 cm

The object distance is 15 cm

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

-1/30 – 1/15 = (- 15 - 30)/450

= -45/450

= 1/10

= 1/f

F = - 10 cm

The focal length is 10 cm.

If the object is moved 10 cm towards the mirror the image will be of the nature virtual and erect and the size will be enlarged.

Answer

(i) Diagram

Marking ∠r and x

(ii) _an_g = 3/2

∴ _gn_a = 1/_an_g = 1/(3/2) = 2/3

Alternately, C_air/C_glass = 3/2

∴ C_glass/C_air = 2/3

Answer

Diverging lens/concave lens

 Focal length = -20 cm (lens is concave, hence f is –ve)

Power = P

= 1/f

= 100/-20 cm

= -5D

Answer

f = - 10 cm, u = - 20 cm, v = ?

1/v = 1/f + 1/u

= 1/-10 + 1/(-20)

= -2 -1/20

= -(3/20)

v = -20/3 cm

m = h₁/h₀ = -v/u

= h₁ = -v/u × h₀

= -(-20)/(3 × 20) × -5

= 5/3

= 1.6 cm

Image is virtual and diminished.

Answer

Power of a lens is the degree of convergence of light rays achieved by a lens.

Lens L₁ : f₁ = 100/P₁ = 100/+10 = + 10 cm; Convex lens

Lens L₂ : f₂ = 100/P₂ = 100/+5 = +20 cm: Convex lens

Lens L₃ : f₃ = 100/P₃ = 100/-10 = -10 cm; Concave lens L₂ will from a virtual and magnified image of an object placed at 15 cm from the convex lens because concave lens can never from virtual and magnified image of an object and convex lens form such image only when the object is placed between the optical centre and principal focus of the convex lens.

Answer

(a) Convex lens

(b) Focal length of the lens is 2 m.

Distance of candle flame from the lens is 4 m.

(c) Ray Diagram :

Answer

Given,

Object distance, u = -40 cm

Image distance, v = 40 cm

Using lens formula, 1/f = 1/v – 1/u = 1/40 – 1/(-40)

Or Focal length,

f = +20 cm 

The Positive sign of focal length shows that it is a convex lens.

Thus, the image is real, inverted and same size as object.

If candle flame is shifted 25 cm towards the lens, then;

u = -(40 – 25) = -15 cm

then, 1/20 = 1/v – 1/(-15) = 1/v + 1/15

or, v = -60 cm

 m = v/u = -60/-15

= 4

Thus the image will be virtual, erect and enlarged. For this you can refer to the following diagram,

Answer

(a) (i) P = 1/f, f = 100/5 = 20 cm

Object at 18 cm, 22 cm, and 30 cm, image can be magnified.

(ii) At 22 cm and 30 cm, image can be obtained on a screen.

(b) Film projectors and telescopes.

Answer

f = 100 cm

m = v/u = 2

v = 2u

1/f = 1/v – 1/u

⇒ 1/100 = 1/2u – 1/u

⇒ 1/100 = (1-2)/2u

⇒ 1/100 = - 1/2u

∵ 2u = -100

∵ u = -50 cm

50 cm in front of lens.

Answer

Image with magnification -1 means image is inverted and of the same size.

Therefore. Object is at 2F and the image is also at 2F on the other side of the lens.

Therefore. Distance between the object and its image is 4f = 60 cm.

⇒ f = 15 cm

Object distance 2f = 30 cm, if the object is shifted towards the lens by 20 cm, the new object distance.

= 30 cm – 20 cm

= 10 cm

This distance is less than the focal length, and the image formed in this case would be virtual, erect and will form on the same side as the object.

Answer

A concave lens always forms a virtual and erect image on the same side of the object.

Image distance, v = ?

Focal length, f = - 5 cm

Object distance u = -10 cm

1/f = 1/v – 1/u

⇒ 1/v = 1/f + 1/u

= 1/-5 + 1/-10

= 1/-5 – 1/10

= (-1 - 2)/10

= -3/10 or, v = 3.3 cm.

v = - 3.3 cm

(Size of the image)/(Size of the object) = + v/u

h’/h = (-3.3)/(-10)

⇒ h’/6 = 3.3/10

h₁ = (6 × 3.3)/10

= 19.8/10

= 1.98 cm.

Size of the image is 1.98 cm.

Answer

(a) (i) Convex lens = + 50 cm

(ii) Concave lens = 25 cm

(b) 1/f = 1/v – 1/u

 f = 1/-4D = -25 cm 

u = - 100 cm

1/v = 1/f + 1/u

= -1/25 - 1/100

= -5/100

= -1/20

v = -20 cm

Answer

(i) ∠i = 0°, ∠r = 0°

(ii) n_x = c/v_x = (3 × 10⁸)/(1.5 × 10⁸) = 2

Answer

f = 20 cm, u =?, m = 2

m = v/u

⇒ 2 = v/u

⇒ v = 2u

1/f = 1/v – 1/u

⇒ 1/20 = 1/2u – 1/u

⇒ u = -10 cm

⇒ v = -20 cm

Answer

f = - 15 cm, v = - 10 cm, u = ?

1/v – 1/u = 1/f

⇒ 1/-10 – 1/u = 1/-15

⇒ 1/u = -1/10 + 1/15

⇒ u = -30 cm.

Magnification, m = v/u = (-10/-30)

Hence, m = 1/3

Answer

(i)

n₁ > n₂, lens behaves as diverging lens.

(ii)

n₁ = n₂, no refraction occurs.

Answer

The image formed is real (convex lens forms diminished real image)

m = -2/3, v = +12 cm, m = v/u = -2/3

- 2u = 36

⇒ u = -18 cm

⇒ 1/f = 1/v – 1/u

⇒ 1/f = 1/12 – 1/18

= (3 + 2)/36

⇒ f = 36/5 = 7.2 cm

Answer

When n₁ > n₂

Light goes from rarer to denser medium

∴ It diverges.

(ii) When n₁ = n₂

There is no change in the medium

∴ no bending or refraction occurs.

(iii) When n₁ < n₂

Light goes from denser to rarer medium

∴ It converges.

Answer

(a) u = 30 cm, v = 30 cm

This is possible if the object is placed at 2f.

∴ 2f = 30 cm, f = 15 cm

(b) u = 15 cm, v = 70 cm is incorrect. This is because if the object is at focus then image is formed at infinity.

(c) In (iii) case, because object is at centre of curvature.

Answer

(i) Refractive index of alcohol > refractive index of water.

So, alcohol is optically denser than water.

When a ray of light enters from water to alcohol, it bends towards the normal.

(ii) _gμ_D = _aμ_D/_aμ_g = 2.24/1.5

= 1.61

Answer

Answer

Nature of spherical lens = Convex

Given u = - 45 cm, v = + 90 cm, h₁ = + 2 cm

Using lens formula:

1/f = 1/v – 1/u

= 1/90 – 1/-45

= 1/90 + 1/45

= (1 + 2)/90

= 3/90

= 1/30

⇒ f = + 30 cm

Again m = h₂/h₁ = v/u

h₂ = v/u × h₁

 = 90/-45 × 2

 = -4 cm (inverted image)

So, height of image is -4cm. Negative sign indicates that it is formed below the principal axis.