For distant vision = −0.181 m, for near vision = 0.667 m The power P of a lens of focal length f is given by the relation P = 1/f(in metres) (i) Power of the lens used for correcting distant vision = −5.5 D Focal length of the required lens, f =1/P F = 1/-5.5 = -0.181ₘ The focal length of the lens fRead more

For distant vision = −0.181 m, for near vision = 0.667 m
The power P of a lens of focal length f is given by the relation
P = 1/f(in metres)
(i) Power of the lens used for correcting distant vision = −5.5 D
Focal length of the required lens, f =1/P
F = 1/-5.5 = -0.181ₘ
The focal length of the lens for correcting distant vision is −0.181 m.
(ii) Power of the lens used for correcting near vision = +1.5 D
Focal length of the required lens, f =1/P
F = 1/1.5 = +0.667ₘ

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 metres) 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 For more aRead more

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 metres)
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

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. For more answerRead more

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.

Power of a lens, P = 1/f (in meters) Power, P = 1.5d F = 1/1.5 = 10/15 = 0.66m A convex lens has a positive focal length. Hence, it is a convex lens or a converging lens. For more answers visit to website: https://www.tiwariacademy.com/ncert-solutions/class-10/science/chapter-10/

Power of a lens, P = 1/f (in meters)
Power, P = 1.5d
F = 1/1.5 = 10/15 = 0.66m
A convex lens has a positive focal length. Hence, it is a convex lens or a converging lens.

Power of a lens, P = 1/f (in meters) P = -2 D f = -1/2 = -0.5m A concave lens has a negative focal length. Hence, it is a concave lens. For more answers visit to website: https://www.tiwariacademy.com/ncert-solutions/class-10/science/chapter-10/

Power of a lens, P = 1/f (in meters)
P = -2 D
f = -1/2 = -0.5m
A concave lens has a negative focal length. Hence, it is a concave lens.

Object distance, u = −27 cm Object height, h = 7 cm Focal length, f = −18 cm According to the mirror formula, 1/u + 1/v = 1/f 1/v = 1/f – 1/u = -1/18 + 1/27 = -1/54 V = -54 cm The screen should be placed at a distance of 54 cm in front of the given mirror. Magnification, m = - Image Distance/ObjectRead more

Object distance, u = −27 cm
Object height, h = 7 cm
Focal length, f = −18 cm
According to the mirror formula,
1/u + 1/v = 1/f
1/v = 1/f – 1/u = -1/18 + 1/27 = -1/54
V = -54 cm
The screen should be placed at a distance of 54 cm in front of the given mirror.
Magnification, m = – Image Distance/Object Distance = -54/27 = -2
The negative value of magnification indicates that the image formed is real.
Magnification, m = Height of the image / Height of the object = h’/h
h‘ = 7 x (-2) = -14 cm
The negative value of image height indicates that the image formed is inverted.

Object distance, u = −20 cm Object height, h = 5 cm Radius of curvature, R = 30 cm Radius of curvature = 2 × Focal length R = 2f f = 15 cm According to the mirror formula, 1/v + 1/u = 1/f 1/v = 1/f – 1/u = 1/15 + 1/20 = 4+3/60 = 7/60 V =8.57 The positive value of v indicates that the image is formedRead more

Object distance, u = −20 cm
Object height, h = 5 cm
Radius of curvature, R = 30 cm
Radius of curvature = 2 × Focal length
R = 2f f = 15 cm
According to the mirror formula,
1/v + 1/u = 1/f
1/v = 1/f – 1/u = 1/15 + 1/20 = 4+3/60 = 7/60
V =8.57
The positive value of v indicates that the image is formed behind the mirror.
Magnification, m = – Image Distance/Object Distance = -8.57/-20 = 0.428
The positive value of Magnification indicates that the image is formed is virtual.
Magnification, m = Height of the image / Height of the object = h’/h
h’ m x h = 0.428 x 5 = 2.14 cm
The positive value of image height indicates that the image formed is erect.
Therefore, the image formed is virtual, erect, and smaller in size.

Magnification produced by a mirror is given by the relation Magnification, m = Image height (H₁) / Object height (H0) 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 imaRead more

Magnification produced by a mirror is given by the relation

Magnification, m = Image height (H₁) / Object height (H0)
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.

Focal length of concave lens (OF1), f = −15 cm Image distance, v = −10 cm According to the lens formula, 1/v – 1/u = 1/f 1/u = 1/v – 1/f = -1/10 – 1/(-15) = -1/10 + 1/15 = -5/150 U = -30 cm The negative value of u indicates that the object is placed 30 cm in front of the lens. Focal length of convexRead more

Focal length of concave lens (OF1), f = −15 cm
Image distance, v = −10 cm
According to the lens formula,
1/v – 1/u = 1/f
1/u = 1/v – 1/f = -1/10 – 1/(-15) = -1/10 + 1/15 = -5/150
U = -30 cm
The negative value of u indicates that the object is placed 30 cm in front of the lens.

Focal length of convex mirror, f = +15 cm
Object distance, u = −10 cm
According to the mirror formula,
1/v – 1/u = 1/f
1/v = 1/f – 1/u = 1/15 + 1/10 = 25/150
V = 6 cm
The positive value of v indicates that the image is formed behind the mirror.
Magnification, m = – Image Distance / Object Distance = – v/u = -6/-10 = +0.6
The positive value of magnification indicates that the image formed is virtual and erect.

## A person needs a lens of power –5.5 dioptres for correcting his distant vision. For correcting his near vision he needs a lens of power +1.5 dioptre. What is the focal length of the lens required for correcting

For distant vision = −0.181 m, for near vision = 0.667 m The power P of a lens of focal length f is given by the relation P = 1/f(in metres) (i) Power of the lens used for correcting distant vision = −5.5 D Focal length of the required lens, f =1/P F = 1/-5.5 = -0.181ₘ The focal length of the lens fRead more

For distant vision = −0.181 m, for near vision = 0.667 m

The power P of a lens of focal length f is given by the relation

P = 1/f(in metres)

(i) Power of the lens used for correcting distant vision = −5.5 D

Focal length of the required lens, f =1/P

F = 1/-5.5 = -0.181ₘ

The focal length of the lens for correcting distant vision is −0.181 m.

(ii) Power of the lens used for correcting near vision = +1.5 D

Focal length of the required lens, f =1/P

F = 1/1.5 = +0.667ₘ

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## Define 1 dioptre of power of a lens.

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 metres) 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 For more aRead more

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 metres)

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

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## 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?

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. For more answerRead more

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.

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## 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?

Power of a lens, P = 1/f (in meters) Power, P = 1.5d F = 1/1.5 = 10/15 = 0.66m A convex lens has a positive focal length. Hence, it is a convex lens or a converging lens. For more answers visit to website: https://www.tiwariacademy.com/ncert-solutions/class-10/science/chapter-10/

Power of a lens, P = 1/f (in meters)

Power, P = 1.5d

F = 1/1.5 = 10/15 = 0.66m

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

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## Find the focal length of a lens of power – 2.0 D. What type of lens is this?

Power of a lens, P = 1/f (in meters) P = -2 D f = -1/2 = -0.5m A concave lens has a negative focal length. Hence, it is a concave lens. For more answers visit to website: https://www.tiwariacademy.com/ncert-solutions/class-10/science/chapter-10/

Power of a lens, P = 1/f (in meters)

P = -2 D

f = -1/2 = -0.5m

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

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## 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 focussed image can be obtained? Find the size and the nature of the image.

Object distance, u = −27 cm Object height, h = 7 cm Focal length, f = −18 cm According to the mirror formula, 1/u + 1/v = 1/f 1/v = 1/f – 1/u = -1/18 + 1/27 = -1/54 V = -54 cm The screen should be placed at a distance of 54 cm in front of the given mirror. Magnification, m = - Image Distance/ObjectRead more

Object distance, u = −27 cm

Object height, h = 7 cm

Focal length, f = −18 cm

According to the mirror formula,

1/u + 1/v = 1/f

1/v = 1/f – 1/u = -1/18 + 1/27 = -1/54

V = -54 cm

The screen should be placed at a distance of 54 cm in front of the given mirror.

Magnification, m = – Image Distance/Object Distance = -54/27 = -2

The negative value of magnification indicates that the image formed is real.

Magnification, m = Height of the image / Height of the object = h’/h

h‘ = 7 x (-2) = -14 cm

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

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## 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.

This answer was edited.Object distance, u = −20 cm Object height, h = 5 cm Radius of curvature, R = 30 cm Radius of curvature = 2 × Focal length R = 2f f = 15 cm According to the mirror formula, 1/v + 1/u = 1/f 1/v = 1/f – 1/u = 1/15 + 1/20 = 4+3/60 = 7/60 V =8.57 The positive value of v indicates that the image is formedRead more

Object distance, u = −20 cm

Object height, h = 5 cm

Radius of curvature, R = 30 cm

Radius of curvature = 2 × Focal length

R = 2f f = 15 cm

According to the mirror formula,

1/v + 1/u = 1/f

1/v = 1/f – 1/u = 1/15 + 1/20 = 4+3/60 = 7/60

V =8.57

The positive value of v indicates that the image is formed behind the mirror.

Magnification, m = – Image Distance/Object Distance = -8.57/-20 = 0.428

The positive value of Magnification indicates that the image is formed is virtual.

Magnification, m = Height of the image / Height of the object = h’/h

h’ m x h = 0.428 x 5 = 2.14 cm

The positive value of image height indicates that the image formed is erect.

Therefore, the image formed is virtual, erect, and smaller in size.

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## The magnification produced by a plane mirror is +1. What does this mean?

This answer was edited.Magnification produced by a mirror is given by the relation Magnification, m = Image height (H₁) / Object height (H0) 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 imaRead more

Magnification produced by a mirror is given by the relation

Magnification, m = Image height (H₁) / Object height (H0)

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.

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## 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.

This answer was edited.Focal length of concave lens (OF1), f = −15 cm Image distance, v = −10 cm According to the lens formula, 1/v – 1/u = 1/f 1/u = 1/v – 1/f = -1/10 – 1/(-15) = -1/10 + 1/15 = -5/150 U = -30 cm The negative value of u indicates that the object is placed 30 cm in front of the lens. Focal length of convexRead more

Focal length of concave lens (OF1), f = −15 cm

Image distance, v = −10 cm

According to the lens formula,

1/v – 1/u = 1/f

1/u = 1/v – 1/f = -1/10 – 1/(-15) = -1/10 + 1/15 = -5/150

U = -30 cm

The negative value of u indicates that the object is placed 30 cm in front of the lens.

Focal length of convex mirror, f = +15 cm

Object distance, u = −10 cm

According to the mirror formula,

1/v – 1/u = 1/f

1/v = 1/f – 1/u = 1/15 + 1/10 = 25/150

V = 6 cm

The positive value of v indicates that the image is formed behind the mirror.

Magnification, m = – Image Distance / Object Distance = – v/u = -6/-10 = +0.6

The positive value of magnification indicates that the image formed is virtual and erect.

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