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
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.
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 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.
For more answers visit to website:
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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.
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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