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.
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.
For more answers visit to website:
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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.
For more answers visit to website:
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The magnification produced by a plane mirror is +1. What does this mean?
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.
For more answers visit to website:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-10/science/chapter-9/