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.
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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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.
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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.
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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