an object is placed at a distance of 15 cm from a concave lens of focal length 30 cm list four characteristics nature position and size of the image formed by the lens.....
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the image will be between focus and optical centre dimnished virtual and erect
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given
object distance U = -15 cm
focal lenght , f = -30 cm
postion of image =V
let size of object = h
size of image = h'
from lens formula
1/v -1/u = 1/f
1/v + 1/15 = -1/30
1/v = (-1-2)/30 = -3/30
[1] position, V = -10
[2] size of object = h'
from maginification
m= v/u = 10/15 = 2/3
also m= h'/h = 2/3
h' = 2/3×h
that means height of image is 2/3 times the height of object
[3] nature of image
since m is +ve
hence object is erect and virtual
[4] since image distance is in -ve
hence image will form on the same side of the object...
hope it will help u....
object distance U = -15 cm
focal lenght , f = -30 cm
postion of image =V
let size of object = h
size of image = h'
from lens formula
1/v -1/u = 1/f
1/v + 1/15 = -1/30
1/v = (-1-2)/30 = -3/30
[1] position, V = -10
[2] size of object = h'
from maginification
m= v/u = 10/15 = 2/3
also m= h'/h = 2/3
h' = 2/3×h
that means height of image is 2/3 times the height of object
[3] nature of image
since m is +ve
hence object is erect and virtual
[4] since image distance is in -ve
hence image will form on the same side of the object...
hope it will help u....
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