the focal length of a lens suggested to a person with hypermetropia is 100cm.find the distance of near point and power of lens?
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i) The distance of near point:
If the distance of near point is'd' and focal length is 'f then the relation between
them f = 25d/d-25 (f in cm)
Given focal length (f) = 100 cm
f = 25d/d-25
=====> 100 = 25d/d-25
d-25 = 25d/100
===> d-25 = d/4
====> 4(d-25) = d
===> 4d-100 = d
===> 4d-d = 100
===> 3d = 100
===> d = 100/3 = 33.33 cm
ii) Power of lens :
Given that; f = 100 cm = 1m
Power = 1/ f in metres diopters
= 1/100 cm = 1/1m D = 1D
If the distance of near point is'd' and focal length is 'f then the relation between
them f = 25d/d-25 (f in cm)
Given focal length (f) = 100 cm
f = 25d/d-25
=====> 100 = 25d/d-25
d-25 = 25d/100
===> d-25 = d/4
====> 4(d-25) = d
===> 4d-100 = d
===> 4d-d = 100
===> 3d = 100
===> d = 100/3 = 33.33 cm
ii) Power of lens :
Given that; f = 100 cm = 1m
Power = 1/ f in metres diopters
= 1/100 cm = 1/1m D = 1D
Sunny03:
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