a person is unable to see an object placed within 50cm from the eye. mention the defect and calculate the power of the lens that should be used to correct the defect. assume that the normal near point of the eye is 25cm. please explain how you got the answer as well. (really confusing as both answers could apply to this question)
Answers
Answered by
4
The defect is hypermetropia
v=-25cm
u=-50cm
by using lens formula
1/f=1/v-1/u
1/f=-1/25-(-1/50)
1/f=-1/25+1/50
f=-50cm
P=100/f(in cm)
=100/-50
= -2Dioptre
hope it will help you...
In myopia we use u= infinity means =0 always.....so dnt be confuse...
nimamenon:
but for hypermetropia power should be positive because we use convex lens
power pe dhyan mt doo
u should know signs of F V and U
power has to be positive for convex lens thats the rule how can i not pay attention to that?
if signs of F v and u are correct then power will also be correct
and they asked for power of the lens therefore sign must be correct
this is what i was confused about
u r confusing me nw
ugh
??
Answered by
4
Defect is hypermetropia....
For the person who is not able to see objects nearer than 50 cm:
we use convex lens to create a virtual object of an object at 25 cm at 50 cm. The objects at distances between 25 cm and 50 cm will result in virtual images placed erect at distances more than 50 cm. So the person will see them clearly.
v = -50 cm u = -25 cm then focal length of convex lens is:
1/v - 1/u = -1/50 + 1/25 = 1/50 => f = 50 cm and power is +2 Dioptre.
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