Finding Distance of Retina from Lens: Power from 40D to 44D

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The discussion focuses on calculating the distance of the retina from the lens based on the accommodation power of a boy, which ranges from 40D to 44D. Using the lens formula, when the object is at infinity, the focal length is determined to be 2.5 cm, leading to a calculated distance of 2.5 cm for the retina from the lens. The poster questions whether this calculation is sufficient or if they need to verify it by considering the near point, which is not provided in the problem. It is suggested that assuming a near point of 25 cm could provide additional insight, though it is not necessary for the initial calculation. The conclusion emphasizes that the initial calculation appears correct based on the given information.
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Homework Statement


The accomodation power of a boy whose far point is at infinity can vary from 40D to 44D.
Using this data find the distance of the retina from the lens.


Homework Equations


Power of lens,P = 1/f(in meters)
f = focal length
lens formula, 1/v - 1/u = 1/f

The Attempt at a Solution



when object is at infinity,
f=1/40 = 2.5 cm
by lens formula,

1/v-(-1/\infty) = 1/2.5
=>1/v - 0 = 1/2.5
=>v=2.5 cm

thus, distance = 2.5cm

Have i done this question right? And do i have to confirm the answer by putting the object at the near point or this is enough?
 
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You can't double check because the question doesn't tell you what the near point is.Just to satisfy your curiosity you may wish to assume that the near point is at 25 cm,this is taken to be the average value.
 
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