Question

What is near point and far point of a myopic eye and also what is the near point and far point of a hypermetropic eye??​

Answer 1

1):The patient accommodates 8D beyond far point, thus his near point is 1/6+8=0.07m (7.1cm). If we compare the case with emmetropic eye with accommodation amplitude of 8D, the near point is 1/8=0.12m(12cm). So, yes, near point of myopic person is different from an emmetropic one.
2):Farsightedness. For a normal vision (clear vision from near-point 25 cmto infinity), when the ciliary muscles are relaxed, an object at the near point is in focus. For a far-sighted (hyperopic), the relaxed eye focuses only within a finite distance from a certain point, called the near point toward infinity.

ok hope it may be helpful

Answer 2

Answer:So to find the focal length, We need to use lens makers formula,

We know that , the formula is ,

1/f = 1/v - 1/u,

Normal person's Near point is 25cm,

Hypermetropic person usually can't see near objects, So what they have to do is, Use convex lens and make the object's image fall at 25 cm and let the lens of eye to make the image as object,

So what we have to do here is, Make the near point of Hypermetropic person to 25cm,

So here object distance = Near point of Hypermetropic person,

Image distance = -25cm(- indicates direction),

=> Object distance = u = -40cm (Direction),

Focal length of Convex lens is always positive !,

Now Inputting all the values to find focal length in cm,

=> 1/f = -(1/25) -(-1/40),

=> 1/f = 1/40 - 1/25,

=> 1/f = (5-8)/200

=> 1/f = -3/200 cm,

=> f = -200/3 cm, Converting it into m,

=> f = -200/3/100 = -2/3 m,

So therefore focal length = -2/3 m,

Now calculating power,

Power = 1/Focal length in m ,

Units = D , Dioptre,

=> Power = 1/-2/3 = -3/2D,

Therefore power = -3/2 D, and focal length = -2/3 m,

Hope you understand, Have a great day