Question

14. The near point and far point of a person are a
60 cm and 300 cm respectively. Find the power of
lens he should use while reading at 30 cm. With
this lens on the eye, what maximum distance is
clearly visible?​

Answer 1

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