Question

if a person in a various problems uses spectacles of power +1.0 dioptric what is the nearest distance of distinct vision for him?

Answer 1

Power=1/(Focal legth (in m))

here,

Power=+1.0

therefore,

focal length=1m

At the nearest point of distinct vision, all objects from infinite dist. form an image there.

therefore,

by using the relation:

1/f=1/v-1/u

1=1/v-1/(infinity)

so,

v=1m <--- this is the dist. of nearest vision

Answer 2

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∵ Near point = 75 cm and Power of lens = + 1 D

∴    P = 100 / f ( in cm )

f = 100 / P

= 100 / 1

 

= 100 cm

The distance of distinct vision v can be found as below.

Using lens formula here ,

1/f  = 1/v - 1/u

⇒ 1/v = 1/f + 1/u

= 1 / 100 + 1 / - 75  =  3 - 4 / 300

= - 1 / 300

v = ₋300 cm

=  - 3m  ( in meters )