the near point of a hypermetropic person is 75cm if the person uses eyes glasses having power +1D, calculate distance of distinct vision for him.

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Solution

P $=$ 1D
f $=$ 1m $=$ 100cm
u $=$ $-$ 25
Lens formula: 1 $/$ f $=$ 1 $/$ v $-$ 1 $/$ u
1 $/$ 100 $=$ 1 $/$ v $+$ 1 $/$ 25
1 $/$ 100 $-$ 1 $/$ 25 $=$ 1 $/$ v
1 $/$ v $=$ $-$ 3 $/$ 100
v $=$ $-$ 33.3cm

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the near point of a hypermetropic person is 75cm if the person uses eyes glasses having power +1D, calculate distance of distinct vision for him.

P=1D ​f=1m =100cm u=-25 Lens formula: 1/f=1/v-1/u 1/100=1/v+1/25 1/100 -1/25 =1/v 1/v=-3/100 v=-33.3cm

P $=$ 1D
f $=$ 1m $=$ 100cm
u $=$ $-$ 25
Lens formula: 1 $/$ f $=$ 1 $/$ v $-$ 1 $/$ u
1 $/$ 100 $=$ 1 $/$ v $+$ 1 $/$ 25
1 $/$ 100 $-$ 1 $/$ 25 $=$ 1 $/$ v
1 $/$ v $=$ $-$ 3 $/$ 100
v $=$ $-$ 33.3cm