A hypermetropic person has to use a lens of power +5 D to normalise his vision. The near point of the hypermetropic eye is

Question

A hypermetropic person has to use a lens of power +5 D to normalise his vision. The near point of the hypermetropic eye is (A) 1 m (B) 1.5 m (C) 0.5 m (D) 0.66 m

Tardigrade Admin · Accepted Answer

We have, f=(1/P)=(1/5) m =20 cm Now, (1/v)-(1/u)=(1/20) (from lens formula) Or (1/v)-(1/-25)=(1/20) or (1/v)=(1/20)-(1/25) Or (1/v)=(5-4/100) or (1/v)=(1/100) Or d=100 cm =1 m

Q. A hypermetropic person has to use a lens of power +5D to normalise his vision. The near point of the hypermetropic eye is

1 m

1.5 m

0.5 m

0.66 m

We have, f=P1​=51​m=20cm Now, v1​−u1​=201​ (from lens formula) Or v1​−−251​=201​ or v1​=201​−251​ Or v1​=1005−4​ or v1​=1001​ Or d=100cm=1m

Q. A hypermetropic person has to use a lens of power $+ 5 D$ to normalise his vision. The near point of the hypermetropic eye is