Question

the power of A lens used to remove the myopic defect of eye is 0.66D.the far point of this eye is.​

Answer 1

p = 1÷f

0.66=1÷f

0.66×f=1

f=1÷0.66

f= 1×100÷0.66×100

f= 100÷66

f= 1.51cm

here is ur answer

Answer 2

I think it should be the option a