For near point of 20 cm, a hypermetropic of +4D has to exercise accommodation of -: (A) +4.0 D, (B) +9.0 D, (C) +10.0 D, (D) 0 D
Answer is B. Ans- B +9.0 D A=P-R where P-is the refractive power of the eye when fully accommodated and R-is the accommodative power of the eye at rest. Hypermetropes have their far point behind eye. Hence, R= -4 Applying this to the formula= A= 100/20 - (-4)= 5 + 4 = +9 D.