Question

the near point of a hypermetropic eye is 1 metre. what is the power of the lens required to correct this defect. assume that the near point of the normal eye is 25 CM​

Answer 1

$<b>$

$1 \div f = 1 \div \: v - 1 \div u \\ \\ 1 \div f = 1 \div - 100 - 1 \div - 25 \\ \\ 1 \div f = - 3 \div - 100 \\ f = 33.3cm$

Power

$p = 100 \div f \\ p = 100 \div 33.3 \\ p = 3.00(approx)$

$\boxed{P=3D}$


HOPE IT HELPS

Answer 2

Near point(u) = - 25 cm

H. Near point(v) =-100cm

P=1/f

1/f=1/v-1/u

,, =1/-100 - 1/-25

,, = -1+4/100

1/f=3/100

P=100*3/100 (in cm)

P=+3 diaptore