For a normal eye, the distance between the retina and the cornea eye lens is 2.5 cm. If least converging power of eye lens is 30 D, then find converging power of cornea of eye.

Question

For a normal eye, the distance between the retina and the cornea eye lens is 2.5 cm. If least converging power of eye lens is 30 D, then find converging power of cornea of eye. (A) 20 D (B) 30 D (C) 10 D (D) 5 D

Tardigrade Admin · Accepted Answer

P =(100/ v )-(100/ u ) P =(100/2.5)-(100/-∞) P =40 D P = P 1+ P 2 40=30+ P 2 P 2=10 D

Q. For a normal eye, the distance between the retina and the cornea eye lens is $2.5 c m$ . If least converging power of eye lens is $30 D$ , then find converging power of cornea of eye.

20 D

30 D

10 D

5 D

$P = \frac{100}{v} - \frac{100}{u}$
$P = \frac{100}{2.5} - \frac{100}{- \infty}$
$P = 40 D$
$P = P_{1} + P_{2}$
$40 = 30 + P_{2}$
$P_{2} = 10 D$

Q. For a normal eye, the distance between the retina and the cornea eye lens is 2.5cm. If least converging power of eye lens is 30D, then find converging power of cornea of eye.

20 D

30 D

10 D

5 D

P=v100​−u100​ P=2.5100​−−∞100​ P=40D P=P1​+P2​ 40=30+P2​ P2​=10D

Q. For a normal eye, the distance between the retina and the cornea eye lens is $2.5 c m$ . If least converging power of eye lens is $30 D$ , then find converging power of cornea of eye.

$P = \frac{100}{v} - \frac{100}{u}$
$P = \frac{100}{2.5} - \frac{100}{- \infty}$
$P = 40 D$
$P = P_{1} + P_{2}$
$40 = 30 + P_{2}$
$P_{2} = 10 D$