Originally Posted by **sa-ri-ga-ma** You can take the object at infinity.
Then use the formula 1/f1 + 1/f2 = 1/F
Here f1 is the focal length of the normal eye and F is the focal length of defective eye. And f2 is the focal length of the correcting lens. |

So taking the object at infinity,for the defective eye,and applying sign convention(taking the distances measured in the direction of the rays from the object as negative,while those measured in the direction opposite to the rays as positive),

1/F = -1/2.3 - 1/infinity

1/F = -1/2.3cm = -0.435cm

taking the object at infinity,for the normal eye,

1/f1 = -1/2.4 - 1/infinity

1/f1 = -0.417cm

Then using the formula suggested,

1/f2 = 1/F - 1/f1

1/f2 = -0.435 -(-0.417)

I get f2 = -55.56cm,

since the focal length is negative ,wouldn't the correcting lens be a converging lens?