Question

For the following calculations, these assumptions have been made.

The distance between the front of the cornea of the eye to the retina = 0.025m (2.5cm)

The focusing distance for a normal adult eye = 0.25m - ∞.

Q1: Calculate the power of the eye (composite of cornea and lens) where:

u = 1.0m        v = 0.025m

Q2: An object at 10.0m focuses 2mm in front of the retina (myopic eye). Calculate the power of the retina/lens combination of the defective eye. What is the power of the corrective lens required to bring the focus back to the retina surface?

Q3: An object at 0.25m focuses 2mm behind the retina surface (hypermetropia). Calculate the power of the defective eye and the power of the corrective lens required to bring the focus back to the retina surface.

Answer 1

The concept used in this question is lens formula , as the distance between cornea and retina is given so i used this to calculate the image distance by subtracting the distance between tthe cornea and retina from the distance infront of the retina , after applying this concept i deduced the values .

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