(a) The near point of a hypermetropic person is at `75 cm` from the eye. What is the power of the lens required to enable him to read clearly a book held at `25 cm` from the eye ?
(b) In what way does the corrective lens help the person above ? Does the lens magnify objects held near the eye ?
( c) The person above prefers to remove his spectacles while looking at the sky. Explain why ?

Explanation

(a) Here, `v = - 75 cm`,
`u = - 25 cm, f = ?`
`(1)/(f) = (1)/(v) - (1)/(u) = - (1)/(75) + (1)/(25) = (- 1 + 3)/(75) = (2)/(75)`
`f = (75)/(2)cm = 37.5 cm`
`P = (100)/(f) = (100 xx 2)/(75) = (8)/(3) = 2.66 D`
The corrective lens is convex.
(b) The corrective lens produces a virtual image (at `75 cm`) of teh object (at `25 cm`). As angular size of this image is same as that of the object, the lens does not magnify the object. It simply brings the object to the near point of the eye from where eye lens focussed it on the retina.
( c) A hypermetropic eye may have normal far point i.e., || rays from infinity may be focussed on the retina of the shortened eye ball. Wearing spectacles of converging lenses (used for near vision) will amount to more converging power than needed for parallel rays. Therefore, distant objects may get focussed in front of the retina , appearing blurred. That is why the person prefers to remove the spectacles while looking at the sky.