A child while playing with his father’s spectacles burnt a hole in a piece of paper by focusing a small image of the sun on it.

i. What defect of vision his father is suffering from?


ii. Write two causes for this defect.


iii. Draw a ray diagram to show image formation by the defective eye.


iv. Draw a ray diagram showing corrected eye using proper lens.


OR


A person cannot see objects beyond 80 cm from his eye while a person with normal eyesight can see object easily placed up to 160 cm from the eye. Find the nature, the focal length and the power of the correcting lens.


i. The defect is Myopia or near-sightedness.


Explanation: -


As the rays from sun are getting focused onto the paper and burning it, this is indicating that the far point of the eye has shifted from infinity to a few meters and this is clear sign of Myopia as person is able to nearer objects but cannot see far objects.


ii. This defect may arise due to a.) excessive curvature of the eye lens which reduces the focal length of the eye thus the rays coming from infinity get focused much before reaching the retina.


or


b.) elongation of the eyeball which shifts the retina from the focal point of the eye lens thus leading to myopia.



iii.


The point O’ is the actual position of the object and O is the far point of the defected eye the rays from O are getting focused much before reaching the retina due to the aforesaid reason(s).



iv.


The defect is corrected by using a spectacle consisting concave lens of appropriate focal length so as to increase the focal length of the eye lens(the far point has to be shifted) as shown in the diagram.


OR


The person is suffering from myopia, the correcting lens is a concave lens which is of diverging nature. The focal length of the correcting lens can be easily calculated by converting the problem into a simple lens formula problem,


The data given is:


the image distance is, ,


the object distance is, ,


we know that the lens formula is,


, where,


v is the image distance, u is the object distance and f is the focal length of the lens.


Calculations:


So, after substituting the data we have,





So, we get . ,


The power is the reciprocal of the focal length in meters, so



P=-0.625 D.


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