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A person cannot see an object lying beyond 80 cm, whereas a normal person can easily see an object at a distance of 160 cm. What will be the focal length and nature of the lens used to rectify this defect?

A. Convex lens, [tex]f = 20 \, \text{cm}[/tex]
B. Concave lens, [tex]f = 20 \, \text{cm}[/tex]
C. Convex lens, [tex]f = -20 \, \text{cm}[/tex]
D. Concave lens, [tex]f = -20 \, \text{cm}[/tex]

Answer :

Final answer:

To correct farsightedness, a convex lens with a focal length of 20 cm (Choice a) would be used to converge light rays, allowing the person to see objects clearly beyond 80 cm.

The correct option is, Convex lens, f = 20 cm, which is, (a).

Explanation:

The condition described is farsightedness (presbyopia or hyperopia), where a person can see distant objects but cannot clearly see objects that are close. To correct this vision defect, a convex lens is used, which converges the light rays before they enter the eye, effectively increasing the overall converging power of the eye's lens system.

Since a normal sighted person can see up to 160 cm and the farsighted person can only see up to 80 cm, we need to determine the focal length of the corrective lens that allows the person to see the object at 160 cm clearly. Applying the lens formula (1/f = 1/v - 1/u) and using the appropriate sign conventions where the object distance (u) is -80 cm (negative because it is on the same side of the lens as the incoming light) and we want the image distance (v) to be at least -160 cm (again negative), we can find the required focal length (f).

Considering the information provided and the correction required for farsightedness, the correct choice is: a) Convex lens, f = 20 cm.

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