Question

With a near point of 15.0 cm, a nearsighted professor can focus on relatively nearby objects. Yet, with a far point 62.0 cm out, he cannot clearly see distant objects without optical ald. (a) Calculate the focal length (in cm) of a contact lens that would enable this professor to clearly view very distant objects. (Hint: Allow p to approach Infinity in the thin-lens equation). (b) What power (in diopters) describes this contact lens? diopters (c) When this contact lens is in use, what is the professor's near point (in cm)? cm (d) which statement best describes the consequences of correcting distance vision, when it comes to viewing nearby objects? Contacts or glasses for nearsightedness make it harder to see very close objects. Contacts or glasses for nearsightedness make it easier to see very close objects, too. Contacts or glasses for nearsightedness have no effect upon how you see very close objects.

Answer 1

The solution for this question is given below

Far point, n = 62.0cm. Near point, y= 15.0cm. (a) If object at very large distance is to be seen, then, Vo=D & image should be formed at the near point to be seen , so, vi = -x. Now, from lens equation = 2 = 6 - 2 = =-62 cm] (b) PB +62 X20- 2 00 =)P =-1,6101 (c) For this professor's eye, the near point is y = 15 cm. This means than when an object is placed some where its image should be formed at y before eyes for professor to see so,Vi = -15cm So, I = 1 - 1 f Vi Vo = 62 = -15 to >> Vo = -19.78 cm / Any object nearer than this cannot be seen clearly. (d) So, contacts or glasses for nearsightedness make it harda to see very close objects

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