A person\'s old prescription corrected her vision when her nearpoint was at 60cm

Question

A person's old prescription corrected her vision when her nearpoint was at 60cm, but now she needs new glasses with a power of2.67D. (Assume n = 2)
a) what is the power of her old lenses? b) where is her new near point located?
Please show All yourcalculation(with numbers) and the reasons behind them in astep-by-step fashion. Thank you and I'll ratefast!
a) what is the power of her old lenses? b) where is her new near point located?
Please show All yourcalculation(with numbers) and the reasons behind them in astep-by-step fashion. Thank you and I'll ratefast!

Explanation / Answer

Given : taking actual near point as object distance and changed nearest point of accomodation as imagedistance        1 /f = 1 /p + 1/ q       1 /f = 1/25.0cm + 1 / -60cm since the image is virtual taking negative for qvalue from the above relation we can solve for f     f = 43 cms positive values of f indicates , converging lense (1)power of old lense p = 1/ f (2) given the new focal length f ' = 100 /2.67 =37.5cm        and new nearest point 'q' ' can be determined as          1/ p + 1 /q'= 1 /f'      p = 25cms from the abv relation we can solve for q' >60cm I hope it helps you and changed nearest point of accomodation as imagedistance        1 /f = 1 /p + 1/ q       1 /f = 1/25.0cm + 1 / -60cm since the image is virtual taking negative for qvalue from the above relation we can solve for f     f = 43 cms positive values of f indicates , converging lense (1)power of old lense p = 1/ f (2) given the new focal length f ' = 100 /2.67 =37.5cm        and new nearest point 'q' ' can be determined as          1/ p + 1 /q'= 1 /f'      p = 25cms from the abv relation we can solve for q' >60cm I hope it helps you

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