Answer:
Explanation:
The refractive index of the lens is
d = 1.570 dioptres
Therefore, her focal length is
f = 1 / d
f = 1 / 1.570
f = 0.6369 m
f = 63.69cm
The distance between the newspapers and her eyes is 42cm
Distance from her eye to the glasses is 2cm
We want to calculate her near point?
So, the distance between the newspaper and the glass
do = 42 - 2 = 40cm
The distance between the glass and the virtual image formed is given as
di = f•do / (do —f)
di = 63.69 × 40 / (40 — 63.69)
di = 2547.6cm / -23.69
di = —107.54cm
The relation use is obtain from the lens equation and the negative shows that virtual image formed.
The near point from her eye is
107.54 + 2 = 109.54cm
The near point from her eye when she wears the eyeglasses that is 2.00 cm from her eyes should be considered as the 109.54cm.
Since
The refractive index of the lens is
d = 1.570 dioptres
Now
her focal length is
f = 1 / d
f = 1 / 1.570
f = 0.6369 m
f = 63.69cm
Since The distance between the newspapers and her eyes is 42cm
And,
Distance from her eye to the glasses is 2cm
So, the difference is
= 42 - 2
= 40cm
Now
The distance between the glass and the virtual image formed is
di = f•do / (do —f)
di = 63.69 × 40 / (40 — 63.69)
di = 2547.6cm / -23.69
di = —107.54cm
Now near point should be
= 107.54 + 2
= 109.54cm
hence, The near point from her eye should be considered as the 109.54cm.
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