Question

25.
i) The power of concaye lens used for correcting a myopic eye is -0.6 D. find the far
point of the eye.
ii) The power of convex lens used to correct a hypermetropic eye is + 0.8D. Find the
near point of the eye. Assume the least distance of distinct vision to be 25 cm.


I need answers please fast​

Answer 1

Answer:

i)

Far point is shifted to 1.66 m

ii)

Near point is shifted to 31.25 cm

Explanation:

Part i)

Power of the lens is given as P = -0.6 D

now by the formula we know that

$P = \frac{1}{F}$

$-0.6 = \frac{1}{F}$

$F = -1.66 m$

now we know by lens formula

$\frac{1}{d_o} + \frac{1}{d_i} = \frac{1}{f}$

so we will have

$d_i = 1.66 m$

So the far point is shifted to 1.66 m

Part ii)

Power of the lens is given as P = +0.8 D

now by the formula we know that

$P = \frac{1}{F}$

$0.8 = \frac{1}{F}$

$F = 1.25 m$

now we know by lens formula

$\frac{1}{d_o} + \frac{1}{d_i} = \frac{1}{f}$

$\frac{1}{0.25} + \frac{1}{d_i} = \frac{1}{1.25}$

$d_i = -0.3125 m$

So the near point is shifted to 31.25 cm

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Topic : Eye defect

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