The near point of a hypermetropic eye is 1 metre. What is the power of the lens required to correct this defect? Assume that the near point of the normal eye is 25 cm.

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Solution

Here, u = -25 cm = -0.25 m
v = -1 m
Now, $P = \frac{1}{f} = \frac{1}{v} - \frac{1}{u} = \frac{1}{- 1} - \frac{1}{- 0.25}$ $= - 1 + 4 = + 3 D$

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The near point of a hypermetropic eye is 1 metre What is the power of the lens required to correct this defect? (Assuming that the near point of a normal eye as 25cm)

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The near point of a hypermetropic eye is 1 metre. What is the power of the lens required to correct this defect? Assume that the near point of the normal eye is 25 cm.

Here, u = -25 cm = -0.25 m v = -1 m Now, P=1f=1v−1u=1−1−1−0.25 =−1+4=+3D

Here, u = -25 cm = -0.25 m
v = -1 m
Now, $P = \frac{1}{f} = \frac{1}{v} - \frac{1}{u} = \frac{1}{- 1} - \frac{1}{- 0.25}$ $= - 1 + 4 = + 3 D$