The diameter of the eye-ball of a normal eye | vedclass

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(D) For a normal eye,the image distance $v$ is equal to the diameter of the eyeball,$v = 2.5\, cm = 0.025\, m$.When viewing distant objects (at infini

Vedclass · Accepted Answer

$44\, D$ to $40\, D$

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(D) For a normal eye,the image distance $v$ is equal to the diameter of the eyeball,$v = 2.5\, cm = 0.025\, m$.When viewing distant objects (at infinity),the eye is relaxed. Using the lens formula $\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$:$P = \frac{1}{f} = \frac{1}{0.025} - \frac{1}{-\infty} = 40\, D$.When viewing objects at the near point $(u = -25\, cm = -0.25\, m)$,the eye is under maximum strain:$P = \frac{1}{f} = \frac{1}{0.025} - \frac{1}{-0.25} = 40 + 4 = 44\, D$.Thus,the power of the eye lens varies from $44\, D$ to $40\, D$.

The diameter of the eye-ball of a normal eye is about $2.5\, cm$. The power of the eye lens varies from

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