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Q12PE

Expert-verifiedFound in: Page 956

Book edition
1st Edition

Author(s)
Paul Peter Urone

Pages
1272 pages

ISBN
9781938168000

**A student’s eyes, while reading the blackboard, have a power of 51.0 D. How far is the board from his eyes?**

The board is 1 m far from the student's eyes.

**The unit of measurement for the optical power of a lens or curved mirror is the dioptre of power of a lens, which is equal to the reciprocal of focal length.**

The power of a student’s eye is P = 51.0 D

The lens-to-retina distance is, \({{\rm{d}}_{\rm{i}}} = {\rm{2}}{\rm{.00 cm}}\left( {\frac{{1\;{\rm{m}}}}{{100\;{\rm{cm}}}}} \right) = {\rm{0}}{\rm{.02 m}}\).

The power of the eyes of a normal person can be calculated using the formula,

$P=\left(\frac{1}{{d}_{\circ}}+\frac{1}{{d}_{i}}\right)\phantom{\rule{0ex}{0ex}}51.0D=\left(\frac{1}{0.02m}+\frac{1}{{d}_{\circ}}\right)\phantom{\rule{0ex}{0ex}}51.0D=51.0D+\frac{1}{{d}_{\circ}}\phantom{\rule{0ex}{0ex}}\frac{1}{{d}_{\circ}}=1.0D$

This result in the value of

$\frac{1}{{d}_{\circ}}=1.0D\left(\frac{1{m}^{-1}}{1D}\right)\phantom{\rule{0ex}{0ex}}{d}_{\circ}=1m$

Therefore, the board is 1 m from the student's eyes.

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