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The near point and far point of a child are at $10 cm$ and $100 cm$ . If the retina is $2.0 cm$ behind the eye lens, what is the range of the power of the eye lens?

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50 D to 30 D

75 D to 55 D

60 D to 51 D

51 D to 60 D

answer is C.

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Detailed Solution

The range of the power of the eye lens is

60 D to 51 D

.
The focal length of a lens is calculated using the given formula

\frac{1}{f} = \frac{1}{v} - \frac{1}{u}

If the object placed at a nearby position can form the image of it at the retina,

u (object distance) = - 10 cm

v (image distance) = 2 cm

\frac{1}{f} = \frac{1}{2} - \frac{1}{- 10}

\Rightarrow \frac{1}{f} = \frac{5 + 1}{10}

\Rightarrow \frac{1}{f} = \frac{6}{10}

\Rightarrow f = \frac{10}{6}

\Rightarrow f = \frac{5}{3} cm

And the power of the lens is

P = \frac{100}{f}

\Rightarrow P = \frac{100 \times 3}{5}

\Rightarrow P = + 60 D

When the image is formed at retina if the object is placed beyond the far point,

u = - 100 cm

and

v = 2 cm

\frac{1}{f} = \frac{1}{2} - \frac{1}{- 100}

\Rightarrow \frac{1}{f} = \frac{50 + 1}{100}

\Rightarrow \frac{1}{f} = \frac{51}{100}

\Rightarrow f = \frac{100}{51} m

power, P = \frac{100}{f}

\Rightarrow P = \frac{51 \times 100}{100}

\Rightarrow P = 51 D

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