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The distance between retina and eye lens is $2.5 cm$ and most of refraction takes place by crystalline lens only then minimum power of lens is $D$ and maximum power of lens is $D$ ?

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Detailed Solution

The distance between retina and eye lens is

2.5 cm

and most of refraction takes place by crystalline lens only then the minimum power of lens is

40 D

and the maximum power of lens is

44 D

.
Given,

D = 2.5 cm,

(a) Minimum power of lens:
Power of accommodation

' D^{'}

can be calculated using formula:

D = \frac{1}{f}

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

…(1)
Where,

f = focal length  of lens

v = near point of normal person (i . e . 25 cm)

u = given near point of person (i . e . 2.5 cm)

By substituting the values in equation(1) we get,

= > \frac{1}{f} = \frac{1}{2.5} - \frac{1}{25}

= > ⸫ \frac{1}{f} = 0.04 cm

= > ⸫ f = \frac{1}{0.04} cm  or

= > f = 0.025 m

= > ⸪ Power (D) = \frac{1}{f}

= > ⸫ D = \frac{1}{0.025} = 40 D

⸫ Minimum Power = 40 D

(b)Maximum Power of lens:
⸪

Power of accommodation of normal person = 4 D

= > ⸫ Maximum Power = Minimum Power + Power of Accommodation

= > Maximum power = 40 + 4

= > ⸫ Maximum power = 44 D

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