Calculate the power of accommodation of a person having normal vision. Assume diameter of his eye to be 25 mm.
(Near point = 25 cm, Far point = ∞)
(Near point = 25 cm, Far point = ∞)
→ When the object is at near point (25 cm):
u = -25 cm
v = +2.5 cm
f = ?
v = ?
Finding focal length ‘f’ using lens formula
1/f = 1/v - 1/u
= 1/2.5 - (1/-25)
= (10+1)/25
= 11/25 cm
f = 25/11 cm = 2.27 cm or 0.0227 m
→ P = 1/f (in m)
= 1/0.0227
P = 44 D
→ When the object is at far point (∞):
u = -∞ cm
v = +2.5 cm
f = ?
v = ?
Finding focal length ‘f’ using lens formula
1/f = 1/v - 1/u
= 1/2.5 - (1-/∞)
= 1/2.5 (since 1/∞ = 0)
f = 2.5 cm or 0.025 m
→ P = 1/f (in m)
= 1/0.025
P = 40 D
-
Therefore the range of power of accommodation of a normal human eye lens is 44 D to 40 D for objects placed at near objects to distant objects
→ When the object is at near point (25 cm):
u = -25 cm
v = +2.5 cm
f = ?
v = ?
Finding focal length ‘f’ using lens formula
1/f = 1/v - 1/u
= 1/2.5 - (1/-25)
= (10+1)/25
= 11/25 cm
f = 25/11 cm = 2.27 cm or 0.0227 m
→ P = 1/f (in m)
= 1/0.0227
P = 44 D
→ When the object is at far point (∞):
u = -∞ cm
v = +2.5 cm
f = ?
v = ?
Finding focal length ‘f’ using lens formula
1/f = 1/v - 1/u
= 1/2.5 - (1-/∞)
= 1/2.5 (since 1/∞ = 0)
f = 2.5 cm or 0.025 m
→ P = 1/f (in m)
= 1/0.025
P = 40 D
-
Therefore the range of power of accommodation of a normal human eye lens is 44 D to 40 D for objects placed at near objects to distant objects
