A myopic person uses spectacles of power -.5 D. what is the distance of the far point from the eye?
Answers
As f = 1/p
So f=1/.5 =2m
So far point is 2m
Explanation:
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Class 12
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>>Defects of Vision and Correction
>>A myopic adult has a far point at 0.1m .
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A myopic adult has a far point at 0.1m. His power of accommodation is 4 diopters
(i) What power lenses are required to see distant objects
(ii) What is his near point without glasses?
(iii) What is his near point with glasses? (Take the image distance from the lens of the eye to the retina to be 2cm)
Medium
Solution
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(i) power of lens required to see clearly the object placed at infinity.
U=−∞
f
1
=
v
1
−
u
1
⇒
f
1
=
−10
1
−
∞
1
;P=
f
1
m
⇒
f
1
=
10
1
;P=
−0.1
1
m
(ii) when no corrective lens used: Let powers of eye when object is at far point, near point are P
f
and P
n
respectively and power
∴ P
n
=P
f
+P
a
When object is at far point its clear image is formed at retina 2cm from eye lens
∴ u=1−cm=−0.1m,V=2cm=0.02m
If f is focal length of eye lens focused at far point then
f
1
=
v
1
−
u
1
=
0.02
1
−
(−0.1)
1
f
1
=50+10=60
P
f
=60D
∴ P
n
=P
f
+P
a
=60+4=64D
Let the near point be x
n
u=x
n
(v=2cm=0.02m)
v
1
−
u
1
=
f
1
0.02
1
+
x
n
1
=Power (P
n
)
50+
x
n
1
=64
x
n
1
=64−50=14D
Near point without glass x
n
=
14
1
m=
14
100
cm=7cm (approx.)
(iii) When used corrective lens: When corrective lens is used then eye can see the object at infinity. Power of eye lens in this situation is P
∞
u=∞ and v=2cm=0.02m
f
1
=
v
1
−
u
1
P
∞
=
0.02
1
−
∞
1
=50
P
∞
=50+0
P
∞
=50D
If P
n
= power of eye at near point when corrective lens is used
P
n
=P
∞
+P
a
=50+4=54D
Let near point in this situation is x
n
u=−x
n
m
v=+2cm=0.02m
f
1
=54
v
1
−
u
1
=
f
1
0.02
1
+
x
n
1
=54 (all distance are in m)
50+
x
n
1
=54(
x
n
1
=4)
x
n
=
4
1
m=0.25m