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Ahat [919]
3 years ago
5

Darcy suffers from farsightedness equally severely in both eyes. The focal length of either of Darcy's eyes is 19.8 mm in its mo

st accommodated state (i.e, when the eye is focusing on the closest object that it can clearly see) What lens strength (a.k.a., lens power) of contact lenses should be prescribed to correct the farsightedness in Darcy's eyes? (Assume the lens-to retina distance of Darcy's eyes is 2.00 cm, and the contact lenses are placed a negligibly small distance from the front of Darcy's eyes)
i. 1.64D
ii. 2.98 D
iii. 2.19 D
iv. 3.49 D
v. 1.37 D
Physics
1 answer:
bogdanovich [222]3 years ago
8 0

Answer:

Explanation:

Power of defective eye = (1 / near point )+ (1 / lens-retina distance )

1 / .0198 m  =  (1 / near point )+ (1 / .02 m  )

= 50.50 = (1 / near point )+ 50

0.50 = 1 / near point

near point = 1 / .5 = 2m

= 200 cm .

for near point at 25 cm , convex lens will be required.

u = - 25 cm , v = - 200 cm ,

1 / v - 1 / u = 1 /f for lens

- 1 / 200 +  1 / 25 = 1 / f

= .04 - .005 = 1 /f

.035 = 1 / f

f = 28.57 cm = 0.2857 m

power = 1 /f

= 1 / 0.2857

= 3.5 D .

i v ) option is correct .

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Answer:

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Explanation:

Given that,

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f = \mu N

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Answer:

Explanation:

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Explanation:

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Vector ~A has a negative x-component 3.07 units in length and a positive y-component 3.17 units in length. When a vector ~B = b1
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Given that A = -3.07i + 3.17j and B = b1i + b1j and C = A + B = 0i + 4.43j

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