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3 years ago

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Mary's glasses have +4.4D converging lenses. This gives her a near point of 20cm. What is the location of her near point when sh

e is not wearing her glasses?
Express your answer with the appropriate units.

1 answer:

VikaD [51]3 years ago

3 0

Answer:

The position of near point is 1.68 m.

Explanation:

Given that,

Power = + 4.4 D

Object distance = 20 cm

We need to calculate the focal length

Using formula of power

$P =\dfrac{1}{f}$

$f=\dfrac{1}{P}$

$f=\dfrac{1}{4.4}$

$f=0.227\ m$

We need to calculate the image distance

Using formula of lens

$\dfrac{1}{f}=\dfrac{1}{v}-\dfrac{1}{u}$

$\dfrac{1}{v}=\dfrac{1}{f}+\dfrac{1}{u}$

Put the value into the formula

$\dfrac{1}{v}=\dfrac{1}{0.227}-\dfrac{1}{0.20}$

$\dfrac{1}{v}=-\dfrac{135}{227}$

$v=-1.68\ m$

Hence, The position of near point is 1.68 m.

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Read 2 more answers

A diverging lens with a focal length of 14 cm is placed 12 cm to the right of a converging lens with a focal length of 21 cm. An

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Answer:

-0.233 m left of diverging lens and ( 0.12 - 0.233 ) = -113 m left of conversing

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0.023 m right of diverging lens

Explanation:

given data

focal length f2 = 14 cm = -0.14 m

Separation s = 12 cm = 0.12 m

focal length f1 = 21 cm = 0.21 m

distance u1 = 38 cm

to find out

final image be located and Where will the image

solution

we find find image location i.e v2

so by lens formula v1 is

1/f = 1/u + 1/v ...............1

v1 = 1/(1/f1 - 1/u1)

v1 = 1/( 1/0.21 - 1/0.38)

v1 = 0.47 m

and

u2 = s - v1

u2 = 0.12 - 0.47

u2 = -0.35

so from equation 1

v2 = 1/(1/f2 - 1/u2)

v2 = 1/(-1/0.14 + 1/0.35)

v2 = -0.233 m

so -0.233 m left of diverging lens and ( 0.12 - 0.233 ) = -113 m left of conversing

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for Separation s = 45 cm = 0.45 m

v1 = 1/(1/f1 - 1/u1)

v1 =0.47 m

and

u2 = s - v1

u2 = 0.45 - 0.47 =- 0.02 m

so

v2 = 1/(1/f2 - 1/u2)

v2 = 1/(-1/0.14 + 1/0.02)

v2 = 0.023

so here 0.023 m right of diverging lens

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