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krek1111 [17]
3 years ago
7

The closest distance a book can be read from a pair of reading eyeglasses (Power = 1.55 dp) is 26.0 cm. What is the near distanc

e?(Assume a distance between the eyeglasses and the eyes to be 3.00 cm)
Physics
1 answer:
Mnenie [13.5K]3 years ago
5 0

Answer:

The image distance is 20.0 cm.

Explanation:

Given that,

Power = 1.55 dp

Distance between book to eye = 26.0+3.00=29.0 cm

We need to calculate the focal length

Using formula of focal length

f = \dfrac{1}{P}

Put the value into the formula

f=\dfrac{1}{1.55}

f=0.645\ m

f=64.5\ cm

We need to calculate the image distance

Using lens formula

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

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

Put the value into the formula

\dfrac{1}{v}=\dfrac{1}{64.5}-\dfrac{1}{-29}

\dfrac{1}{v}=\dfrac{187}{3741}

v=20.0\ cm

Hence, The image distance is 20.0 cm.

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where m is the mass of the pendulum, g  = 10 m/s2 is the constant gravitational acceleration, h = 0.05 is the vertical it travels, v is the pendulum velocity at the bottom, which we are trying to solve for.

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