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ANEK [815]
3 years ago
12

A 35-year-old patent clerk needs glasses of 50-cm focal length to read patent applications that he holds 25 cm from his eyes. Fi

ve years later, he notices that while wearing the same glasses, he has to hold the patent applications 40 cm from his eyes to see them clearly.
What should be the focal length of new glasses so that he can read again at 25cm?
Physics
1 answer:
Natali5045456 [20]3 years ago
6 0

Answer:

28.57 cm

Explanation:

We are given that

Focal length,f=50 cm

Distance of application from his eyes,s=40 cm

\frac{1}{s}+\frac{1}{s'}=\frac{1}{f}

\frac{1}{40}+\frac{1}{s'}=\frac{1}{50}

\frac{1}{s'}=\frac{1}{50}-\frac{1}{40}=\frac{4-5}{200}

s'=\frac{200}{-1}=-200 cm

s=25 cm

Substitute the values

\frac{1}{25}-\frac{1}{200}=\frac{1}{f'}

\frac{8-1}{200}=\frac{1}{f'}

\frac{1}{f'}=\frac{7}{200}

f'=\frac{200}{7}=28.57 cm

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Answer:

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Explanation:

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Also let the orbital period of the planet be T_p and its mean distance from the sun be R_p.

Equation (2) therefore implies the following;

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We make the period of the planet T_p the subject of formula as follows;

T_p^2=\frac{T_e^2R_p^3}{R_e^3}\\T_p=\sqrt{\frac{T_e^2R_p^3}{R_e^3}\\}................(4)

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Recall that the orbital period of the earth is about 365.25 days, hence;

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