Answer:
The refractive power of the contact lenses is 2.75 D
Explanation:
Given that,
Image distance v = -80.0 cm
Object distance u= -25 cm
We need to calculate the refractive power
Using lens's formula
[tex]\dfrac{1}{f}=\dfrac{1}{v}-\dfrac{1}{u}[/tex]
Where, v = Image distance
u = Object distance
[tex]\dfrac{1}{f}=\dfrac{1}{-80}-\dfrac{1}{-25}[/tex]
[tex]\dfrac{1}{f}=\dfrac{11}{400}\ cm^{-1}[/tex]
[tex]\dfrac{1}{f}=\dfrac{11}{4}\ m^{-1}[/tex]
The refractive power of the contact lenses
[tex]D=\dfrac{1}{f}[/tex]
[tex]D=\dfrac{11}{4}[/tex]
[tex]D=2.75\ diopters[/tex]
Hence, The refractive power of the contact lenses is 2.75 D.
The refractive power of the contact lenses that will enable her to comfortably see objects at distance of 25.0 cm from her eyes is 2.75 D
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We will solve this problem using following formula:
[tex]\large{ \boxed {\frac{1}{s_o} + \frac{1}{s_i} = \frac{1}{f}}}[/tex]
where:
so = distance of object from lens
si = distance of image from lens
f = focal length of lens
Let's tackle the problem!
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Given:
distance of object = s_o = 25.0 cm = 0.25 m
distance of image = s_i = -80.0 cm = -0.80 m ( virtual image )
Unknown:
refractive power = P = ?
Solution:
[tex]\frac{1}{s_o} + \frac{1}{s_i} = \frac{1}{f}[/tex]
[tex]\frac{s_o + s_i}{s_o s_i} = P[/tex]
[tex]\large { \boxed {P = \frac{s_o + s_i}{s_o s_i}} }[/tex]
[tex]P = \frac{0.25 - 0.80}{0.25 (-0.80)}[/tex]
[tex]P = \frac{-0.55}{-0.2}[/tex]
[tex]P = 2.75 \texttt{ D}[/tex]
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The refractive power of the contact lenses that will enable her to comfortably see objects at distance of 25.0 cm from her eyes is 2.75 D
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Grade: High School
Subject: Physics
Chapter: Light
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Keywords: Light , Mirror , Magnification , Ray , Diagram , Image , Real , Virtual
