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nlexa [21]
3 years ago
9

A 80-cm-wide object seen through an optical device appears inverted and 50 cm wide. What is the magnification of the optical dev

ice?
A. 1.3
B. 0.63
C. -0.63
D. 2.5
E. -1.3
Physics
1 answer:
Arte-miy333 [17]3 years ago
6 0
The magnification is the ratio between the size of the image h_i and the size of the original object h_o:
M= \frac{h_i}{h_o}
The sign of the magnification contains another piece of information: if it's positive, then the image is upright, if it's negative, the image is inverted. In our problem, the image is inverted, so the magnification will have a negative sign, and it will be
M=(-) \frac{50 cm}{80 cm}=-0.63
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Answer:

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Explanation:

Given;

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m1c∆T1 = m2c∆T2 + m2Lf

m1c(T1-T) = m2c(T-T2) + m2Lf

m1cT1 - m1cT = m2cT - m2cT2 + m2Lf

m1cT + m2cT = m1cT1 + m2cT2 - m2Lf

T(m1c + m2c) = m1cT1 + m2cT2 - m2Lf

T = (m1cT1 + m2cT2 - m2Lf)/(m1c + m2c)

Substituting the values;

T = (m1cT1 + m2cT2 - m2Lf)/(m1c + m2c)

T = (0.89×4186×31 + 0.075×4186×0 - 0.075×33.5 × 10^4)/(0.89×4186 + 0.075×4186)

T = 22.37 °C

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