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3 years ago

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A microscope has an objective lens with a focal length of 14.0mm . A small object is placed 0.80mm beyond the focal point of the

objective lens.Part A: At what distance from the objective lens does a real image of the object form? Express your answer to two significant figures and include the appropriate units.Part B: What is the magnification of the real image? Express your answer using two significant figures.Part C: If an eyepiece with a focal length of 2.5 cm is used, with a final image at infinity, what will be the overall angular magnification of the object? Express your answer using two significant figures.

1 answer:

Nikolay [14]3 years ago

3 0

Answer:

A. 260 mm

B. - 18

C. 175

Explanation:

A

The expression for the lens equation is

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

$\frac{1}{14.0mm} = \frac{1}{14.80 mm} + \frac{1}{v}$

$\frac{1}{v} = \frac{1}{14.0} - \frac{1}{14.80}$

v = 259 mm

= 260 mm or 26 cm (to 2 s.f)

check the attached files for additional solution

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Sixth question (bottom right):
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Learn more about gravitational force:

brainly.com/question/1724648

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For this case we have the following system with the forces on the figure attached.

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Part b

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And the work donde by the spring on this case is given by:

$W_{spring}= -\frac{1}{2} Kx^2 =-\frac{1}{2} 200 N/m (0.1m)^2=-1 J$

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