Question #207988


An amoeba is 0.305cm away from the 0.300cm focal length objective lens

of a microscope. (Use the thin lens formula and the magnification

formula)


A. Where is the image formed by the objective lens?


B. What is the image magnification?


C. An eyepiece with a 2.00cm focal length is placed 20.0cm from the

objective. Where is the final image?





1
Expert's answer
2021-06-17T15:27:51-0400

Gives

Object distance do=0.305cmd_o=0.305cm

Focal length


fo=0.300f_o=0.300

We know that

Part(a)

1f=1di+1do\frac{1}{f}=\frac{1}{d_i}+\frac{1}{d_o}

1di=1f1do\frac{1}{d_i}=\frac{1}{f}-\frac{1}{d_o}

1di=10.30010.305\frac{1}{d_i}=\frac{1}{0.300}-\frac{1}{0.305}

di=18.3cmd_i=18.3cm

Part(b)

Magnification (m) =dido=-\frac{d_i}{d_o}

m=18.30.305=60m=-\frac{18.3}{0.305}=-60

Part(c)

fo=2.00cmf_o=2.00cm

do=2018.3=1.70cmd'_o=20-18.3=1.70cm

We know that

1di=1fo1d0\frac{1}{d'_i}=\frac{1}{f_o}-\frac{1}{d'_0}

1di=12.0011.70=334\frac{1}{d'_i}=\frac{1}{2.00}-\frac{1}{1.70}=-\frac{3}{34}

di=11.33cmd'_i=-11.33cm


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