1.An object is 1.0 cm tall and its erect image is 4.0 cm tall. What is the exact magnification?
2.When using the Lens Equation, a virtual image has a
a. positive object distance
b. negative image distance
c. positive image distance
3.The scale at the bottom of the diagram is calibrated in cm. Find the focal length of the lens (in cm). (Do not include units with the answer.)
Answer
(1) Magnification
=\frac{\text { Size of imay }}{\text { Size of object }}=\frac{4.0 \mathrm{~cm}}{1.0 \mathrm{~cm}}=4cm
(2) Lens epuatien virtual Imaye have
Megative image distance
(3) Lens formula =\frac{1}{d_{0}}+\frac{1}{d i}=\frac{1}{d f}
Given
position of lense =31 \mathrm{~cm}
pasition of object =11 \mathrm{~cm}
position of image =51 \mathrm{~cm}
\therefore do ob'ject distama form lers =31-11=20 \mathrm{~cm}
d i= imase distance form \operatorname{len} 1=5 \mid-31=20 \mathrm{~cm}
\operatorname{lon} \frac{1}{20}+\frac{1}{20}=\frac{1}{f}
\frac{1}{f}=\frac{2}{20}=\frac{1}{10}
f=10 \mathrm{~cm}
