Question

4. A 1.5 m tall man is standing 2m away from the concave lens (remember f is negative) of a peephole with a focal length of 3.0 cm. a) What is the distance to the image in centimeters? b) What is the magnification of the image in meters?

Answer 1

Here,

Size of object = 1.5 m;

Object distance (u) = - 2m = -200cm

focal length (f) = - 3.0 cm

Image distance (v) = ?

Using lens formula

`\begin{gathered} (1)/(f)=\text{ }(1)/(v)-(1)/(u); \\ (1)/(-3)=(1)/(v)-(1)/(-200); \\ (1)/(-3)=\text{ }(1)/(v)\text{ +}(1)/(200) \\ (1)/(v)=\text{ -}(1)/(3)-(1)/(200)=-\text{ }(203)/(600) \\ v=\text{ - }(600)/(203)=\text{ -2.96 cm} \end{gathered}`

Now magnification is given by

`Magnification=\text{ }(v)/(u)=\text{ }(2.96)/(200)=0.0148`

Final answer is :-

Image distance = - 2.96 cm & magnification = 0.0148