Question

1.  Len DeHand places a converging lens of focal length 5.0 cm, a distance of  3.0 cm from a dime with diameter(height) of 1.8 cm.  Calculate…
2. the image distance;
3. the image diameter;
4. the magnification.  
5. What type of image is this?
6. How is the lens being used?

Answer 1

1.) Here,

u= -3 cm

f = + 5 cm

v= ?

2.) Using lens formula,

$\dfrac{1}{f}=\dfrac{1}{v}-\dfrac{1}{u}$

$\dfrac{1}{v}=\dfrac{1}{f}+\dfrac{1}{u}\\ \\\\ v=\dfrac{f\times u}{f+u}=\dfrac{-15}{2}=-7.5\ cm$

Image will be 7.5 cm from the lens and on the same side as object.

3.) Diameter of object $h_o=1.8\ cm$

We know that $\dfrac{h_i}{h_o}=\dfrac{v}{u}$

$h_i=\dfrac{v\times h_o}{u}=\dfrac{-7.5\times 1.8}{-3}= 4.5 \ cm$

So, the diameter (height ) of image will be 4.5 cm

4.) Magnification $m= \dfrac{h_i}{h_o}=\dfrac{4.5}{1.8}=2.5$

5.) Image is virtual , erect and enlarged.

6.) Lens is used here as a magnifying glass.