Question

An object is placed

(i) 12cm

(ii) 4cm from a converging (convex) lens of focal length 6cm. Calculate the image position and the magnification in each case and draw sketches illustrating the formation of the image.

Answer 1

$(i). Object\:distance\:\left(O\right)=12cm$

$f=focal\:length$

$\:v=image\:distance\:$

$u=\:obect\:distance$

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

$\frac{1}{v}=\frac{1}{f}-\frac{1}{u}\:$

$\frac{1}{v}=\frac{1}{6cm}-\frac{1}{12cm}$

$v=\frac{1}{\frac{1}{12cm}}=12cm$

Image distance is equal to 12cm.

$Magnification\left(M\right)\:=\frac{v}{u}\:\:\:$

$M=\frac{12cm}{12cm}=1$

The size of the image is equal to that of the object.

$(ii). Object\:distance\:\left(O\right)=4cm$

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

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

$\frac{1}{v}=\frac{1}{6cm}-\frac{1}{4cm}$

$v=\frac{1}{\frac{-1}{12cm}}=-12cm\:$

The negative sign on the image distance indicates that the image is formed behind the mirror, virtual and upright.

$Magnification\left(M\right)\:=\frac{v}{u}\:\:$

$M=\frac{-12cm}{4cm}\:\:=-3\:$

The negative sign on the value of magnification indicates that the object decreases in size to form an image during magnification.