Answer to Question #164512 in Optics for ajay

Question

Tom and Jerry are placed at a distance of 32 cm. Between them a convex lens is placed having a focal length of 15 cm in such a way that images of both Tom and Jerry way are formed at same place. What can be the possible distance of Tom from lens?

Answer 1

Since both images are formed at the same place, one image will real and another will be virtual.

Let image of Tom is real.

Image of O1(Tom): $\dfrac{1}{v_1}-\dfrac{1}{-x}=\dfrac{1}{15}\Rightarrow \dfrac{1}{v_1}=\dfrac{1}{15}- \dfrac{1}{x}$

Image of O2(Jerry):$\dfrac{1}{v_2}-\dfrac{1}{-(32-x)}=\dfrac{1}{15}\Rightarrow\dfrac{1}{v_2}=\dfrac{1}{15}-\dfrac{1}{(32-x)}$

$v_1$ is positive and $v_2$ is negative

$v_1=-v_2$

$\dfrac{1}{v_1}= - \dfrac{1}{v_2}$

$\Rightarrow \dfrac{1}{15}-\dfrac{1}{x}= -[\dfrac{1}{15}-\dfrac{1}{(32-x)}]$

$\Rightarrow \dfrac{1}{15}+\dfrac{1}{32-x}=\dfrac{2}{15}\Rightarrow \dfrac{32}{x(32-x)}= \dfrac{2}{15}$

$\Rightarrow \dfrac{16}{x(32-x)}= \dfrac{1}{15}$

$\Rightarrow 240= 32x- x^2$

$\Rightarrow x^2-32x+240=0$

$\Rightarrow(x-20)(x-12)=0$

$\Rightarrow x= 20cm, 12cm$

Hence, the distance of Tom from lens is either 20cm or 12cm.