Question #164512

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?


1
Expert's answer
2021-02-22T11:10:10-0500




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): 1v11x=1151v1=1151x\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):1v21(32x)=1151v2=1151(32x)\dfrac{1}{v_2}-\dfrac{1}{-(32-x)}=\dfrac{1}{15}\Rightarrow\dfrac{1}{v_2}=\dfrac{1}{15}-\dfrac{1}{(32-x)}


v1v_1 is positive and v2v_2 is negative

v1=v2v_1=-v_2


1v1=1v2\dfrac{1}{v_1}= - \dfrac{1}{v_2}


1151x=[1151(32x)]\Rightarrow \dfrac{1}{15}-\dfrac{1}{x}= -[\dfrac{1}{15}-\dfrac{1}{(32-x)}]


115+132x=21532x(32x)=215\Rightarrow \dfrac{1}{15}+\dfrac{1}{32-x}=\dfrac{2}{15}\Rightarrow \dfrac{32}{x(32-x)}= \dfrac{2}{15}


16x(32x)=115\Rightarrow \dfrac{16}{x(32-x)}= \dfrac{1}{15}


240=32xx2\Rightarrow 240= 32x- x^2

x232x+240=0\Rightarrow x^2-32x+240=0

(x20)(x12)=0\Rightarrow(x-20)(x-12)=0

x=20cm,12cm\Rightarrow x= 20cm, 12cm

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




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