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): v11−−x1=151⇒v11=151−x1
Image of O2(Jerry):v21−−(32−x)1=151⇒v21=151−(32−x)1
v1 is positive and v2 is negative
v1=−v2
v11=−v21
⇒151−x1=−[151−(32−x)1]
⇒151+32−x1=152⇒x(32−x)32=152
⇒x(32−x)16=151
⇒240=32x−x2
⇒x2−32x+240=0
⇒(x−20)(x−12)=0
⇒x=20cm,12cm
Hence, the distance of Tom from lens is either 20cm or 12cm.
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