A) 40 cm
B) 60 cm
C) 25 cm
D) 15 cm
Correct Answer: A
Solution :
Let us first locate the image \[{{I}_{1}}\] of source S formed by the lens. Here, \[u=-12\,\,cm\] and \[f=15\,cm\] We have, \[\frac{1}{v}-\frac{1}{u}\,=\frac{1}{f}\,\Rightarrow \] \[\frac{1}{v}=\frac{1}{f}+\frac{1}{u}=\frac{1}{15}-\frac{1}{12}\] \[\Rightarrow \,v=-60\,cm\] \[\therefore \,\,P{{I}_{1}}\,=60\,cm\] the image \[{{I}_{1}}\] acts as the object for the mirror. The mirror forms an image \[{{I}_{2}}\] of the object \[{{I}_{1}}\]. this image \[{{I}_{2}}\] then act as the object for the lines and the final beam comes out parallel to the principal axis. Clearly \[{{I}_{2}}\] must be at the focus of the lens. We have \[{{I}_{1}}{{I}_{2}}={{I}_{1}}P+P{{I}_{2}}=60\,cm\,+15\,cm\,=75\,cm\] Suppose, the distance of the mirror from \[{{I}_{2}}\] is x cm For there reflection from the mirror, f = - 20 cm \[u=M{{I}_{1}}\,=-(75+x)\,cm\] Using \[\frac{I}{v}+\frac{I}{u}\,=\frac{1}{f}\,\] we get, \[\frac{1}{x}+\,\frac{1}{75+x}=\frac{1}{20}\] \[\Rightarrow \,\,\frac{75+2x}{(75+x)x}\,=\frac{1}{20}\,\,\,\Rightarrow \,{{x}^{2}}+35\,x-1500\,=0\] \[\Rightarrow \,\,(x+60)\,(x-25)\] this gives \[x=25\] or - 60 As the negative sign has no physical meaning, only positive sign should be taken. Taking x = 25, the separation between the lens and the mirror is: \[P{{I}_{2}}\,+x=15\,\,cm\,+25\,cm\,=40\,cm\]
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