A) \[+80\,cm\]
B) \[-80\,cm\]
C) \[-24\,cm\]
D) \[-100\,cm\]
Correct Answer: D
Solution :
Let the radius of left and right side of the lens be \[\Rightarrow \,y={{x}^{2}}\,+\frac{1}{x}+1;\] and \[y(-2)\,=\frac{9}{2}\] respectively. Then, for the lens in air, we have \[\Rightarrow \] \[=\int\limits_{2/3}^{1}{f(x)-g(x)\,dx=\frac{1}{2}}\] The focal lengths of the two lenses formed by liquid on the left and right side are respectively given by \[\frac{{{(x+1)}^{2}}}{\pi }\,+\frac{{{y}^{2}}}{3}=1\] and \[\therefore \,\,(P{{S}_{1}}+P{{S}_{2}})\,=2\sqrt{\pi }\] the combined focal length f of the system is given by \[|z|\,=4\,\,i.e.,\,\,|z|\,=4\sqrt{2}\] \[\frac{x}{1}=\frac{y}{-2}=\frac{z}{1}\] \[d=\frac{|\vec{v}\times \vec{c}|}{|\vec{c}|}\,=\frac{\sqrt{|\vec{v}{{|}^{2}}|\vec{c}{{|}^{2}}-{{(\vec{v}.\vec{c})}^{2}}}}{|\vec{c}|}\]
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