• # question_answer As shown in figure, a convergent lens is placed inside a cell filled with liquid. The lens has focal length $+20\,cm$ when in air and its material has refractive index $1.50$. If the liquid has refractive index $1.60$, the focal length of the system is: A) $+80\,cm$                                        B) $-80\,cm$ C) $-24\,cm$                                         D) $-100\,cm$

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 $|z|\,=4\,\,i.e.,\,\,|z|\,=4\sqrt{2}$ $\frac{x}{1}=\frac{y}{-2}=\frac{z}{1}$ $d=\frac{|\vec{v}\times \vec{c}|}{|\vec{c}|}\,=\frac{\sqrt{|\vec{v}{{|}^{2}}|\vec{c}{{|}^{2}}-{{(\vec{v}.\vec{c})}^{2}}}}{|\vec{c}|}$