Physics - Geometrical Optics Question with Solution | TestHub

$\frac{1}{\mathrm{f}}=\left(\frac{3}{2}-1\right)\left(\frac{1}{{\mathrm{R}}_1}+\frac{1}{{\mathrm{R}}_2}\right)$ $\frac{1}{\mathrm{f}}=\frac{1}{2}\left(\frac{1}{{\mathrm{R}}_1}+\frac{1}{{\mathrm{R}}_2}\right)$ $\frac{1}{{\mathrm{f}}^{\prime }}=\left(\frac{\frac{3}{2}}{4/3}-1\right)\left(\frac{1}{{\mathrm{R}}_1}+\frac{1}{{\mathrm{R}}_2}\right)$ $\frac{1}{{\mathrm{f}}^{\prime }}=\left[\frac{9}{8}-1\right]\left[\frac{1}{{\mathrm{R}}_1}+\frac{1}{{\mathrm{R}}_2}\right]$ $\frac{1}{{\mathrm{f}}^{\prime }}=\frac{1}{8}\left(\frac{1}{{\mathrm{R}}_1}+\frac{1}{{\mathrm{R}}_2}\right)=\frac{1}{4\mathrm{f}}$ ${\mathrm{f}}^{\prime }=4\mathrm{f}$ \section*{Initial :} \begin{center}

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$$\mu =\frac{3}{2}$$

$\frac{1}{\mathrm{f}}=\frac{1}{\mathrm{v}}+\frac{1}{\mathrm{d}}$ $\frac{1}{\mathrm{v}}=\frac{1}{\mathrm{f}}-\frac{1}{\mathrm{d}}\Rightarrow \mathrm{v}=\frac{\mathrm{fd}}{\mathrm{d}-\mathrm{f}}$ Image is inverted, real and magnified.

$$\therefore v>0\Rightarrow d>f$$

and $\frac{\mathrm{f}}{\mathrm{d}-\mathrm{f}}>1$ $\mathrm{f}>\mathrm{d}-\mathrm{f}$ 2f $>$ d Final $\frac{1}{4\mathrm{f}}=\frac{1}{\mathrm{v}}+\frac{1}{\mathrm{d}}$ $\frac{1}{\mathrm{v}}=\frac{1}{4\mathrm{f}}-\frac{1}{\mathrm{d}}$ ${\mathrm{v}}^{\prime }=\frac{4\mathrm{fd}}{\mathrm{d}-4\mathrm{f}}<0\&\left|\frac{{\mathrm{v}}^{\prime }}{{\mathrm{u}}^{\prime }}\right|>1$ ∴ image is virtual \& magnified.