$\triangle\text{ABC}\sim\triangle\text{A}_1\text{B}_1\text{C}$
$\Rightarrow\frac{\text{A}_1\text{B}_1}{\text{AB}}=\frac{\text{A}_1\text{C}}{\text{AC}}=\frac{(+\text{v})+(-\text{R)}}{(-\text{R})-(-\text{u)}}\ ...(1)$
$\triangle\text{ABP}\sim\triangle\text{A}_1\text{B}_1\text{P}$
$\Rightarrow\frac{\text{A}_1\text{B}_1}{\text{AB}}=\frac{\text{A}_1\text{P}}{\text{AP}}=\frac{+\text{v}}{-\text{u}}\ ...(2)$
$(1)=(2)$
$\Rightarrow\frac{\text{v}-\text{R}}{-\text{R}+\text{u}}=\frac{\text{v}}{-\text{u}}$
$\Rightarrow-\text{uv}+\text{uR}=-\text{vR}+\text{uv}$
$\Rightarrow\text{uR}+\text{vR}=2\text{uv}$
$\div\text{ by }\text{uvR}$
$\Rightarrow\frac{1}{\text{v}}+\frac{1}{\text{u}}=\frac{2}{\text{R}}$
$\because\text{R}=2\text{F}$
$\therefore\frac{1}{\text{v}}+\frac{1}{\text{u}}=\frac{1}{\text{f}}$
$=\frac{0.5}{20}=\frac{5}{200}=\frac{1}{40}$
$\therefore\text{f}=40\text{cm}$
Now, $\frac{1}{\text{f}}=\frac{1}{\text{v}}-\frac{1}{\text{v}}$
$\Rightarrow\text{v}=\frac{\text{fu}}{\text{f}+\text{u}}=\frac{40\times-30}{40-30}$
$\Rightarrow\text{v}=\frac{-40\times30}{10}=-120\text{cm}$
Image is virtual, erect and enlarged in front of lens 120cm away.
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$\text{y}=\text{A}\text{e}^{-\Big(\frac{\text{x}}{\text{a}}+\frac{\text{t}}{\text{T}}\Big)^2}.$