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A bi-convex lens is formed with two thin plano-convex lenses as shown in the figure. Refractive index $n$ of the first lens is 1.5 and that of the second lens is 1.2. Both the curved surfaces are of the same radius of curvature $R=14 \mathrm{~cm}$. For this bi-convex lens, for an object distance of $40 \mathrm{~cm}$, the image distance will be
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$40.0 \mathrm{~cm}$
According to lens maker's formula
$\frac{1}{f}=(n-1)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)$
For the first lens : $n=1.5, R_1=+14 \mathrm{~cm}, R_2=\infty$
$\therefore \quad \frac{1}{f_1}=(1.5-1)\left(\frac{1}{14}-\frac{1}{\infty}\right)=\frac{0.5}{14}$
For the second lens : $n=1.2, R_1=\infty, R_2=-14 \mathrm{~cm}$
$\therefore \quad \frac{1}{f_2}=(1.2-1)\left(\frac{1}{\infty}-\frac{1}{-14}\right)=\frac{0.2}{14}$
The focal length of the bi-convex lens is
$\frac{1}{f}=\frac{1}{f_1}+\frac{1}{f_2}=\frac{0.5}{14}+\frac{0.2}{14}=\frac{0.7}{14}=\frac{1}{20}$
According to thin lens formula; $\frac{1}{v}-\frac{1}{u}=\frac{1}{f}$
$\Rightarrow \frac{1}{v}-\frac{1}{-40}=\frac{1}{20}$ or $v=40 \mathrm{~cm}$
$\frac{1}{f}=(n-1)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)$
For the first lens : $n=1.5, R_1=+14 \mathrm{~cm}, R_2=\infty$
$\therefore \quad \frac{1}{f_1}=(1.5-1)\left(\frac{1}{14}-\frac{1}{\infty}\right)=\frac{0.5}{14}$
For the second lens : $n=1.2, R_1=\infty, R_2=-14 \mathrm{~cm}$
$\therefore \quad \frac{1}{f_2}=(1.2-1)\left(\frac{1}{\infty}-\frac{1}{-14}\right)=\frac{0.2}{14}$
The focal length of the bi-convex lens is
$\frac{1}{f}=\frac{1}{f_1}+\frac{1}{f_2}=\frac{0.5}{14}+\frac{0.2}{14}=\frac{0.7}{14}=\frac{1}{20}$
According to thin lens formula; $\frac{1}{v}-\frac{1}{u}=\frac{1}{f}$
$\Rightarrow \frac{1}{v}-\frac{1}{-40}=\frac{1}{20}$ or $v=40 \mathrm{~cm}$
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