A biconvex thick lens is constructed with glass $(\mu=1.50)$ Each of the surfaces has a radius of 10cm and the thickness at the middle is 5cm. Locate the image of an object placed far away from the lens.
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$\text{R}_1=\text{R}_2=10\text{cm}, \ \text{t}=5\text{cm}, \ \text{u}=-\infty$
For the first refraction, (at A)
$\frac{\mu_{\text{g}}}{\text{v}}-\frac{\mu_{\text{a}}}{\text{u}}=\frac{\mu_{\text{g}}-\mu_{\text{a}}}{\text{R}_1}$
$\Rightarrow \frac{1.5}{\text{v}}-\Big(-\frac{1}{\infty}\Big)=\frac{1.5-1}{10}$
$\Rightarrow \frac{1.5}{\text{v}}=\frac{0.5}{10}$
$\Rightarrow\text{v}=30\text{cm}$
Again, for $2^{nd}$ surface, u = (30 - 5) = 25cm (virtual object)
$R_2 = -10cm$
So, $\frac{1}{\text{v}}-\frac{15}{25}=\frac{-0.5}{-10}\Rightarrow\text{v}=9.1\text{cm.}$
So, the image is formed 9.1cm further from the $2^{nd}$ surface of the lens.
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