## Related questions with answers

An equiconvex lens having spherical surfaces of radius $10 \mathrm{~cm}$, a central thickness of $2 \mathrm{~cm}$, and a refractive index of $1.61$ is situated between air and water $(n=1.33)$. An object $5 \mathrm{~cm}$ high is placed $60 \mathrm{~cm}$ in front of the lens surface. Find the cardinal points for the lens and the position and size of the image formed.

Solution

VerifiedThe description of an image means how much the image is larger/smaller than real object . so the image dimension is given by

$\begin{align*} h_{_i}=m h_{_o} \tag{1} \end{align*}$

So all we have to do is to find the term $m$ and to find $m$ we need to find $s_{_i}$ and $s_{_o}$ so let's start with the equation of first focal distance

$\begin{align*} \frac{1}{f_{1}}=\frac{n_{L}-n^{\prime}}{n R_{2}}-\frac{n_{L}-n}{n R_{1}}-\frac{\left(n_{L}-n\right)\left(n_{L}-n^{\prime}\right)}{n n_{L}} \frac{t}{R_{1} R_{2}} \tag{2} \end{align*}$

Where it is given that :

- (1) The lens index $n_{_L}=1.61$
- (2) The index of first medium is air's $n=1$
- (3) The first radius $R_{_1}=10.0\;\mathrm{cm}$
- (4) The second radius $R_{_2}=-10\;\mathrm{cm}$
- (5) The thickness $t=2.0\;\mathrm{cm}$
- (6) The index of second medium is water's $n=1.33$

So substitution in (2) yields

$\begin{align*} \dfrac{1}{f_{_1}} =\dfrac{1.61-1.33}{-10} - \dfrac{1.61-1}{10}-\dfrac{(1.61-1.33)(1.61-1)}{1.61}\dfrac{2}{(10)(-10)} &&\Rightarrow&& \boxed{f_{_1}=-11.51\;\mathrm{cm}} \end{align*}$

The second focal distance

$\begin{align*} f_{_2}=-\dfrac{n'}{n} f_{_1} \tag{3} \end{align*}$

So substitution in (3) yields

$\begin{align*}f_{_2}=-\dfrac{1.33}{1}(-11.51) &&\Rightarrow && \boxed{f_{_2}=15.31\;\mathrm{cm}} \end{align*}$

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