Question

An equiconvex lens having spherical surfaces of radius 10 cm10 \mathrm{~cm}, a central thickness of 2 cm2 \mathrm{~cm}, and a refractive index of 1.611.61 is situated between air and water (n=1.33)(n=1.33). An object 5 cm5 \mathrm{~cm} high is placed 60 cm60 \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

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

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

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

1f1=nLnnR2nLnnR1(nLn)(nLn)nnLtR1R2\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 nL=1.61n_{_L}=1.61
  • (2) The index of first medium is air's n=1n=1
  • (3) The first radius R1=10.0  cmR_{_1}=10.0\;\mathrm{cm}
  • (4) The second radius R2=10  cmR_{_2}=-10\;\mathrm{cm}
  • (5) The thickness t=2.0  cmt=2.0\;\mathrm{cm}
  • (6) The index of second medium is water's n=1.33n=1.33

So substitution in (2) yields

1f1=1.611.33101.61110(1.611.33)(1.611)1.612(10)(10)f1=11.51  cm\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

f2=nnf1\begin{align*} f_{_2}=-\dfrac{n'}{n} f_{_1} \tag{3} \end{align*}

So substitution in (3) yields

f2=1.331(11.51)f2=15.31  cm\begin{align*}f_{_2}=-\dfrac{1.33}{1}(-11.51) &&\Rightarrow && \boxed{f_{_2}=15.31\;\mathrm{cm}} \end{align*}

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