# Calculate the radius of curvature of an equi-concave lens of refractive index $1.5$, when it is kept in a medium of refractive index $1.4$, to have a power of $-5\; D$ ?

The radius can be taken $'R'$    [ or, Let the radius be $'R'$]

then, $\inline R_{1}=-R$ and $\inline R_{2}=R$

According to question

Power of Lens, = -5D

Refractive index

$\\n_2=1.5\\n_1=1.4$

we know the lens makers formula  -

$P=\frac{1}{f}=(\frac{n_2}{n_1}-1)\left ( \frac{1}{R_{1}}-\frac{1}{R_{2}} \right )$

On putting the values we have -

$\\-5m^{-1}=(\frac{1.5}{1.4}-1)(\frac{-1}{R}-\frac{1}{R})\\\\\Rightarrow 5=\frac{0.2}{1.4R}\\\\\Rightarrow R=\frac{0.2}{7}=0.0286 m=2.86 cm$

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