Question

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 $$-5D$$?

Answer 1

Using lens maker formulae for given equal concave lens,
$$\dfrac{1}{f} = \left(\dfrac{n_2}{n_1} - 1\right) \left(\dfrac{1}{-R}-\dfrac{1}{R}\right)$$
$$\Rightarrow \dfrac{1}{f} = \left(\dfrac{n_2}{n_1} - 1\right) \left(-\dfrac{2}{R}\right)$$
Where $$n_2$$ and $$n_1$$ are refreactive index of the lens and medium, respectively where
$$n_2 = 1.5$$ & $$n_1 = 1.4$$
Power of lens $$-5D$$
Focal length $$f = \dfrac{1}{-5}\times 100 = -20cm$$
Putting all the values in equation $$1$$,
$$\Rightarrow \dfrac{1}{-20} = \left(\dfrac{1.5}{1.4} - 1\right) \left(-\dfrac{2}{R}\right)$$
$$\Rightarrow \dfrac{1}{20} = \left(\dfrac{0.1}{1.4}\right) \left(\dfrac{2}{R}\right)$$
$$\Rightarrow R = \dfrac{4}{1.4} = 2.86cm$$