A thin glass lens of refractive index `mu_2=1.5` behaves as an interface between two media of refractive indices `mu_1=1.4 and mu_3=1.6` respectively. Determine the focal length of the lens for the shown arrangement of radius of curvature of both surfaces `20 cm.`

Explanation

Using `mu_2/v-mu_1/u=(mu_2-mu_1)/R`, twice with `u=oo`,
we have
`1.5/(v_1)=(1.5-1.4)/(+20)`….(i)
`1.6/(v_2)-1.5/(v_1)=(1.6-1.5)/(-20)`…..(ii)
Solving Eqs.`(i)` and `(ii)`, we get `f=v_2=oo`,
i.e. the system behaves like a glass plate.