A planoconvex lens has thickness of 4cm. When placed on a...

In first case refraction of the rays, takes place from a plane surface, so we can use ${d}_{app} = \dfrac{d\text{actual}}{μ}$

$3 = \dfrac{4}{μ} ⇒ \mu = \dfrac{4}{3}$

where, ${d}_{app }$ and ${d}_{actual } =$ apparent and real depth, respectively.

Now, in second case, as from Fig. (b), refraction takes place from a spherical surface, so

$\dfrac{\mu2}{v} - \dfrac{\mu1}{u} = \dfrac{{μ}_{2} - {μ}_{1}}{Rm}$

or $\dfrac{1}{( - 25 / 8 )} - \dfrac{4 / 3}{( - 4 )} = \dfrac{1 - 4 / 3}{- R}$

or $\dfrac{1}{3R} = \dfrac{1}{3} - \dfrac{8}{25} = \dfrac{1}{75}$

$R = 25cm$

Now, using Lens' Makers formula, to calculate focal length

$\dfrac{1}{f} = \left(\mu -1\right)\left(\dfrac{1}{R1} - \dfrac{1}{R2}\right)</p><p>\dfrac{1}{f} = \left(\dfrac{4}{3} -1\right)\left(\dfrac{1}{\infty} - \dfrac{1}{( - 25 )}\right) = \dfrac{1}{75}$

$f = 75cm$