In the given figure the radius of curvature of curved face in the plano- convex and the plano-concave lens is $\displaystyle 15$ cm each. The refractive index of the material of the lenses is $\displaystyle 1.5$ . Find the final position of the image formed.

Using the formula $\displaystyle \dfrac{1}{f} = (\mu - 1)\left( \dfrac{1}{R_1} - \dfrac{1}{R_2} \right)$ Focal length of plano-concave lens $\displaystyle = -30$ cm As object is at $\displaystyle \infty$, its real image will be formed at its focus i.e $\displaystyle +30$ cm $\displaystyle v_1 = +30$ cm $\displaystyle = +10$ cm\[\frac{1}{f_2} = \left( \frac{1}{v_2} - \frac{1}{u_2} \right) \]\[\frac{1}{-30} = \left( \frac{1}{v_2} - \frac{1}{10} \right) \]$\displaystyle \therefore v_2 = 15$ cm