A biconvex lens separates two media of refractive indices 1.3 and 1.7. the refractive index of the lens is 1.5 and the radii of curvature of the two sides of the lens are $r_1=10cm$ and $r_2=60cm$ .
The medium of refractive index 1.3 extends to 78 cm from the lens and that of refractive index 1.7 extends of 34 cm from the lens. A luminous object O is at a distance of 144 cm from the lens. Find the position of the final image from the lens.

Sure, here are the step by step solutions: 1. Refraction at surface S1: We use the single surface refraction equation: $\frac{n_3}{v}-\frac{n_1}{u}=\frac{n_3-n_1}{R_1}$ where $n_1=1.1$ (refractive index of air), $n_3=1.3$ (refractive index of the first medium), $u=-66\text{ cm}$ (object distance), $R_1=\infty$ (radius of curvature of surface S1). Plugging in the values, we get: $\frac{1.3}{v}-\frac{1.1}{-66}=\frac{1.3-1.1}{\infty }$ Simplifying, we get: $v=-78\text{ cm}$ Therefore, the image formed by surface S1 is located at a distance of -78 cm from the surface, which means it is located to the left of the surface. 2. Refraction through the lens: We use the general thin lens equation: $\frac{n_3}{v^{\prime }}-\frac{n_1}{u^{\prime }}=\frac{n_3-n_1}{R_1}+\frac{n_2-n_3}{R_2}$ where $n_1=1.1$ (refractive index of air), $n_2=1.5$ (refractive index of the lens), $n_3=1.7$ (refractive index of the second medium), $u^{\prime }=-156\text{ cm}$ (object distance), $R_1=10\text{ cm}$ (radius of curvature of the left surface of the lens), $R_2=-60\text{ cm}$ (radius of curvature of the right surface of the lens). We know that the image formed by surface S1 is located at a distance of -78 cm from the surface, which means it acts as an object for the lens. Therefore, the object distance for the lens is: $u^{\prime \prime }=-78\text{ cm}-(-78\text{ cm})=0$ Plugging in the values, we get: $\frac{1.7}{v^{\prime }}-\frac{1.3}{-156}=\frac{1.5-1.3}{10}+\frac{1.7-1.5}{-60}$ Simplifying, we get: $v^{\prime }=340\text{ cm}$ Therefore, the image formed by the lens is located at a distance of 340 cm to the right of the lens. 3. Refraction at surface S4: We use the single surface refraction equation: $\frac{n_2}{v^{\prime \prime }}-\frac{n_3}{u^{\prime \prime }}=\frac{n_2-n_3}{R_2}$ where $n_2=1.5$ (refractive index of the lens), $n_3=1.7$ (refractive index of the second medium), $u^{\prime \prime }=340-34=306\text{ cm}$ (object distance), $R_2=-60\text{ cm}$ (radius of curvature of the right surface of the lens). Plugging in the values, we get: $\frac{1.5}{v^{\prime \prime }}-\frac{1.7}{306}=\frac{1.5-1.7}{-60}$ Simplifying, we get: $v^{\prime \prime }=198\text{ cm}$ Therefore, the final image is formed at a distance of 198 cm to the right of surface S4.