MCQ
Given a thin convex lens (refractive index $\mu_2$ ), kept in a liquid (refractive index $\mu_1, \mu_1<\mu_2$ ) having radii of curvature $\left|R_1\right|$ and $\left|R_2\right|$. Its second surface is silver polished. Where should an object be placed on the optic axis so that a real and inverted image is formed at the same place?
- A$\frac{\mu_1\left| R _1\right| \cdot\left| R _2\right|}{\mu_2\left(\left| R _1\right|+\left| R _2\right|\right)-\mu_1\left| R _1\right|}$
- ✓$\frac{\mu_1\left| R _1\right| \cdot\left| R _2\right|}{\mu_2\left(\left| R _1\right|+\left| R _2\right|\right)-\mu_1\left| R _2\right|}$
- C$\frac{\mu_1\left| R _1\right| \cdot\left| R _2\right|}{\mu_2\left(2\left| R _1\right|+\left| R _2\right|\right)-\mu_1 \sqrt{\left| R _1\right| \cdot\left| R _2\right|}}$
- D$\frac{\left(\mu_2+\mu_1\right)\left|R_1\right|}{\left(\mu_2-\mu_1\right)}$
