JEE Advanced 2023 — Paper 1 — Question 25

$\mathbf{2 2 0}$ or $\mathbf{8 0}$ or $\mathbf{1 5 0}$ and you can explore other possibilities also Case I:If $\mathrm{M}_{1}$ is placed at distance $\left(20+\frac{80}{7}\right) \mathrm{cm}$ from lens, the rays retrace its path and image will be formed at S . $\therefore \mathrm{n}=220$ From mirror $\frac{1}{\mathrm{~V}}+\frac{1}{-10}=\frac{1}{-12} \Rightarrow \mathrm{~V}=60 \mathrm{~cm}$ Refraction from lens $\frac{1}{\mathrm{~V}}-\frac{1}{-80}=\frac{1}{10}$ $\mathrm{V}=\frac{80}{7} \mathrm{~cm}$

Case II: Consider reflection from $M_{2}$ then refraction from lens. Image is at $\frac{80}{7} \mathrm{~cm}$ left of lens if $\mathrm{M}_{1}$ is placed at

position of this image by lens, rays reflect back and final image is formed at $\mathrm{S} \therefore \mathrm{n}=80$. Case III: First consider refraction from lens then reflection from $M_{1}$ if image due to this reflection is formed at

$\frac{80}{7} \mathrm{~cm}$ left of the lens, then image after refraction with lens and reflection with $\mathrm{M}_{2}$ will be formed at $S$. Then

distance between $L$ and $M_{1}=10+\frac{80}{7}=\frac{150}{7}=\frac{n}{7} \Rightarrow n=150$