Q142 / 160 · TS EAMCET 2022 (19 Jul Shift 2) · Physics
A spherical glass is attached to a rigid wall as shown in the figure. An observer located at point $\mathrm{O}$ is looking at a point $\mathrm{A}$ on the wall. The refractive index of the glass is 1.5 and that of air is 1.0. The distances are $\mathrm{OA}=8 \mathrm{~cm}, \mathrm{XA}=3$ $\mathrm{cm}$. If the radius of curvature of spherical glass surface is $R$ $=5 \mathrm{~cm}$, then the apparent distance of $\mathrm{A}$ from observer $\mathrm{O}$ is
A spherical glass is attached to a rigid wall as shown in the figure. An observer located at point $\mathrm{O}$ is looking at a point $\mathrm{A}$ on the wall. The refractive index of the glass is 1.5 and that of air is 1.0. The distances are $\mathrm{OA}=8 \mathrm{~cm}, \mathrm{XA}=3$ $\mathrm{cm}$. If the radius of curvature of spherical glass surface is $R$ $=5 \mathrm{~cm}$, then the apparent distance of $\mathrm{A}$ from observer $\mathrm{O}$ is
- A6.5 cm
- B8.5 cm
- C7.0 cm
- Correct answerD7.5 cm
Solution
At convex surface
$\begin{aligned} & \frac{\mu_2}{\mathrm{~V}}-\frac{\mu_1}{\mathrm{u}}=\frac{\mu_2-\mu_1}{\mathrm{R}} \\ & \Rightarrow \quad \frac{1}{\mathrm{~V}}-\frac{1.5}{-3}=\frac{1-1.5}{-5} \\ & \Rightarrow \quad \frac{1}{\mathrm{~V}}+0.5=+0.1 \\ & \Rightarrow \quad \frac{1}{\mathrm{~V}}=-0.4 \\ & \Rightarrow \quad \mathrm{V}=-2.5 \mathrm{~cm} \\ & \quad \text { So, OI }=5+2.5=7.5 \mathrm{~cm}\end{aligned}$
$\begin{aligned} & \frac{\mu_2}{\mathrm{~V}}-\frac{\mu_1}{\mathrm{u}}=\frac{\mu_2-\mu_1}{\mathrm{R}} \\ & \Rightarrow \quad \frac{1}{\mathrm{~V}}-\frac{1.5}{-3}=\frac{1-1.5}{-5} \\ & \Rightarrow \quad \frac{1}{\mathrm{~V}}+0.5=+0.1 \\ & \Rightarrow \quad \frac{1}{\mathrm{~V}}=-0.4 \\ & \Rightarrow \quad \mathrm{V}=-2.5 \mathrm{~cm} \\ & \quad \text { So, OI }=5+2.5=7.5 \mathrm{~cm}\end{aligned}$
Hence, the correct answer is option: 'D'.