Q143 / 160 · TS EAMCET 2017 · Physics
Consider a light source placed at a distance of $1.5 \mathrm{~m}$ along the axis facing the convex side of a spherical mirror of radius of curvature $1 \mathrm{~m}$. The position $\left(s^{\prime}\right)$, natwre and magnification (m) of the image are
- Correct answerA$s^{\prime}=0.375 \mathrm{~m}$, virtual, upright, $m=0.25$
- B$s^{\prime}=0.375 \mathrm{~m}$, real, inverted, $m=025$
- C$s^{\prime}=3.75 \mathrm{~m}$, virtual, inverted, $m=2.5$
- D$s^{\prime}=3.75 \mathrm{~m}$, real, upright, $m=2.5$
Solution
Given,
Distance of light source $(u)=-1.5 \mathrm{~m}=-\frac{-3}{2} \mathrm{~m}$
Radius of curvature of mirror $(R)=1 \mathrm{~m}$
of We know that,
$$
\frac{1}{f}=\frac{1}{V}+\frac{1}{u} \quad\left[\because f=\frac{R}{2}\right]
$$
$\therefore \quad \frac{2}{R}=\frac{1}{v}+\frac{1}{u} \Rightarrow \frac{2}{1}=\frac{1}{v}+\frac{1}{\frac{-3}{2}}$
$$
\begin{array}{r}
\frac{2}{1}=\frac{1}{v}-\frac{2}{3} \Rightarrow \frac{1}{v}=2+\frac{2}{3} \\
\frac{1}{v}=\frac{6+2}{3} \Rightarrow \frac{1}{v}=\frac{8}{3} \Rightarrow v=\frac{3}{8}=0.375 \mathrm{~m}
\end{array}
$$
The image is virtual.
Magnification $(m)=\frac{-v}{u}=\frac{\frac{-3}{8}}{\frac{-3}{2}}=\frac{1}{4}=0.25$
Hence, the correct answer is option: 'A'.