Q129 / 200 · NEET 2011 (Screening) · Physics
A biconvex lens has a radius of curvature of magnitude $20 \mathrm{~cm}$. Which one of the following options describe best the image formed of an object of height $2 \mathrm{~cm}$ placed $30 \mathrm{~cm}$ from the lens?
- AVirtual, upright, height $=0.5 \mathrm{~cm}$
- Correct answerBReal, inverted, height $=4 \mathrm{~cm}$
- CReal, inverted, height $=1 \mathrm{~cm}$
- DVirtual, upright, height $=1 \mathrm{~cm}$
Solution
In general we have assumed $\mu=1.5$
So, $f=20 \mathrm{~cm}$
$$
\begin{aligned}
\frac{1}{f} & =\frac{1}{y}+\frac{1}{u} \\
\frac{1}{20} & =\frac{1}{y}+\frac{1}{30} \\
\frac{1}{y} & =\frac{1}{20}-\frac{1}{30}=\frac{10}{600} \\
y & =60 \mathrm{~cm} \\
\frac{h_i}{h_0} & =2 \\
h_i & =2 \times \mid h_o \\
h_i & =4 \mathrm{~cm}
\end{aligned}
$$
Here, Image is real, inverted, magnified field and height of image is $4 \mathrm{~cm}$.
Hence, the correct answer is option: 'B'.