Question

An equi-convex lens of $\mu =1.5$ and $R=20\text{cm}$ is cut into two equal parts along its axis. Two parts are then separated by a distance of $120\text{cm}$ (as shown in figure). An object of height $3\text{mm}$ is placed at a distance of $30\text{cm}$ to the left of the first half lens. The final image will form at

• A. $120\text{cm}$ to the right of first half lens, $3\text{mm}$ in size and inverted.
• B. $150\text{cm}$ to the right of first half lens, $3\text{mm}$ in size and erect.
• C. $150\text{cm}$ to the right of first half lens, $6\text{mm}$ in size and erect.
• D. $120\text{cm}$ to the right of first half lens, $3\text{mm}$ in size and inverted

Answer 1

The correct option is B $150\text{cm}$ to the right of first half lens, $3\text{mm}$ in size and erect.

Given that,
Refractive index of equi-convex lens, $\mu =1.5$
Radius of curvature of lens, $R=20\text{cm}$

Sign convention: Direction along incident ray is taken as $+ve$.

The focal length of the lens is

$\frac{1}{f}=(\frac{{\mu }_2}{{\mu }_1}-1)(\frac{1}{R_1}-\frac{1}{R_2})$

$\Rightarrow \frac{1}{f}=(\frac{1.5}{1}-1)(\frac{1}{20}-\frac{1}{-20})$

$\Rightarrow f=+20\text{cm}$

Thus, focal length of both the lens are $+20\text{cm}$.

From question, object is placed at $u_1=-30\text{cm}$ from the left lens.

Using lens formula,

$\frac{1}{v}-\frac{1}{u}=\frac{1}{f}$

$\Rightarrow \frac{1}{v_1}-\frac{1}{-30}=\frac{1}{+20}$

$\therefore v_1=+60\text{cm}$

$m_1=\frac{v_1}{u_1}=\frac{+60}{-30}=-2$

Distance between image and right lens $=120-60=60\text{cm}$

For second lens, $u_2=-60\text{cm}$

$\frac{1}{v_2}-\frac{1}{-60}=\frac{1}{+20}$

$\Rightarrow v_2=+30\text{cm}$

$\therefore m_2=\frac{v_2}{u_2}=\frac{+30}{-60}=-\frac{1}{2}$

The final magnification is
$m=m_1{m}_2=1$

Final image is at $30\text{cm}$ to the right of the second lens and $150\text{cm}$ to the right of the first lens and $m=1$ means image will be form of equal height of object.

Hence,
image height = object height$=3\text{mm}$.

Therefore, option $\left(b\right)$ is correct.