Question

A converging lens of focal length$20cm$ is placed between an object and a concave mirror of focal length $10cm$ as shown in figure. The final image is

• A. Coinciding with object, enlarged, Inverted real
• B. Coinciding with object, same size, erect, real
• C. Coinciding with object, same size, Inverted, Vertical
• D. Coinciding with object, same size, Inverted, real

Answer 1

The correct option is D

Coinciding with object, same size, Inverted, real

Step 1. Given data

Distance of image formed by the lens, $v_1=?$

Distance of image from a concave mirror, $\begin{array}{rcl}{v}_2& =& ?\end{array}$

The focal length of the concave mirror, $f_2=-10cm$ [Negative sign due to sign convention]

Distance of object from the lens, $u_1=-40cm$ [Negative sign due to sign convention]

The focal length of a convex lens, $f_1=20cm$

Step 2. Calculating the distance of the image formed by the lens

Using Lens formula:

$\frac{1}{v_1}-\frac{1}{u_1}=\frac{1}{f_1}$

Putting values in the above equation, we get

$\begin{array}{rcl}\frac{1}{v_1}-\left(-\frac{1}{40}\right)& =& \frac{1}{20}\end{array}$

$\Rightarrow$ $\begin{array}{rcl}\frac{1}{v_1}& =& \frac{1}{20}-\frac{1}{40}\end{array}$

$\Rightarrow$ $\begin{array}{rcl}\frac{1}{v_1}& =& \frac{1}{40}\end{array}$

$\begin{array}{rcl}\therefore v_1& =& 40cm\end{array}$

Step 3. Calculating distance of image formed by concave mirror:

The image formed by the convex lens works as an object for the concave mirror:

Distance of image by the convex lens from the concave mirror is, $\begin{array}{rcl}{v}_1\text{'}& =& 60-40=20\mathrm{cm}\end{array}$

Distance of the object from the concave mirror$\begin{array}{rcl}{u}_2& =& -v_1\text{'}=-20cm\end{array}$ [Sign convention: (Negative sign) object distance measured opposite to the light ray direction]

Distance of image from a concave mirror, $\begin{array}{rcl}{v}_2& =& ?\end{array}$

Using mirror formula,

$\frac{1}{v_2}+\frac{1}{u_2}=\frac{1}{f_2}$

Putting all values, we get:

$\begin{array}{rcl}\frac{1}{v_2}+\left(-\frac{1}{20}\right)& =& \left(-\frac{1}{10}\right)\\ \frac{1}{v_2}& =& \frac{1}{20}-\frac{1}{10}\\ \frac{1}{v_2}& =& \left(-\frac{1}{20}\right)\\ \therefore v_2& =& -20cm\end{array}$

The ray diagram for the given case is:

1. From the figure, it can be seen that the image of the object $\left(AB\right)$ is real and inverted $(A\text{'}B\text{'})$ and it is of the same size as the object.
2. This image $(A\text{'}B\text{'})$ is at a distance of $20cm$ from the concave mirror, which is formed at the center of the curvature of the mirror.
3. Thus, the mirror will form the image $(A\text{'}B\text{'})$ at the same place as $(A\text{'}\text{'}B\text{'}\text{'})$ and will be of the same size.
4. Now, due to refraction from the lens, the final image $(A\text{'}\text{'}B\text{'}\text{'})$ will be formed at $AB$ and of the same size as the object $\left(AB\right)$.

Hence, the correct option is (D).