Question

Focal length of concave lens shown in figure is $60\text{cm}$. For an object placed at ${}^{\prime }{\text{c}}^{\prime }$ on principal axis, find image position and its size.

• A. $v=-20\text{cm},h_i=2\text{mm}$
• B. $v=-40\text{cm},h_i=3\text{mm}$
• C. $v=20\text{cm},h_i=3\text{mm}$
• D. $v=40\text{cm},h_i=2\text{mm}$

Answer 1

The correct option is A $v=-20\text{cm},h_i=2\text{mm}$

For the given situation,
$u=-30\text{cm}.f=-60\text{cm}$

Using the lens formula, we have
$\frac{1}{v}-\frac{1}{-30}=\frac{1}{-60}$

Solving this equation, we get
$v=-20\text{cm}$

Further ,

Lateral magnification , $m=\frac{v}{u}=\frac{-20}{-30}=+\frac{2}{3}$

Since, the magnification is positive, the image formed is erect.

From the figure, we can deduce that the height $bc$ is $h_0=2\text{mm}$ above the principal axis.

Therefore, its image $b^{\prime }{c}^{\prime }$ will have $h_i=2\times \left(\frac{2}{3}\right)$ or $\frac{4}{3}\text{mm}$, above the principal axis.

Similarly, height of $ac$ is $h_0=1\text{mm}$ below the principal axis.

Therefore, its image $a^{\prime }{c}^{\prime }$ will have $h_i=1\times \left(\frac{2}{3}\right)$ or $\frac{2}{3}\text{mm}$ below the principal axis.

The final image is as shown in the figure below.


Therefore, the height of the image is $a^{\prime }{b}^{\prime }$ :
$h_i=2\text{mm}$

Hence, option (a) is the correct answer.