First slide

Ray optics

Question

An object is placed at a distance of 40cm from a concave mirror of focal length 15cm. If the object is displaced through a distance of 20cm towards the mirror, the displacement of the image will be

Moderate

A

30 cm away from the mirror
B

36 cm away from the mirror
C

30 cm towards the mirror
D

36 cm towards the mirror

Solution

Using mirror formula,
$\begin{array}{l} \frac{1}{f} = \frac{1}{v_{1}} + \frac{1}{u}; - \frac{1}{15} = \frac{1}{v_{1}} - \frac{1}{40} \Rightarrow \frac{1}{v_{1}} = \frac{1}{- 15} + \frac{1}{40} \\ v_{1} = - 24 cm \end{array}$
When object is displaced by 20 cm towards mirror.
Now, $u_{2} = - 20 cm$
$\begin{array}{l} \frac{1}{f} = \frac{1}{v_{2}} + \frac{1}{u_{2}}; \frac{1}{- 15} = \frac{1}{v_{2}} - \frac{1}{20} \Rightarrow \frac{1}{v_{2}} = \frac{1}{20} - \frac{1}{15} \\ v_{2} = - 60 cm \end{array}$
So, the image will be shift away from mirror by (60-24)cm=36 cm

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