# Selina Solutions Class 9 Physics Solutions for Exercise 7C in Chapter 7 - Chapter 7 Reflection Of Light

Question 49 Exercise 7C

A concave mirror forms a virtual image of size twice that of the object placed at a distance5cm from it. Find: (a) the focal length of the mirror (b) position of the image.

Given:

Distance of the object (u) = 5 (negative), m = 2, focal length = ?

We know that

\begin{array}{l} \mathrm{m}=-\frac{v}{u} \\ \Rightarrow 2=-\frac{v}{u} \\ \Rightarrow \mathrm{v}=-10 \end{array}

The image is formed 10cm behind the mirror.

To find the focal length;

We know that\begin{array}{r} \frac{1}{u}+\frac{1}{v}=\frac{1}{f} \\ \Rightarrow \frac{1}{f}=\frac{1}{10}-\frac{1}{5} \\ \Rightarrow f=-10 \mathrm{cm} \end{array}

Therefore, the focal length of the mirror is 10cm

Video transcript
"hello students welcome to lido q a video session i am saf your science tutor and question for today is a concave meter forms a virtual image of size twice that of an object placed at a distance of 5 centimeter from it find a the focal length of a mirror and be position of the image first of all let us see what is the distance of the object distance of the object u is given that is phi and distance is always negative magnification m that can be said to be 2 because it is given that virtual image of size twice that of object hence we say magnification m is equal to 2 our question is what is focal length we know magnification m is equal to v minus p upon u let us substitute the value of m that is 2 and value of u that is minus v upon u is 5 so you have now pi into 2 that is v is equal to minus 10 now once you got the v we can say here that the image is formed 10 centimeter behind the mirror now to find the focal length we have a formula so let us continue here one upon u plus 1 upon v is equal to 1 upon f so we know that 1 upon f will be equal to let us substitute u as well as v 1 upon u was u was minus 5 and v was we got v to be equal to 1 upon 10 now as you know we have placed b as a 10 because image image distance is always positive so here we have to consider this signs so 1 upon f here you take you can take lcm 2 plus 1 that is minus 2 plus 1 upon 10 hence f is equal to we will get minus 10 centimeter from the sign you can see that it is minus focus is minus so it is on book the focus focal point is on the side of object that is in front of the mirror therefore we can say that the focal length of the mirror is 10 centimeter that is our solution for the given question if you have any queries you can drop it in comment section and subscribe to lido for more q a thank you for watching"
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