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Selina Solutions Class 9 Physics Solutions for Exercise 7C in Chapter 7 - Chapter 7 Reflection Of Light
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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.
Answer:
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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