Question

a concave mirror has a radius of curvature of 0.4 m. Find the position and size of the image of an object 0.2 m high placed o.8 m in front of the mirror

Answer 1

We can use mirror formula to determine the position and the height of the image formed by the concave mirror of the object.

(1/v)-(1/u)=(1/f)

Now here we have to determine the value of v and the value of u and f are given.

Here u is the distance of the object from the mirror which is given 0.8 m and f is the focal length of the mirror which is half of the length of the radius of curvature so f is equals to 0.2 m.

So, we can write,
(1/v)=(1/0.2)+(1/0.8)
(1/v)=(5/0.8)
v=(4/25) m =0.16m

to find the height of the image we can use the magnification formula which is given as.

m=(Hi/Ho) = -(v/u)
here m is equals to magnification.
Hi is equals to height of the image.
Ho is equals to height of the object.
v is the distance of the image from the mirror.
u is the distance of the object from the mirror

So, we can write,

Hi= -(0.16/0.8)x(0.2)
= - 1m
negative represents that the image will be formed in a downward direction to Di principal line.

Answer 2

Explanation:

Radius of curvature, R = 0.4 m

So, focal length, f = 0.4/2 = -0.2 m

Object distance, u = -0.8 m

Height of object, h = 0.2 m

Using mirror's formula :

\dfrac{1}{f}=\dfrac{1}{u}+\dfrac{1}{v}

f

1

=

u

1

+

v

1

\dfrac{1}{v}=\dfrac{1}{f}-\dfrac{1}{u}

v

1

=

f

1

−

u

1

\dfrac{1}{v}=\dfrac{1}{-0.2}-\dfrac{1}{-0.8}

v

1

=

−0.2

1

−

−0.8

1

v = -0.26 m

So, the image distance is 0.26 m

The magnification of mirror is given by :

m=\dfrac{-v}{u}=\dfrac{h'}{h}m=

u

−v

=

h

h

′

h'=\dfrac{-vh}{u}h

′

=

u

−vh

h'=\dfrac{-(-0.26)\times 0.2}{-0.8}h

′

=

−0.8

−(−0.26)×0.2

h' = -0.065 m

h' = -0.07 m

Hence, the size of the image is 0.07 m