Question

an object 0.04m high placed at a distance of 0.08m from concave mirror of radius of curvature0.4m find position nature size of the image formed ​

Answer 1

Answer:

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

Answer 2

Answer:

Now, M=-1 upon 0.4= (4 upon 15 ) upon 0.8

or -1= 0.4 multiple 4( 0.8 ) upon 15

= (1:6) (0.8) upon 15

= -0.085m