Question

an object of length 25 m is placed at 25 cm and from a concave mirror of radius of curvature 50 cm . find nature position of maqnification of image and draw ray diagram​

Answer 1

Given,

Height of object =5cm

Position of object, u=−25cm

Focal length of the lens, f=10cm

Position of image, v=?

We know that,

v1−u1=f1

v1+251=101

v1=101−251

So,

v1=50(5−2)

That is,

v1=503

So,

v=350=16.66cm

Thus, distance of image is 16.66cm on the opposite side of lens.

Now, magnification =uv

That is,

m=−2516.66=−0.66

Also,

m=h

Answer 2

Answer:

An object of 5cm tall is placed at a distance of 10cm from a concave mirror of radius of curvature 30cm

To find: the nature, position and size of the image and draw the ray diagram to represent the above case.

Solution:

According to the given criteria,

Let height of the image be h

i

and

Height of the object, h

o

=5cm

Object distance, u=−10cm

Radius of curvature, R=−30cm

We know, focal length, f=

2

R

=

2

30

=−15cm

Now applying mirror formula, we get

f

1

=

v

1

+

u

1

⟹

v

1

=

f

1

−

u

1

⟹

v

1

=

−15

1

−

−10

1

⟹

v

1

=

150

−10+15

=

150

5

⟹v=30cm

Position of the image is behind the mirror at the distance of 30 cm

Magnification, m=−

u

v

=

h

o

h

i

⟹−

−10

30

=

5

h

i

⟹h

i

=3×5=15cm

Nature of the image formed: virtual, erect (as magnification is positive) and enlarged.