Question

Q.1 An object is placed at a distance of 20 cm from a concave mirror of focal length 15 cm.

(A) What distance from the mirror a screen should be placed to get a sharp image?

(B) What is the nature of the image?​

Answer 1

Answer:

According to the question:

Object distance, u=−30 cm

Focal length, f=−15 cm

Let the Image distance be v.

By mirror formula:

v

1

+

u

1

=

f

1

[4pt]

⇒

v

1

+

−30 cm

1

=

−15 cm

1

[4pt]

⇒

v

1

=−

−30 cm

1

+

−15 cm

1

[4pt]

⇒

v

1

=

30 cm

1−2

[4pt]

⇒

v

1

=−

30 cm

1

[4pt]

∴v=−30 cm

Thus, screen should be placed 30 cm in front of the mirror (Centre of curvature) to obtain the real image.

Now,

Height of object, h

1

=2 cm

Magnification, m=

h

1

h

2

=−

u

v

Putting values of v and u:

Magnification m=

2 cm

h

2

=−

−30 cm

−30 cm

⇒

2 cm

h

2

=−1

[4pt];

⇒h

2

=−1×2 cm=−2 cm

Thus, the height of the image is 2 cm and the negative sign means the image is inverted.

Thus real, inverted image of size same as that of object is formed.

The diagram shows image formation.

solution