An object 4 cm in size is placed 60 cm in front of a concave mirror of focal length 15 cm. At what distance from the mirror should a screen be placed in order to obtain a sharp image? What will be the nature and the size of the image formed? Draw a ray diagram to show the formation of the image in this case.
Answers
Explanation:
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.
