Question

am object 2cm in size is placed 30cm in front of a concave mirror of focal length 15cm. At what distance from the mirror should a screen be placed in order to obtain a sharp image formed? Draw a ray diagram to show the formation of the image in this case.​

Answer 1

Explanation:

Measure distance BC. It will be found to be equal to 6 cm. Thus, object is placed at a distance of 6 cm × 5 = 30 cm from the lens. A concave lens of focal length 15 cm forms an image 10 cm from the lens

Answer 2

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.