in a Concave mirror f = 10m, u = 15 , find V and on. (Derive it)
Answers
Answer:
a) For a concave mirror, the focal length (f) is negative.
When the object is placed on the left side of the mirror, the object distance (u) is negative
For image distance v we can write
v
1
+
u
1
=
f
1
v
1
=
f
1
−
u
1
...(1)
The object lies between f and 2f.
2f<u<f
2f
1
>
u
1
>
f
1
2f
−1
<
u
−1
<
f
−1
f
1
−
2f
1
<
f
1
−
u
1
..(2)
from eq. 1
2f
1
<
v
1
<0 ; v is negative
2f>v
−v>−2f
Therefore, the image lies beyond 2f.
b) For a convex mirror, the focal length (f) is positive
When the object is placed on the left side of the mirror, the object distance (u) is negative
For image distance v, we have the mirror formula
v
1
+
u
1
=
f
1
v
1
=
f
1
−
u
1
Using equal to (2)
v
1
<0
v>0
Thus, the image is formed on the back side of the mirror.
Hence, a convex mirror always produces a virtual image, regardless of the object distance.
c) For a convex mirror, the focal length (f) is positive
When the object is placed on the left side of the mirror, the object distance (u) is negative
For image distance v, we have the mirror formula
v
1
+
u
1
=
f
1
v
1
=
f
1
−
u
1
but u<0
therefore
v
1
>
f
1
v<f
Hence, the image formed is diminished and is located between the focus (f) and the pole
d) For a concave mirror, the focal length (f) is negative
When the object is placed on the left side of the mirror, the object distance (u) is negative
It is placed between the focus (f) and the pole
therefore f>u>0
f
1
<
u
1
<0
f
1
−
u
1
For image distance v, we have the mirror formula
v
1
+
u
1
=
f
1
v
1
=
f
1
−
u
1
v
1
<0
v>0
The image is formed on the right side of the mirror. Hence, it is a virtual image
For u<0 and v>0, we can write:
u
1
>
v
1
so v>u
magnification m=
u
v
>1
Hence, the formed image is enlarged.
Explanation: