The distance between a concave mirror and an object placed in front of it is 1.0m. If the raduis of curvature of the mirror is 4.0m, the image formed will be
2.00m behind the mirror
2.00m in front of the mirror
1.25m in front of the mirror
1.25m behind the mirror
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the focal lenght of a concave mirror is always positive, i dont know when this people Discovered their own.
FCCP. = focal lent of concave is positive.
FCVN= focal length of convex is negative
We are given:
- Object distance \( u = -1.0 \, \text{m} \) (negative because it's in front of the mirror).
- Radius of curvature \( R = 4.0 \, \text{m} \).
The **focal length** \( f \) of a concave mirror is:
\[
f = \frac{R}{2} = \frac{4.0}{2} = 2.0 \, \text{m}
\]
Using the mirror formula:
\[
\frac{1}{f} = \frac{1}{u} + \frac{1}{v}
\]
Substitute the values:
\[
\frac{1}{2.0} = \frac{1}{-1.0} + \frac{1}{v}
\]
Simplify the equation:
\[
\frac{1}{2.0} = -1.0 + \frac{1}{v}
\]
\[
\frac{1}{v} = 1.0 + 0.5 = 1.5
\]
\[
v = \frac{1}{1.5} \approx 0.67 \, \text{m}
\]
This gives us the image distance \( v = 0.67 \, \text{m} \). Since \( v \) is **positive**, the image is formed in **front of the mirror**.
So the correct answer is **C. 1.25 m in front of the mirror**.
1/f=1/v + 1/u
1/-2 =1/v-1/1
1/v =1/-2 +1/1 =1+2/2 =3/2
1/v =3/2 . then u cross mutiply it
to give 2*1=3v
then u divide both side by 3 to give
v=2/3 =0.67m approximately 1m
