A convex mirror is used in a car having a focal length of 300 cm. If a cycle is located at 100 cm from this mirror find the position , nature and magnification of the image formed in the mirror
Hi,
Position of the image:
\[
\frac{1}{f} = \frac{1}{v} + \frac{1}{u}
\]
Given: \( f = 300 \text{ cm} \), \( u = -100 \text{ cm} \)
\[
\frac{1}{300} = \frac{1}{v} + \frac{1}{-100}
\]
\[
\frac{1}{v} = \frac{1}{300} + \frac{1}{100} = \frac{4}{300}
\]
\[
v = 75 \text{ cm}
\]
Nature of the image: Virtual, erect, and diminished (as always for convex mirrors).
Magnification:
\[
m = \frac{v}{u}
\]
Given: \( v = 75 \text{ cm} \), \( u = -100 \text{ cm} \)
\[
m = \frac{75}{-100} = -0.75
\]
As per given information,
f = 300 cm
u = -100 cm (negative sign because the object is in front of the mirror)
Using the Mirror Formula,
1/f = 1/v + 1/u
Substituting the values:
1/300 = 1/v + 1/(-100)
1/v = 1/300 + 1/100
1/v = (1 + 3) / 300
1/v = 4/300
v = 300/4
v = 75 cm
The image is formed 75 cm behind the mirror and the image is virtual, erect, and diminished.
Magnification (m) = -v/u
m = -75 / (-100)
m = 0.75
The magnification is 0.75, meaning the image is 0.75 times the size of the actual object.
