The magnification of the image of an object placed in front of a convex mirror is \(\frac{1}{3}\). If the radius of curvature of the mirror is 24cm, what is the distance between its object and its image?
8cm
16cm
24cm
32cm
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F=r/2=24/2=12cm
M=1/3 , F=12cm
To find magnification of the object , we use M=f/u-f
1/3=12/u-12
36=u-12
U=36+12
U=48cm
To find magnification of the image we use M=V-f/f
1/3=V-12/12
12=3(v-12)
12=3v-36
12+36=3v
48=3v
V=16cm
Now, the distance between the object and its image will be U-V
U=48cm V=16cm
48cm-16cm
=32cm
f=½ r , f=½×24 , f= 12
M=v/u
1/3=v/u (cross multiply)
u= 3v
We can now use this formula
1/f=1/u+1/v
1/12=1/3v+1/v (lcm of 3v and v is 3v)
1/12 = 1+3/3v
1/12= 4/3v (cross multiply)
12×4=3v, = 48=3v ( divide both sides by 3)
v=16
Soo...
u=3v, =3(16)
u=48
And the question said distance between objects and it's image
i.e u-v
48-16
Ans= 32
(Option D)
Hope it helped 😁👊
The answer is B, don't 4get that focal length of a convex mirror is always negative
F=r/2=24/2=12cm
M=1/3
M=F/U-F
1/3=12/U-12
U-12=36
U=36+12
U=48Cm
M=V/U
1/3=V/48
3V=48
divide both side by 3
V=16Cm
distance between object and image =2*16
=32Cm
