Numerical solutions: reflection at curved mirror

At what position an object be placed in front of a concave mirror of radius of curvature 0.4 m so that an erect image of magnification 3 be produced?
Solution:
Given,
Object distance (u) =?
Radius of curvature (R) = 0.4 m
Magnification (m) = 3
We know,
Focal length (f) = [latex]\frac{R}{2} = \frac{0.4}{2} = 0.2[/latex] m

Now, magnification (m) = [latex]\frac{v}{u}[/latex]

Or, 3 = [latex]\frac{v}{u}[/latex]

Or, v = 3u
Since, image is erect, v = -3u
Now, using mirror formula,
[latex]\frac{1}{f} = \frac{1}{u} + \frac{1}{v}[/latex]

Or, [latex]\frac{1}{0.2} = \frac{1}{u} – \frac{1}{3u}[/latex]

Or, [latex]\frac{1}{0.2} = \frac{3 – 1}{3u}[/latex]

Or, [latex]\frac{1}{0.2} = \frac{2}{3u}[/latex]

Or, 3u = 0.4
Or, u = [latex]\frac{0.4}{3}[/latex]

= 0.133 m = 13.3 cm

A pole 4 m long is laid along the principal axis of a convex mirror of focal length 1 m. The end of the pole nearer the mirror is 2m from it. Find the length of image of the pole.
Solution:
Given,
Length of pole (l) = 4m
Focal length of convex mirror (f) = -1 m
Nearer end:
Object distance (u) = 2m

Using mirror formula,
[latex]\frac{1}{f} = \frac{1}{u} + \frac{1}{v}[/latex]

Or, [latex]-\frac{1}{1} = \frac{1}{2} + \frac{1}{v_1}[/latex]

Or, [latex]\frac{1}{v_1} = -1 – \frac{1}{2} = – \frac{3}{2}[/latex]

Or, v₁ = [latex]-\frac{2}{3}[/latex] = – 0.667 m
Again, far end:
Object distance (u’) = 6 m
Using mirror formula,
[latex]\frac{1}{f} = \frac{1}{u} + \frac{1}{v_2}[/latex]

Or, [latex]-\frac{1}{1} = \frac{1}{6} + \frac{1}{v_2}[/latex]

Or, [latex]\frac{1}{v_2} = -1-\frac{1}{6} = -\frac{7}{6}[/latex]

Or, v₂ = [latex]-\frac{6}{7}[/latex] = – 0.857 m

Now, length of image of pole ([latex]\Delta v[/latex]) = v₂ – v₁
      = – 0.857 – ( – 0.667)
      = – 0.857 + 0.667
      = – 0.19 m.