Since, the object distance is u. Let us consider the two ends of the object be at distance \({ u }_{ 1 }=u-L/2\) and \({ u }_{ 2 }=u+L/2\), respectively, so that \(|{ u }_{ 1 }-{ u }_{ 2 }|=L\). Let the image of the two ends be formed at \({ v }_{ 1 } \ and \ { v }_{ 2 }\), respectively so that the image length would be
 \({ L }^{ ' }=|{ v }_{ 1 }-{ v }_{ 2 }|\)
Applying mirror formula, we have
\(\frac { 1 }{ u } +\frac { 1 }{ v } =\frac { 1 }{ f } \ or\ v=\frac { fu }{ u-f } \)
On shoving, the positions of two images are given by
\({ v }_{ 1 }=\frac { f\left( u-L/2 \right) }{ u-f-L/2 } ,{ v }_{ 2 }=\frac { f\left( u+L/2 \right) }{ u-f+L/2 } \)
For length, substituting these values in Eq. (i), we have
 \({ L }^{ ' }=|{ v }_{ 1 }-{ v }_{ 2 }|=\frac { { f }^{ 2 }L }{ \left( u-f \right) ^{ 2 }-{ L }^{ 2 }/4 } \)
Since, the object is short and kept away from focus, we have \({ L }^{ 2 }/4<<\left( u-f \right) ^{ 2 }\)
Hence, finally, \({ L }^{ ' }=\frac { { f }^{ 2 } }{ \left( u-f \right) ^{ 2 } } L\)
This is the required expression of length of an image.

Given, magnification, m = +3, R = -36cm
Object distance, u=?
Let  u = -x
\(m=\frac { { h }_{ 2 } }{ { h }_{ 1 } } =\frac { +v }{ -u } =3 \ \Rightarrow v=-3u \ \Rightarrow v=3x\)
Applying mirror formula, we have
\(\frac { 1 }{ u } +\frac { 1 }{ v } =\frac { 1 }{ f } =\frac { 2 }{ R } \ \Rightarrow \frac { 1 }{ -x } +\frac { 1 }{ 3x } =\frac { 2 }{ -36 }\)
\( \Rightarrow \ \frac { -3+1 }{ 3x } =\frac { -1 }{ 18 } \ \Rightarrow 3x=36\ \Rightarrow x=12cm \ or \ u=-12cm\)

Radius of curvature of objectrive mirror, 
R₁ = 220 mm
\({ f }_{ 1 }=\frac { { R }_{ 1 } }{ 2 } =\frac { 220 }{ 2 } =110 \ mm\) 
Radius of curvature of secondary mirrors, R₂ = 140 mm
\({ f }_{ 2 }=\frac { { R }_{ 2 } }{ 2 } =\frac { 140 }{ 2 } =70 \ mm\) 
Distance between two mirrors, d = 20 mm from objective mirror.
Now, for secondary mirror, u = f₁ - d = 110 - 20
= 90 mm
From mirror formula,
\(\frac { 1 }{ v } +\frac { 1 }{ u } =\frac { 1 }{ { f }_{ 2 } } \Longrightarrow \frac { 1 }{ v } =\frac { 1 }{ { f }_{ 2 } } -\frac { 1 }{ u } \) 
\( =\frac { 1 }{ 70 } -\frac { 1 }{ 90 } \Longrightarrow v=\frac { 630 }{ 2 } =315\ mm\)
i.i., final image will be at 31.5 cm to the right of secondary mirror.

Consider the ray diagram shown below.
  
The point image I₁ due to L₁ is the focal point. Now. due to converging lens L₂, let the final image formed be I which is a point image, hence the wavefront for this image will be of spherical symmetry.

When we are considering a point of sound wavw. The disturbance due to the source propagates in spherical symmetry, that is, in all directions.
       
The formation of wavefront is in accordance with Huygen's principle. So, Huygen's principle is valid for longitudinal sound waves also.