Position of image
Object distance u = − 15 cm (to the left of mirror )
Image distance v = ?
Focal length f = − 10 cm (concave mirror) Using mirror formula.
\(\frac { 1 }{ v } +\frac { 1 }{ u } =\frac { 1 }{ f } \)
\(\frac { 1 }{ v } +\frac { 1 }{ -15 } =\frac { 1 }{ -10 } \)
\(\frac { 1 }{ v } -\frac { 1 }{ 15 } =\frac { -1 }{ 10 } \)
\(\frac { 1 }{ v } =\frac { -1 }{ 10 } +\frac { 1 }{ 15 } \)
\(\frac { 1 }{ v } =\frac { -1 }{ 30 } \)
∴ Image distance v = -30cm (negative sign indicates that the image is on the left side of the mirror)
∴ Position of image is 30 cm in front of the mirror
2.Nature of image: Since the image is in front of the mirror it is real and inverted.
3.Size of image: To find the size of the image, we have to calculate the magnification
\(m=\frac { -v }{ u } \)
Object distance u = − 15cm
Image distance v = − 30 cm 
\(m=\frac { -(-30) }{ (-15) } \)
m = − 2
We know that,\(m=\frac { { h }_{ 2 } }{ { h }_{ 1 } } \)
Here, height of the object h₁ = 1 cm
-2 =\(\frac { { h }_{ 2 } }{ 1 } \)
h₂ = -2\(\times\) 1
= -2 cm
The height of image is 2 cm (negative sign shows that the image is formed below the principal axis).

The spherical mirror equation is \(\frac { 1 }{ f } =\frac { 1 }{ { u } } +\frac { 1 }{ v } \)
u = distance of object.
v = distance of image.
f = focal length of spherical mirror.

A ray of light going towards the centre of curvature is reflected back along the same path. It is a rule for construction of image formed by spherical mirrors.

Height of object h₁ = 2 cm
Height of real image h₂ = -3 cm
Magnification   m =\(\frac { { h }_{ 2 } }{ { h }_{ 1 } } \)
= \(\frac { -3 }{ 2 } \)
= -1.5
We know that,  m = \(\frac { -v }{ u } \)
here, object distance u = - 16 cm 
Substituting the value, we get
 -1.5 = - \(\frac { v }{ (-16) } \)
-1.5 =  \(\frac { v }{ (-16) } \)
v = 16 \(\times\) -1.5
v = -24 cm
The position of image is 24 cm in front of the mirror (negative sign indicates that the image is on the left side of the mirror).

Deviation in the path of light is due to change of velocity of light in the different medium

 Due to refraction.

\({ }_{1} \mu_{2}=\frac{\sin i}{\sin r}=\frac{\sin 45^{\circ}}{\sin 30^{\circ}}=\frac{1 / \sqrt{2}}{1 / 2}=\sqrt{2}=1.414\)