Speed of light in water \(=\frac{\text { speed of light in vacuum }}{\text { refractive index }}\)
=\(\frac{3\times {10}^{8}}{1.33}\)
= \(2.25\times {10}^{8}m/s\)

u = -12 cm, m = -4
\(m=\frac{-v}{u}\)
\(\therefore v=-\left(-4\right)\times \left(-12\right)\)
= -48 cm. 
Image is formed 48 cm in front of mirror.

Object distance, u = - 50 cm
Since the image distance is lesser than object distance and a virtual image is formed. It is a concave lens,
Image distance, v = - 10 cm
To find:
Focal length of the lens, f=?
\(\frac { 1 }{ f } =\frac { 1 }{ v } -\frac { 1 }{ u } \)
\(\frac { 1 }{ f } =\frac { -1 }{ 10 } -\left( \frac { 1 }{ -50 } \right) \)
\(\frac { 1 }{ f } =\frac { -1 }{ 10 } +\frac { 1 }{ 50 } \)
\(\frac { 1 }{ f } =\frac { -50+10 }{ 500 } \)
\(\frac { 1 }{ f } =\frac { -40 }{ 500 } \)
f=\(\frac { -500 }{ 40 } \)
∴ f =-12.5 cm
It is a diverging. ∵ it is concave lens.

Object distance, u =25 cm
virtual image distance, v =-10 cm
To find: Magnification, m = ?
m = \(\frac{\text { Distance of the image }}{\text { Distance of the object }}=\frac{v}{u}\)
= \(\frac { -10 }{ 25 } \) = -0.4
Magnification,
m = -0.4

Power of convex lens, P₁ = + 6 D
Power of concave lens, P₂ = - 4 D
Power of combination of two lenses
P = ?
P = P₁ + P₂ = 6 - 4 = 2D
P = 2D.