\(\cfrac { 1 }{ F } =\cfrac { 1 }{ { f }_{ 1 } } +\cfrac { 1 }{ { f }_{ 2 } } \) 
\(\cfrac { 1 }{ 50 } =\cfrac { 1 }{ 20 } +\cfrac { 1 }{ { { f }_{ 2 } } } \quad f=-\cfrac { 100 }{ 3 } cm\) 
\({ P }_{ 2 }=\cfrac { 100 }{ -\frac { 100 }{ 3 } } \) 
D = -3D.

Formula:
Refractive index = \(A+\cfrac { B }{ { \lambda }^{ 2 } } \) 
where λ is the wavelength.
Power of a lens \(P=\cfrac { 1 }{ f } =\left( { n }_{ g }-1 \right) \left( \cfrac { 1 }{ { { R }_{ 1 } } } -\cfrac { 1 }{ { R }_{ 2 } } \right) \) 

Clearly, the power of a lens (n_g - 1). This implies that the power of a lens decreases with the increase in wavelength \(\left( P\infty \cfrac { 1 }{ { \lambda }^{ 2 } } nearly \right) \) The plot is shown in the figure alongside.

For face AB, ∠i = 0°,∠r = ·0° the ray will pass through AB undeflected 
Now, at face AC

Formula.
Here,\({ \quad i }_{ c }={ sin }^{ -1 }\left( \cfrac { 1 }{ \mu } \right) \) 
= \({ sin }^{ -1 }\left( \cfrac { 2 }{ 3 } \right) ={ sin }^{ -1 }\left( 0.66 \right) \) 
∠i on face AC is 30^o which is less than ∠i_e.
Hence, the ray get refracted.
And, applying Snell's law at face AC
\({ sin30 }^{ o }\times \cfrac { 3 }{ 2 } =sinr\times 1\) 
\(\Rightarrow sinr=\cfrac { 1 }{ 2 } \times \cfrac { 3 }{ 2 } \) 
\(\Rightarrow r={ sin }^{ -1 }\left( \cfrac { 3 }{ 4 } \right) ={ sin }^{ -1 }(0.75)\)
And, dearly r > i, as ray passes from denser to rarer medium.

For the convex lens 
Formula :\(\cfrac { 1 }{ { f }_{ 1 } } =\cfrac { 1 }{ { v }_{ 1 } } -\cfrac { 1 }{ { u }_{ 1 } } \) 
\(\cfrac { 1 }{ +30 } =\cfrac { 1 }{ { v }_{ 1 } } -\cfrac { 1 }{ { u }_{ 1 } } \) 
\(\cfrac { 1 }{ { v }_{ 1 } } =\cfrac { 1 }{ 30 } -\cfrac { 1 }{ 40 } =\cfrac { 1 }{ 120 } \) 
v_I = 120 cm a real image is formed.
On introducing a concave lens
f₂ = 50cm and u₂ = 120 - 20 = +100 cm from the concave lens
\(\cfrac { 1 }{ { f }_{ 2 } } =\cfrac { 1 }{ { v }_{ 2 } } -\cfrac { 1 }{ { u }_{ 2 } } \) 
\(\cfrac { 1 }{ -50 } =\cfrac { 1 }{ { v }_{ 2 } } -\cfrac { 1 }{ +100 } \) 
\(\therefore \cfrac { 1 }{ { v }_{ 2 } } =\cfrac { 1 }{ 580 } +\cfrac { 1 }{ 100 } =\cfrac { 1 }{ 100 } \)
v₂ = -100cm

R = -20 cm and M = -2
Focal length \(f=\cfrac { R }{ 2 } =-10cm\) 
Magnification \(M=\cfrac { -v }{ u } =-2\) 
\(\therefore v=2u\) 
Using mirror formula
\(\cfrac { 1 }{ v } +\cfrac { 1 }{ u } =\cfrac { 1 }{ f } \Rightarrow \cfrac { 1 }{ 2u } +\cfrac { 1 }{ u } =-\cfrac { 1 }{ 10 } \) 
\(\cfrac { 3 }{ 2u } =-\cfrac { 1 }{ 10 } \Rightarrow u=-15\) 
v = 2 (-15)  = -30 cm
(iii) \(\cfrac { 1 }{ v } +\cfrac { 1 }{ u } =\cfrac { 1 }{ f } \) 
Using sign convention for convex mirror we get
f > 0, U < 0
ஃFrom the formula:
\(\cfrac { 1 }{ v } =\cfrac { 1 }{ f } -\cfrac { 1 }{ u } \) 
As f is positive and u is negative, v is always positive, hence the image is always virtual.