\(v=\pm 3u, f=\mp 20 cm\)
Case i :
V = -3u,
\(\frac{1}{f} =\frac{1}{u}+\frac{1}{v} =\frac{1}{u}-\frac{1}{3 u} =\frac{2}{3 u} \)
\(\)3u = 2f
\(u=\frac{2 f}{3}=\frac{2 \times(-20)}{3} \)
\(u=\frac{-40}{3} \mathrm{~cm} \)
Case ii :
v = 3u
\(\frac{1}{f} =\frac{1}{u}+\frac{1}{3 u} \)
\(=\frac{4}{3 u} \)
3u = 4f = -4 x 20
\(u=\frac{-4 \times 20}{3} \)
\(u=\frac{-80}{3} \mathrm{~cm} \)

\(\frac{1}{f}=\frac{1}{v}-\frac{1}{u} \)
\(m=\frac{-v}{u}=-4, f=20 \mathrm{~cm} \text { (Given) } \)
V = 4u
\(\frac{1}{f} =\frac{1}{4 u}-\frac{1}{u} \)
\(=\frac{1-4}{4 u}=\frac{-3}{4 u} \)
\(\frac{1}{f} =\frac{-3}{4 u} \)
4u = -3 x f
\(u=\frac{-3}{4} \times 20=-15 \mathrm{~cm}\)

Amplitudes, a₁ = 5, a₂ = 3
Resultant amplitude,
\(A=\sqrt { { a }_{ 1 }^{ 2 }+{ a }_{ 2 }^{ 2 }+2{ a }_{ 1 }{ a }_{ 2 }cos\varphi } \) 
Resultant amplitude is maximum when,
\(\phi =0,cos0=1,{ A }_{ max }=\sqrt { { a }_{ 1 }^{ 2 }+{ a }_{ 2 }^{ 2 }+{ 2a }_{ 1 }{ a }_{ 2 } } \) 
 \(\\ { A }_{ max }=\sqrt { \left( { a }_{ 1 }+{ a }_{ 2 } \right) ^{ 2 } } =\sqrt { \left( 5+3 \right) ^{ 2 } } =\sqrt { \left( 8 \right) ^{ 2 } } \) 
= 8 units
Resultant amplitude is minimum when
\(\phi =\pi,cos\pi=1,{ A }_{ max }=\sqrt { { a }_{ 1 }^{ 2 }+{ a }_{ 2 }^{ 2 }+{ 2a }_{ 1 }{ a }_{ 2 } } \)
\({ A }_{ max }=\sqrt { \left( { a }_{ 1 }-{ a }_{ 2 } \right) ^{ 2 } } =\sqrt { \left( 5-3 \right) ^{ 2 } } =\sqrt { \left( 2 \right) ^{ 2 } } \)
= 2units
\(I\infty { A }^{ 2 }\) 
\(\cfrac { { I }_{ max } }{ { I }_{ min } } =\cfrac { \left( { A }_{ max } \right) ^{ 2 } }{ \left( { { A }_{ min } } \right) ^{ 2 } } \)
Substituting,
 \(\cfrac { { I }_{ max } }{ { I }_{ min } } =\cfrac { \left( 8 \right) ^{ 2 } }{ \left( 2 \right) ^{ 2 } } =\cfrac { 64 }{ 4 } 16\) (or)
\({ I }_{ max }:{ I }_{ min }=16:1\)

Given data:
λ = 900 nm = 900 x 10^-9 m
λ ₂ = 750 nm = 750 x 10^-9 m
D = 2 m d = 2 nm = 2 x 10^-3 m
Let n^th order bright fringe of λ₁,
Coincides with (n + 1)^th order bright fringe of λ ₂
\(y_{n}=\frac{n \lambda_{1} D}{d}, Y_{n+1}=\frac{(n+1) \lambda_{2} D}{d} \)
\(\frac{n \lambda_{1} D}{d}=\frac{(n+1) \lambda_{2} D}{d} \)
\(n \lambda_{1}=(n+1) \lambda_{2} \)
\(\frac{n+1}{n}=\frac{\lambda_{1}}{\lambda_{2}}=\frac{900 \times 10^{-9}}{750 \times 10^{-9}}=\frac{18}{15}=\frac{6}{5} \)
\(1+\frac{1}{n} =\frac{6}{5} \)
\(\frac{1}{n} =\frac{6}{5}-1=\frac{6-5}{5} \)
\(\frac{1}{n} =\frac{1}{5} \)
n = 5
n + 1 = 6
5^th bright fringe of  λ ₁ coincides with 6^th bright fringe of λ ₂ in the least distance of y
\(y =\frac{n \lambda_{1} D}{d}=\frac{5 \times 900 \times 10^{-9} \times 2}{2 \times 10^{-3}} \)
\(=4500 \times 10^{-6}=4.5 \times 10^{-3} \)
y = 4.5 mm

\(P_{a}=\frac{1}{f_{a}}=\left(\frac{\mu_{g}}{\mu_{a}}-1\right)\left(\frac{1}{R_{1}}-\frac{1}{R_{2}}\right) \)  ..(1)
\(P_{w}=\frac{1}{f_{w}}=\left(\frac{\mu_{g}}{\mu_{w}}-1\right)\left(\frac{1}{R_{1}}-\frac{1}{R_{2}}\right) \)  ...(2)
\(p_{a}=5 D, f_{w}=-100 \text { (Diverging lens) } \)
\(\mu_{\mathrm{g}}=1.5, \mu_{\mathrm{a}}=1 \)
\(5=(1.5-1)\left(\frac{1}{R_{1}}-\frac{1}{R_{2}}\right) \)   ...(3)
\(\left(\frac{1}{R_{1}}-\frac{1}{R_{2}}\right)=\frac{5}{0.5}=\frac{50}{5}=10 \)
\(\frac{1}{f_{w}}=\left(\frac{1.5}{n_{w}}-1\right)\left(\frac{1}{R_{1}}-\frac{1}{R_{2}}\right) \)    ...(4)
\(\frac{-1}{100 \times 10^{-2}}=\left(\frac{1.5}{n_{w}}-1\right)\left(\frac{1}{R_{1}}-\frac{1}{R_{2}}\right) \)
\(-1=\left(\frac{1.5}{n_{w}}-1\right)(10) \Rightarrow \frac{-1}{10}=\frac{1.5}{n_{w}}-1 \)
\(\frac{1.5}{n_{w}}=\frac{-1}{10}+1 \)
\(\frac{1.5}{n_{w}}=\frac{9}{10} \)
\(n_{w}=\frac{1.5 \times 10}{9}=\frac{15}{9}=\frac{5}{3} \)
\(n_{w}=\frac{5}{3} \)