(a) \({ f }_{ o }=\cfrac { 100 }{ 25 } cm=4cm\) 
\({ f }_{ e }=\cfrac { 100 }{ 20 } cm=5cm\) 
Since f₀ therefore the optical instrument is a compound microscope.
(b) \(m=\cfrac { { { v }_{ o } } }{ { u }_{ o } } .\cfrac { D }{ { f }_{ e } } \) 
 

Formula:
Angular fringe width = \(\cfrac { \beta }{ D } =\cfrac { \lambda }{ d } \) 
If apparatus is dipped in water, λ changes to
\({ \lambda }_{ w }=\cfrac { \lambda }{ { n }_{ w } } =\cfrac { \lambda }{ \frac { 4 }{ 3 } } =\cfrac { 3\lambda }{ 4 } \) 
ஃ New angular fringe width
\({ \theta }_{ w }=\cfrac { { \lambda }_{ w } }{ d } \) 
\(\therefore \cfrac { { \lambda }_{ w } }{ \theta } =\cfrac { \left( \frac { 3\lambda }{ 4 } \right) }{ \lambda } =\cfrac { 3 }{ 4 } \) 
\({ \theta }_{ w }=\cfrac { 3 }{ 4 } \theta =\cfrac { 3 }{ 4 } \times { 0.2 }^{ o }=0.15^{ o }\)

Here, m = -20, m_e = 5, ve = -20 cm
For eyepiece, \({ m }_{ e }=\cfrac { { v }_{ e } }{ { u }_{ e } } \) 
\(\Rightarrow 5=\cfrac { -20 }{ 5 } \Rightarrow { u }_{ e }=\cfrac { -20 }{ 5 } =-20cm\) 
Using lens formula,
\(\cfrac { 1 }{ { v }_{ e } } -\cfrac { 1 }{ { u }_{ e } } =\cfrac { 1 }{ { f }_{ e } } =-\cfrac { 1 }{ 20 } +\cfrac { 1 }{ 4 } =\cfrac { 1 }{ { f }_{ e } } \) 
\(\Rightarrow \cfrac { -1+5 }{ 20 } =\cfrac { 1 }{ { f }_{ e } } \Rightarrow { f }_{ e }=5cm\) 
Now, total magnification
m = m_e x m_o
-20 = 5 x m_o ⇒ m_o = -4
Also \(\left| { v }_{ o } \right| +\left| { u }_{ e } \right| =14\) 
\(\left| { v }_{ e } \right| +\left| -4 \right| =14\) 
v_o = 14 - 4 = 10 cm
\({ m }_{ o }=1-\cfrac { { n }_{ o } }{ { f }_{ o } } \Rightarrow -4=1-\cfrac { 10 }{ { f }_{ e } } \)

Formula
We know \({ i }_{ c }=\cfrac { 1 }{ \mu } \)
 \(\mu =\cfrac { 1 }{ { sini }_{ c } } =\cfrac { 1 }{ { sin45 }^{ o } } =\sqrt { 2 } \) 
According to Brewster's law 
\({ i }_{ p }=\mu =\sqrt { 2 } \) 
\(\Rightarrow { i }_{ p }={ tan }^{ -1 }\sqrt { 2 } \) 
= tan^-1(1.414) ≅ 510 40^o
When light is incident at an angle i_p the corresponding angle of refraction 'r' is given by
i_p + r = 90^o
ஃ r = 90^o- (51^o40^') = (38^o 20')

ß₁ = 7.2 mm, ß₂ = 8.1 mm, λ₁ = 600 nm, λ₂ =?
Dividing, we get  \(\beta = \frac{D\lambda}{d}(or)\beta∝\lambda\)
\(\frac{\beta_1}{\beta_2}=\frac{\lambda_1}{\lambda_2}\)
\(\therefore, { \lambda }_{ 2 }=\frac{\beta_2\times \lambda_1}{\beta_1}=\cfrac { 8.1\times10^{-3} \times 600 \times10^{-9}}{ 7.2 \times10^{-3}} \)
\(=\cfrac { 9 \times 600\times10^{-9} }{ 8 } =675\times 10^{-9 }\Rightarrow\lambda_2= 675nm\)