The first refracting ABC forms the image I₁ of the object O. The image I₁ acts as virtual object for the second refracting surface ADC, which forms the real image I as shown in the diagram
For refraction at ABC 
\(\frac { { n }_{ 2 } }{ { v }_{ 1 } } -\frac { { n }_{ 1 } }{ u } =\frac { { n }_{ 2 }-{ n }_{ 1 } }{ { R }_{ 1 } } \)
For refraction at ADC
\(\frac { { n }_{ 1 } }{ v } -\frac { { n }_{ 2 } }{ { v }_{ 1 } } =\frac { { n }_{ 1 }-{ n }_{ 2 } }{ { R }_{ 2 } } \)
Adding equation (i) and equation (ii)
 \(\frac { { n }_{ 1 } }{ v } -\frac { { n }_{ 2 } }{ u } =\left( { n }_{ 2 }-{ n }_{ 1 } \right) \left( \frac { 1 }{ { R }_{ 1 } } -\frac { 1 }{ { R }_{ 2 } } \right) \)
\(\frac { 1 }{ v } -\frac { 1 }{ u } =\left( \frac { { n }_{ 2 } }{ { n }_{ 1 } } -1 \right) \left( \frac { 1 }{ { R }_{ 1 } } -\frac { 1 }{ { R }_{ 2 } } \right) \)
We know, If  \(u=\infty ,v=f\)
\(\frac { 1 }{ v } -\frac { 1 }{ u } =\frac { 1 }{ f } \)
\(\frac { 1 }{ f } =\left( \frac { { n }_{ 2 } }{ { n }_{ 1 } } -1 \right) \frac { 1 }{ { R }_{ 1 } } -\frac { 1 }{ { R }_{ 2 } } \)
\(\frac { 1 }{ f } =\left( { \mu }_{ 21 }-1 \right) \left( \frac { 1 }{ { R }_{ 1 } } -\frac { 1 }{ { R }_{ 2 } } \right) \)
\(\frac { 1 }{ f } =\left( { \mu }_{ 21 }-1 \right) \left( \frac { 1 }{ { R }_{ 1 } } -\frac { 1 }{ { R }_{ 2 } } \right) \)
\(\frac { 1 }{ f } =\left( 1.55-1 \right) \left( \frac { 1 }{ R } -\frac { 1 }{ -R } \right) \)
\(=0.55\times \frac { 2 }{ R } \)
\(R=0.55\times 2\times 20=22 \ cm\) 

\(Here,\quad \frac { \Delta \lambda }{ \lambda } =\frac { 0.032 }{ 100 } ,v=?\)
Since the wavelength of light from a star is shifting towards longer wavelength side, therefore \(\Delta \lambda \)is positive, hence star is moving away from the earth i.e., v is negative.
\(v=\frac { \Delta \lambda }{ \lambda } c=-\frac { 0.032 }{ 100 } \times 3\times { 10 }^{ 8 }\)
\( =-9.6\times { 10 }^{ 4 }m{ s }^{ -1 }\)​​​​​

\(Here, \ v=8.1\times { 10 }^{ 9 }Hz\)
\( \Delta v=2.7\times { 10 }^{ 3 }Hz\)
\(Velocity \ of \ aeroplane, \ v=\frac { 1 }{ 2 } \frac { \Delta v }{ v } \times c\)
\(=\frac { 1 }{ 2 } \times \frac { 2.7\times { 10 }^{ 3 } }{ 8.1\times { 10 }^{ 9 } } \times 3\times { 10 }^{ 8 }=50m/s\)

\(Here,\ D=60 \ cm=0.6 \ m\)
\({ f }_{ 0 }=2.0m,\ { f }_{ e }=1.0cm\)
Distance between stars \(={ 10 }^{ 4 }light \ years\)
\( ={ 10 }^{ 4 }\times (9.46\times { 10 }^{ 15 }m)=9.46\times { 10 }^{ 19 }m\)
Transverse separation of stars \(={ 10 }^{ 10 }m\)
Angle subtended by transverse separation of stars
\(=\frac { { 10 }^{ 10 } }{ 9.46\times { 10 }^{ 19 } } \approx { 10 }^{ -10 }radian\)
Smallest angular separation between two objects that can be resolved by the telescope=limit of resolution of telescope,
\(d\theta =\frac { 1.22\lambda }{ D } =\frac { 1.22\times (6\times { 10 }^{ -7 }) }{ 0.6 } \)
\( =1.22\times { 10 }^{ -6 }rad\)
As the angle subtended by transverse separation of stars is much too small compared to the limit of resolution of the telescope, therefore, the two stars of the binary cannot be resolved by the telescope.

(i) When light travels from water to air,
\(^{ a }{ \mu }_{ w }=\frac { 1 }{ sin\quad C } \)
\(sin\quad C=\frac { 1 }{ ^{ a }{ \mu }_{ w } } =\frac { 1 }{ 1.33 } =0.7518\)
\(\therefore \)   \(C=sin^{ -1 }(0.7518)=48° \ 44'\)
(ii) When light travels from glass into water
\(^{ w }{ \mu }_{ g }=\frac { 1 }{ sin \ C' } , \ i.e., \ \frac { ^{ a }{ \mu }_{ g } }{ ^{ a }{ \mu }_{ w } } =\frac { 1 }{ sin \ C' } \)
or  \(\frac { 1.5 }{ 1.33 } =\frac { 1 }{ sin\quad C' } \)
or \(sin\quad C'=\frac { 1.33 }{ 1.5 } =0.8866\)
\(\therefore \) ​​​​​​​  \(C'=sin^{ -1 }\quad (0.8866)\)
\(=62° \ 27'\)​​​​​​​

\(Here,\quad { \lambda }_{ r }=660nm;\lambda '=?\)
\(For \ diffraction \ minima,\)
\(a \ sin\theta =n\lambda , \ sin\theta =\frac { n\lambda }{ a } \)
\( For \ first \ minima \ of \ red \ light, \ sin\theta =\frac { 1{ \lambda }_{ r } }{ a } \)
\(For \ diffraction \ maxima,\ asin\theta =(2n+1)\frac { \lambda }{ 2 }\)
\(for \ first \ maxima \ of \ \lambda ',\)
\(a \ sin\theta '=\frac { 3\lambda ' }{ 2 } ; \ sin\theta '=\frac { 3\lambda ' }{ 2a } ;\)
\(As \ the \ two \ coincide, \ therefore, \ sin\theta '=sin\theta \)
\(\\ \frac { 3\lambda ' }{ 2a } =\frac { { \lambda }_{ r } }{ a } or\lambda '=\frac { 2 }{ 3 } { \lambda }_{ r }\)
\(or \ \lambda '=\frac { 2 }{ 3 } (660)=440nm\)

Here,  \(i_{ 1 }=45°, \ t=10 \ cm=0.1 \ m\)
\(\mu =1.5,\) lateral shift=?
As \(\mu =\frac { sin \ i_{ 1 } }{ sin \ r_{ 1 } } \)
\(\therefore \) \(sin \ r_{ 1 }=\frac { sin \ i_{ 1 } }{ \mu } =\frac { sin \ 45° }{ 1.5 } =\frac { 0.707 }{ 1.5 } =0.4713\)
\(r_{ 1 }=sin^{ -1 }(0.4713)=28.14°\)
lateral shift\(=\frac { t \ sin(i_{ 1 }-r_{ 1 }) }{ cos \ r_{ 1 } } \)
\(=\frac { 0.1 \ sin \ (45°-28.14°) }{ cos \ 28.14° } \)
\(=\frac { 0.1 \ sin \ 16.86° }{ cos\quad 28.14° } =\frac { 0.1\times 0.2900 }{ 0.8818 } =0.033 \ m\)

Here, 
\(h_{ 1 }=5 \ cm, \ u=-10 \ cm\)
\(f=40 \ cm\)
From 
\(\frac { 1 }{ \upsilon } =\frac { 1 }{ f } -\frac { 1 }{ u } =\frac { 1 }{ 40 } -\frac { 1 }{ -10 } =\frac { 5 }{ 40 } =\frac { 1 }{ 8 }\)
\(\upsilon =8\ cm\)
Image is virtual, erect and is formed 8 cm behind the mirror.
Magnification,
\(m=\frac { h_{ 2 } }{ h_{ 1 } } =\frac { -\upsilon }{ u } =\frac { -8 }{ -10 } =\frac { 4 }{ 5 } \)
\(h_{ 2 }=\frac { 4 }{ 5 } h_{ 1 }=\frac { 4 }{ 5 } \times 5=4\ cm\)
As needle is moved farther away from the mirror, image shifts towards the focus and its size goes on decreasing.

Here, \(\lambda =5000\quad \overset { \circ }{ A } =5\times 10^{ -7 }m\)
\(v=\frac { c }{ \lambda } =\frac { 3\times 10^{ 8 } }{ 5\times 10^{ -7 } } =6\times 10^{ 14 }\) hertz
On reflection, there is no change in wavelength or frequency. Therefore,
\(\lambda '=\lambda =5000\quad \overset { \circ }{ A } \quad ;\quad v'=v=6\times 10^{ 14 }Hz.\)
For reflected ray to be normal to incident ray,
\(i+r=90°\) or \(i+i=90°\)   \((\because r=i)\)
\(\therefore \) \(i=90/2=45°\)​​​

A = 60⁰ ,  \(\delta \)_m = 30⁰
i = e = ¾ A = 450
as A + \(\delta \) = i + e
60 + \(\delta \) = 45 +45
or \(\delta \) = 30⁰
Refractive index,
\(\mu \) = sin a + \(\delta \)m /2/sin A/2 = sin 60⁰+30⁰/2/sin 60⁰/2
= sin 45⁰/sin30⁰ = 1\(\surd 2\) 1/2 = \(\surd 2\) = 1.414

(i) As the lens forms a real iamge, it must be a convex lens.
(ii) From the graph, when u = 20 cm , we have v = 20 cm.
For the convex lens forming a real iamge, u is negative and v and f are positive.
U = -20 cm v = +20cm
Using this lens formula,
1/f = 1/v – 1/u = 1/20 – 1/-20 = 1/10 or f = + 10 cm

Clearly , the fish can see the outside view of the cone with semi vertical angle
But \(\mu \) = 1.sin i_c
or 1/3 = 1/ sin i_c
or sin i_c = 3/4 = 0.75 
\(\theta\)/2  =i_c = sin^-1 (0.75 ) = 48.6⁰