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

Given refractive index of the material of the prism \(\mu =1.5\) 
Critical angle for the material
sin C= \(\frac { 1 }{ \mu } =\frac { 1 }{ 1.5 } =2/3\)
\(\Rightarrow C=sin^{ -1 }\left( \frac { 2 }{ 3 } \right) =42^{ 0 }\)
From the ray diagram, it is clear that angle of incidence i = 30°< C. Therefore the ray incident at the face AC will not suffer total internal reflection and merges out through this face.

(i) From given figure, A = r₁ + r2
As ray QR is parallel to the base BC, then
Therefore, 2r₁(or 2r₂) = A(ii) D = x + y,  r₁ = r₂ = A/2
 D = (i - r₁) + (e - r₂)
 D = (i + e) - (r₁ + r₂)
 D = 2i - A

(i)
 
(ii) The position of image will remain same /unchanged, but the intensity of the image will decrease.

Power of convex lens = 1/0.25 = 4D
Power of concave lens = 1/0.20 = -5D
Power of the combination, P = P₁+P₂ = -1D
Nature : Diverging

While tracing the path of the ray, we should remember that prism bends the incident ray towards its base. 
Refractive index of glass,
\({ \mu }_{ g }=\sqrt { 3 } \)
Since,  i = 0
At the interface AC, we have (according to Snell's law)
\( \frac { sin \ i }{ sin \ r } =\frac { { \mu }_{ g } }{ { \mu }_{ a } }\)
\( But \ sin \ i=sin{ 0 }^{ 0 }=0\)
\(Thus,\ sinr=\frac { { \mu }_{ a }sin \ i }{ { \mu }_{ g } } =0\)
\(Hence,\ r=0\)
This ray pass unrefracted at AC interface and reaches AB interface. Here, we can see angle of incidence becomes 30⁰.
Thus, applying Snell's law,
\(\frac { sin{ 30 }^{ 0 } }{ sin \ e } =\frac { { \mu }_{ a } }{ { \mu }_{ g } } \frac { 1 }{ \sqrt { 3 } } \)     
\(sin \ e=\sqrt { 3 } \times sin{ 30 }^{ 0 }=\frac { \sqrt { 3 } }{ 2 } \)
Thus,  e = 60⁰
Hence, angle of emergence is 60⁰.   

Radius of curvature of objectrive mirror, 
R₁ = 220 mm
\({ f }_{ 1 }=\frac { { R }_{ 1 } }{ 2 } =\frac { 220 }{ 2 } =110 \ mm\) 
Radius of curvature of secondary mirrors, R₂ = 140 mm
\({ f }_{ 2 }=\frac { { R }_{ 2 } }{ 2 } =\frac { 140 }{ 2 } =70 \ mm\) 
Distance between two mirrors, d = 20 mm from objective mirror.
Now, for secondary mirror, u = f₁ - d = 110 - 20
= 90 mm
From mirror formula,
\(\frac { 1 }{ v } +\frac { 1 }{ u } =\frac { 1 }{ { f }_{ 2 } } \Longrightarrow \frac { 1 }{ v } =\frac { 1 }{ { f }_{ 2 } } -\frac { 1 }{ u } \) 
\( =\frac { 1 }{ 70 } -\frac { 1 }{ 90 } \Longrightarrow v=\frac { 630 }{ 2 } =315\ mm\)
i.i., final image will be at 31.5 cm to the right of secondary mirror.

Let \(\theta \) be the angle between the mirrors \({ M }_{ 1 }\) and \({ M }_{ 2 }\)
Since, rays LM and NS are parallel to each other

\(\angle LMN=\angle SNQ=2i\)
\( \Rightarrow \quad 2i_{ 1 }+2i=180^{ \circ }\)
\(\Rightarrow i_{ 1 }+i=90^{ \circ }...............(i)\)
\(Also \ in \ \triangle MBN,\)
\( \theta +\left( 90-i \right) +\left( 90-i \right) =180^{ \circ }\)
\(\Rightarrow \theta =i_{ 1 }+i..............(ii)\)
On comparing Eqs. (i) and (ii), we get
\(​​\theta =90^{ \circ }\)

Here, real depth of the dot under liquid of density \({ d }_{ 1 }\) is h/3. if \({ x }_{ 1 }\) is its apparent depth, when seen from air, then from \({ \mu }_{ 1 }=\frac { h/3 }{ { x }_{ 1 } } \ or \ { x }_{ 1 }=\frac { h }{ 3{ \mu }_{ 1 } } \)
similarly, apparent depths of the dot when seen from air through two other liquids are
\({ x }_{ 2 }=\frac { h }{ 3{ \mu }_{ 2 } } \ and \ { x }_{ 3 }=\frac { h }{ 3{ \mu }_{ 3 } } \)
apparent depths of the dot when seen from air through two other liquids are
\(x={ x }_{ 1 }+{ x }_{ 2 }+{ x }_{ 3 }=\frac { h }{ 3{ \mu }_{ 1 } } +\frac { h }{ 3{ \mu }_{ 2 } } +\frac { h }{ 3{ \mu }_{ 3 } } =\frac { h }{ 3 } \left[ \frac { 1 }{ { \mu }_{ 1 } } +\frac { 1 }{ { \mu }_{ 2 } } +\frac { 1 }{ { \mu }_{ 3 } } \right] \)

On decreasing \(\lambda \), resolving power of microscope increases and on decreasing diameter of objective lens, resolving power of microscope decreases.

Violet colour suffers maximum deviation and red colour suffers the least deviation.

\(\mu \) is minimum for red colour and \(\mu \) is maximum for violet colour.

P = P₁ + P₂  = +3 - 1 = 2D
F = 100/P = 100/2 = 50 cm

\(\frac { 1 }{ { f }_{ v } } =({ \mu }_{ v }-1)\left( \frac { 1 }{ { R }_{ 1 } } -\frac { 1 }{ { R }_{ 2 } } \right) \)
\( \frac { 1 }{ { f }_{ r } } =({ \mu }_{ r }-1)\left( \frac { 1 }{ { R }_{ 1 } } -\frac { 1 }{ { R }_{ 2 } } \right) \)
\(As \ { \mu }_{ v }>{ \mu }_{ r } \ \therefore { f }_{ v }<{ f }_{ r }\)
i.e., focal length decreases.

f = 100/P = 100/-2.5 = -40cm

When refractive index of the liquid is equal to refractive index of the material of the lens = 1.45

\(\mu =\frac { 1 }{ \sin { C } } \)

No, Total internal reflection cannot occur.

Focal length of a plane mirror is infinity.
Power \(P=\frac { 1 }{ f } =\frac { 1 }{ \infty } =Zero\)

This is done to increase the light gathering capacity and hence brightness of image.

20; 13.5 cm

Refraction at convex spherical surface. When object is in rarer medium and image formed is real. 

In \(\triangle AC,i=\alpha +\gamma \) and 
In \(\triangle AIC,\gamma =r+\beta \quad r=\gamma -\beta \)
\(\therefore \) By snell's law n₂ =\(\frac { sini }{ sinr } =\frac { i }{ r } =\frac { \alpha +\gamma }{ \gamma -\beta } \)
or \(\frac { n_{ 2 } }{ n_{ 1 } } =\frac { \alpha +\beta }{ \gamma -\beta } n_{ 2 }\gamma -n_{ 2 }\beta =n_{ 1 }\gamma \)
As \(\alpha ,\beta ,\gamma \) are small and p and N lie close to each other
So, \(\alpha =tan\alpha =\frac { AN }{ NO } =\frac { AN }{ PO } \)
\(\beta =tan\beta =\frac { AN }{ NI } =\frac { AN }{ PI } \)
\(\gamma =tan\gamma =\frac { AN }{ NC } =\frac { AN }{ PC } \)
On using them inEq(1) we get
\(\left( n_{ 2 }-n_{ 1 } \right) \frac { AN }{ PC } =n_{ 1 }\frac { AN }{ PO } +n_{ 2 }\frac { AN }{ PI } \)
\(\frac { n_{ 2 }-n_{ 1 } }{ PC } =\frac { n_{ 1 } }{ PO } +\frac { n_{ 2 } }{ PI } \)
where, PC = +R. radius of curvature
PO = -u, object distance
PI = +v, image distance
So \(\frac { n_{ 2 }-n_{ 1 } }{ R } =\frac { n_{ 1 } }{ -u } +\frac { n_{ 2 } }{ v } \)
\(\frac { n_{ 2 }-n_{ 1 } }{ R } =\frac { n_{ 1 } }{ v } +\frac { n_{ 1 } }{ u } \)
This gives formula for refraction at spherical surface, when object is in rarer medium.

To see objects at infinity, the eye uses its least converging power = (40 + 20) dioptres = 60 dioptres. This gives a rough idea of the distance between the retina and cornea eye lens
So focal length
\( f=\frac { 1 }{ p } \)
\(or \ f=\frac { 1 }{ 60 } m \ = \ \frac { 5 }{ 3 } cm\)
To focus an object at the near point (u = -25 cm) and the retina (v = 5/3 cm) the focal length should be
\(-\frac { 1 }{ u } +\frac { 1 }{ v } =\frac { 1 }{ f }\)
\( \frac { 1 }{ 25 } +\frac { 3 }{ 5 } =\frac { 1 }{ f }\)
\(\frac { 1+15 }{ 25 } =\frac { 1 }{ f }\)
\(orf=\frac { 25 }{ 16 } cm\)

\(First\ case,\)
\(angle \ of \ incidence \ i={ 60 }^{ \circ }\)
\(angle \ of \ refraction \ r={ 35 }^{ \circ }\)
\({ \alpha }_{ { \mu }_{ g } }= \ \frac { 1 }{ { g }_{ { \mu }_{ \alpha } } } =\frac { sin\ i }{ sin \ r } \)
\(Second \ case,\)
\( { \alpha }_{ { \mu }_{ \omega } }=\frac { sin\ { 60 }^{ \circ } }{ sin\ { 47 }^{ \circ } } =1.18\)
\( { \omega }_{ { \mu }_{ g } }= { \alpha }_{ { \mu }_{ g } }\times { \omega }_{ { \mu }_{ \alpha } }\)
\(=\frac { { \alpha }_{ { \mu }_{ g } } }{ { \alpha }_{ { \mu }_{ \omega } } } =\frac { 1.51 }{ 1.18 } =1.28\)
\(Third \ case,\)
\( angle \ of \ incidence \ i={ 45 }^{ \circ }\)
\( angle \ of \ refraction \ r=?\)
\( { \omega }_{ { \mu }_{ g } }=\frac { sin\ i }{ sin\ r } \)
\(\frac { sin \ i }{ sin\ r } =1.28\)
\(sin \ r=\frac { sin \ { 45 }^{ \circ } }{ 1.28 } =0.5525\)
\( sin \ r= \ sin{ 33 }^{ \circ }54'\)
\(r= \ { 33 }^{ \circ }54'\)

For refraction at the first surface
\(\frac { { n }_{ 2 } }{ { v }_{ 1 } } -\frac { { n }_{ 1 } }{ u } =\frac { { n }_{ 2 }-{ n }_{ 1 } }{ { R }_{ 1 } } \)
For the second surface, I₁ acts as a virtual object (located in the denser medium) whose final real image is formed in the rarer medium at I.
so for refraction at this surface, we have
\(\frac { { n }_{ 2 } }{ v } -\frac { { n }_{ 1 } }{ { v }_{ 1 } } =\frac { { n }_{ 2 }-{ n }_{ 1 } }{ { R }_{ 2 } } \)
From the above two equation,  \(\frac { 1 }{ v } -\frac { 1 }{ u } =\left( \frac { { n }_{ 2 } }{ { n }_{ 1 } } -1 \right) \left( \frac { 1 }{ { R }_{ 1 } } -\frac { 1 }{ { R }_{ 2 } } \right) \)
The point, where image of an object, located at infinity is formed, is called the focus F, of the lens and the distance f gives its focal length.
So for \(u=\infty ,v=+f\)
\(\Rightarrow\)  \(\frac { 1 }{ f } =\left( \frac { { n }_{ 2 } }{ { n }_{ 1 } } -1 \right) \left( \frac { 1 }{ { R }_{ 1 } } -\frac { 1 }{ { R }_{ 2 } } \right) \)
(b)

\(\triangle\) ABP is similar to \(\triangle\)A'B'P
So    \(\frac { A'B' }{ AB } =\frac { B'P }{ BP } \)
Nor A'B' = I, AB = O, B'P = + v and BP = - u
So magnification \(m=\frac { I }{ O } =-\frac { v }{ u } \)

The magnifying power of a telescope is equal to the ratio of the visual angle substended at the eye by final image formed at least distance of distinct vision to the visual angle subtended at naked eye by the object at infinity.