(i) Magnifying power of a telescope \((m)=\frac{f_{o}}{f_{e}}\) so the lens of focal length 4 cm is used as an eyepiece
and the lens of focal length 16 cm is used for an objective lens.
(ii) The separation between the two lenses should be increased to 20 cm, because in normal adjustment, the length of a microscope is given by
\( L=f_{o}+f_{e}=16+4=20 \mathrm{~cm} \)
\(\text { (iii) Given: } f_{o}=16, f_{e}=4 \quad \therefore \quad m=\frac{f_{o}}{f_{e}}=\frac{16}{4}=4 \)

Given, focal length, f = +30 cm
Magnification, \(m=\frac{1}{4}, v=?\)
From mirrors formula,
\(\frac{1}{f}=\frac{1}{u}+\frac{1}{v} \quad\left[\because m=-\frac{v}{u} \Rightarrow u=-\frac{v}{m}\right]\)
\( \Rightarrow \quad \frac{1}{f}=-\frac{m}{v}+\frac{1}{v} \)
\(\Rightarrow m=\frac{f-v}{f} \Rightarrow \frac{1}{4}=\frac{30-v}{30} \)
\(\Rightarrow 30=120-4 v \)
\(\Rightarrow \quad v=\frac{90}{4}=+22.5 \mathrm{~cm} \)
As, v is positive, therefore a virtual and erect image will be formed on other side of the object.

Image formed by the lens will be f at focus.

For mirror, U = - 30, f = - 10
According to lens formula
\(\frac { 1 }{ f } =\frac { 1 }{ v } +\frac { 1 }{ u } \Rightarrow \frac { 1 }{ -10 } =\frac { 1 }{ v } -\frac { 1 }{ 30 } \Rightarrow \frac { 1 }{ v } =\frac { 1 }{ 30 } -\frac { 1 }{ 10 } \)
\(\Rightarrow \frac { 1 }{ v } =\frac { 1 }{ 1-3 } \Rightarrow v=-15cm\),

Let d be the diameter of the disc. The sopt shall be invisible, if the incident rays OA and OB suffer total internal reflection.
Let i be the angle of incidence

Using relationship between refractive index and critical angle, then
\(\sin { i=\frac { 1 }{ \mu } }\)
Using geometry and trigonometry,
Now,
\(\frac { d/2 }{ h } =\tan { i } \Rightarrow \frac { d }{ 2 } =h\tan { i } =h[\sqrt { { \mu }^{ 2 }-1{ ] }^{ -1 } }\)
\( [\because \ From \ the \ figure,\ tani=\frac { 1 }{ \sqrt { { \mu }^{ 2 } } -1 } ]\)
\( d=\frac { 2h }{ \sqrt { { \mu }^{ 2 } } -1 } \)
This is required expression of d.

\(Here,\ \upsilon =9\times { 10 }^{ 6 }m/s,\ \lambda =600nm, \ { \lambda }^{ ' }=?\)
\(\Delta \lambda =-\frac { \upsilon }{ c } \lambda =-\frac { 9\times { 10 }^{ 6 } }{ 3\times { 10 }^{ 8 } } \times 600=-18nm\)
\( \therefore \ { \lambda }^{ ' }=\lambda +\Delta \lambda =600-18=582nm\)

To make the reflected ray horizontal, it must be turned further through \(90°-60°=30°\), therefore the mirror must be turned through \(30/2=30°\)

(i) Maximum magnifying power is obtained when the image is the near point (25)
\(\therefore \ v=-25 \ cmf=10 \ cm\)
\(\\ using\frac { 1 }{ v } -\frac { 1 }{ u } =\frac { 1 }{ f } \)
\( -\frac { 1 }{ 25 } -\frac { 1 }{ u } =\frac { 1 }{ f }\)
\(u= -\frac { 50 }{ 7 } =-7.14 \ cm\)
(ii) Magnitude of magnification
\(m=\frac { 25 }{ u } =\frac { 25 }{ 7.14 } =35\)
(iii) magnifying power \(=\left( 1+\frac { D }{ f } \right) =\left( 1+\frac { 25 }{ 10 } \right) =3.5\)
Yes, the magnifying power (when the image is produced at 25 cm) is equal to the magnitude of magnification.

As the speed of light is maximum in vacuum, therefore absolute value of refractive index cannot be less than unity as it is given by the relation \(n=\frac{c}{v}\)

position of object. The colour red is used for danger signals because red light is scattered the least by air molecules. The effect of scattering is inversely, related to the fourth power of the wavelength, i.e. \(I \propto \frac{1}{\sqrt{\lambda^4}}\) of colour, so red light is able to travel the longest distance through fog, rain and the alike.

\(\frac{I_{\min }}{I_{\max }}=\frac{9}{25}=\frac{(a-b)^{2}}{(a+b)^{2}}\\ \Rightarrow \ \frac{a-b}{a+b}=\frac{3}{5}\\ \Rightarrow a = 4b\\ As, \frac{W_{1}}{W_{2}}=\frac{a^{2}}{b^{2}}=\frac{(4 b)^{2}}{b^{2}}=\frac{16}{1}\)

You may think that the image will now show only half of the object, but taking the laws of reflection to be true for all points of the remaining part of the mirror, the image will be that of the whole object. However, as the area of the reflecting surface has been reduced, the intensity of the image will be low (in this case, half).

Wavelength of violet light is smaller than that of red light. Also, angle of minimum deviation,
\(\delta _{ m }=\left( \mu -1 \right) A\Rightarrow \delta _{ m }=\mu \)
As \(\mu _{ R }< \mu _{ V }=\Rightarrow \left( \delta _{ m } \right) _{ E }< \left( \delta _{ m } \right) V\) 
As, deviation is less for red light, hence, angle of deviation decreases.

\(Grating \ element \ =\frac { 1 }{ 5000 } =2\times { 10 }^{ -4 }cm\)

This is because two independent sources of light cannot be coherent, as their relative phases are changing randomly.

ray diagram of a refracting telescope
Magnifying power is the ratio of angle subtended at the eye by the final image to the angle subtended by the object at the eye.
\(m=\frac{f_{o}}{f_{e}}\)
We can increase the magnifying power by:
(i) increasing focal length of an objective.
(ii) decreasing focal length of an eyepiece.
Limitations:
(i) Refracting type telescope suffers from chromatic aberration.
(ii) They have small resolving power.
Advantages of reflecting typestelescope
(i) It is free from chromatic aberration as there is no refraction.
(ii) The aperture of mirror can be kept larger in comparison to the aperture of lenses, because their grinding/polishing is easier and they can easily be provided with mechanical support. With larger aperture the resolving power of telescope increases.

The magnifying power of a telescope is equal to the ratio of the visual angle subtended 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.
Magnification, m = \(\frac{I}{O}=\frac{v_{0}}{u_{0}}=\frac{f_{0}}{u_{0}}\)
\(\Rightarrow \ \frac{I}{100}=\frac{150 \times 10^{-2}}{3 \times 10^{3}}\)
⇒ I = 5 x 10^-2 m = 5 cm

Angle of minimum deviation, \({ \delta }^{ ' }_{ m }\) = 40°
Angle of the prism, A = 60°
Refractive index of water, µ = 1.33
Refractive index of the material of the prism = µ'
The angle of deviation is related to refractive index (µ') as:
\({ \mu }^{ ' }=\frac { sin\frac { (A+{ \delta }_{ m }) }{ 2 } }{ sin\frac { A }{ 2 } } \)
\(=\frac { sin\frac { ({ 60 }^{ o }+{ 40 }^{ o }) }{ 2 } }{ sin\frac { { 60 }^{ o } }{ 2 } } \)\(=\frac { sin{ 50 }^{ o } }{ sin{ 30 }^{ o } } =1.532\)
Hence, the refractive index of the material of the prism is 1.532.
Since the prism is placed in water, let be the new angle of minimum deviation for the same prism.
The refractive index of glass with respect to water is given by the relation:
\({ \mu }_{ g }^{ w }=\frac { { \mu }^{ ' } }{ \mu } =\frac { sin\frac { (A+{ \delta }^{ ' }_{ m }) }{ 2 } }{ sin\frac { A }{ 2 } } \)
=\(sin\frac { (A+{ \delta }^{ ' }_{ m }) }{ 2 } =\frac { { \mu }^{ ' } }{ \mu } sin\frac { A }{ 2 } \)
= \(sin\frac { (A+{ \delta }^{ ' }_{ m }) }{ 2 } =\frac { 1.532 }{ 1.33 } \times sin\frac { { 60 }^{ o } }{ 2 } =\)0.5759
= \(\frac { (A+{ \delta }^{ ' }_{ m }) }{ 2 } ={ sin }^{ -1 }{ 0.5759=35.16 }^{ o }\)o
= \({ 60 }^{ o }+{ \delta }^{ ' }_{ m }={ 70.32 }^{ o }\)
\(\therefore { \delta }^{ ' }_{ m }={ 70.32 }^{ o }-{ 60 }^{ o }={ 10.32 }^{ o }\)
Hence, the new minimum angle of deviation is 10.32°.

(i) Power = +2 dioptre.
(ii) Here, we have f = +12 cm, R₁ = +10 cm, R₂ = -15 cm.
Refractive index of air is taken as unity.
We use the lens formula. The sign convention has to be applied for f, R₁ and R₂.
Substituting the values, we have
\(\frac { 1 }{ 12 } =(n-1)\left( \frac { 1 }{ 10 } -\frac { 1 }{ 15 } \right) \)
This gives n = 1.5.
(iii) For a glass lens in air, n₂ = 1.5, n₁ = 1, f = +20 cm. Hence, the lens formula gives
\(\frac { 1 }{ 20 } =0.5\left( \frac { 1 }{ { R }_{ 1 } } -\frac { 1 }{ { R }_{ 2 } } \right) \)
For the same glass lens in water, n₂ = 1.5, n₁ = 1.33. Therefore \(\frac { 1.33 }{ f } =(1.5-1.33)\left[ \frac { 1 }{ { R }_{ 1 } } -\frac { 1 }{ { R }_{ 2 } } \right] \)
Combining these two equations, we find f = + 78.2 cm.

(a) Wavelength of the light beam, λ₁ = 650 nm
Wavelength of another light beam, λ₂ = 520 nm
Distance of the slits from the screen = D
Distance between the two slits = d
Distance of the n^th bright fringe on the screen from the central maximum is given by the relation,
\(x=n \lambda_{1}\left(\frac{D}{d}\right)\)
For third bright fringe n = 3
\(\therefore x=3 \times 650 \frac{D}{d}=1950\left(\frac{D}{d}\right) n m\) 
(b) Wavelength of the light beam, \(\lambda_1\) =  650 nm
Wavelength of another light beam, λ₂ = 520 nm
Distance of the slits from the screen = D
Distance between the two slits = d
Let the n^th bright fringe due to wavelength λ₂ and (n − 1)^th bright fringe due to wavelength lambda coincide on the screen. We can equate the conditions for bright fringes as:
\(n \lambda_{2}=(n-1) \lambda_{1}\)
520 n = 650 n -650
650 = 130 n
\(\therefore\) n = 5
Hence, the least distance from the central maximum can be obtained by the relation:
\(x=n \lambda_{2} \frac{D}{d}\)
\(=5 \times 582 \frac{D}{d}=2600 \frac{D}{d} n m\) 

Here, distance of the screen from the slit, D = 2m,
a = ?, x = 5 mm = \(5\times { 10 }^{ -3 }m\),
\(\lambda =5000\mathring { A } =5000\times { 10 }^{ -10 }m\)
\(For \ the \ first \ secondary \ minima,\)
\( sin\theta =\frac { \lambda }{ a } =\frac { x }{ D } \)
\(a=\frac { D\lambda }{ x } =\frac { 2\times 5000\times { 10 }^{ -10 } }{ 5\times { 10 }^{ -3 } } =2\times { 10 }^{ -4 }m\)

(i) Merits of Huygens Wave theory of light:
(a) Wave theory correctly predicted that velocity of light in an optically denser medium is less than that in the rarer medium which is in agreement with the experimental results.
(b) On the basis of wave theory phenomenon of reflection, refraction, interference, diffraction, polarization of light could be explained.
Demiritssof Huygens wave theory of light
(a) Huygens wave theory assumes the existence of luminiferous ether. However,experimentally it couldn't be proved.
(b) This theory couldn't explain rectilinear propagation of light.
(ii) (1) Each pointon a given primary wavefront acts as a source of secondary wavelets, sending out disturbances (waves) in all directions in a similar manner as the original source of light does.
(2) The new position of the wavefront at any instant (secondary wavefront) is given by the forward envelope to the secondary wavelets at that instant.
Huygens' construction

Using this principle the laws of reflection and refraction can be verified.
(iii) Primary source of light: It is a real source of light. It generates light itself and sends primary wavefronts in all directions.
Secondary source of light: It is a fictitious source of light presents on the wavefront and sends out secondary waves only in forward direction.

(i) (d): Total internal reflection is the basis for following phenomenon:
(a) Sparkling of diamond.
(b) Optical fibre communication.
(c) Instrument used by doctors for endoscopy.
(ii) (d): Total internal reflection (TIR) is the phenomenon that involves the reflection of all the incident light off the boundary. TIR only takes place when both of the following two conditions are met:The light is in the more denser medium and approaching the less denser medium.The angle of incidence is greater than the critical angle.
(iii) (c) : If incidence of angle, i = critical angle e, then angle of refraction, r = \(90^{\circ}\)
(iv) (b): In optical fibres, core is surrounded by cladding, where the refractive index of the material of the core is higher than that of cladding to bound the light rays inside the core.
(v) (b): From Snell's law, \(\sin C={ }_{1} n_{2}=\frac{v_{1}}{v_{2}}\)
where, c = critical angle = 30° and V₁ and V₂ are speed oflight in medium and vacuum, respectively.
We know that, v² = 3 x 108 m s^-l
\(\therefore \quad \sin 30^{\circ}=\frac{v_{1}}{3 \times 10^{8}} \)
\(\Rightarrow \quad v_{1}=3 \times 10^{8} \times \frac{1}{2} \Rightarrow v_{1}=1.5 \times 10^{8} \mathrm{~ms}^{-1} \)