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Advantages of reflecting telescope over refracting telescope
(i) In reflecting telescope, image formed is free from chromatic aberration defect. So, it is sharper than image formed by a retracting type telescope.
(ii) A mirror is easier to produce with a large diameter, so that it can intercept rays crossing a large area and direct them to the eye-piece.

Since, we know, I_c = I_d = 0.15 A

\(\text { Average power, } \vec{P}=V_{r m s} \times i_{r m s} \times \cos \phi\)
\( \text { Where } \phi=90^{\circ} \\ \therefore \vec{P}=V_{r m s} \times i_{r m s} \times \cos 90^{\circ} \\ =0 \)

Self-inductance is the property of a coil by virtue of which, the coil opposes any change in the strength of the current flowing through it by including an emf in itself.
Its SI unit is henry(H). 

\(here,\ C=45°,\mu =?\)
\(\mu =\frac { 1 }{ \sin { C } } =\frac { 1 }{ \sin { 45° } } =\frac { 1 }{ 1/\sqrt { 2 } } =\sqrt { 2 } \)

Doping is a process of deliberate addition of a desirable impurity in a pure semiconductor to modify its properties in a controlled manner.

The capacitive reactance is
\(X_{C}=\frac{1}{2 \pi v C}=\frac{1}{2 \pi(50 \mathrm{~Hz})\left(15.0 \times 10^{-6} \mathrm{~F}\right)}=212 \Omega\)
The rms current is
\(I=\frac{V}{X_{C}}=\frac{220 \mathrm{~V}}{212 \Omega}=1.04 \mathrm{~A}\)
The peak current is
\(i_{m}=\sqrt{2} I=(1.41)(1.04 A)=1.47 A\)
This current oscillates between +1.47A and -1.47 A, and is ahead of the voltage by π/2.
If the frequency is doubled, the capacitive reactance is halved and consequently, the current is doubled.

Given, focal length of convex lens, f₁ = 30 cm
Focal length of the concave lens, f₂ = -20 cm
Using the formula of combination of lenses,
\(\frac{1}{f}=\frac{1}{f_{1}}+\frac{1}{f_{2}}=\frac{1}{30}-\frac{1}{20}=\frac{2-3}{60}=-\frac{1}{60}\)
\(\Rightarrow\) f = -60 cm
Since, the focal length of combination is negative in nature. so, the combination behaves like a diverging lens, i.e. as a concave lens.

Given, input frequency = 50 Hz
For a half-wave rectifier, the output frequency is equal to the input frequency.
\(\therefore\) Output frequency = 50 Hz
For a full-wave rectifier, the output frequency is twice the input frequency.
\(\therefore\)  Output frequency = 2 x 50 = 100 Hz.

The given diagram shown below

The circuit above can be redrawn as follows

As the p-section is connected to negative terminal of the battery, the diode shown is reverse biased.
(ii) During the first half of input cycle, the upper end of the coil is at positive potential and lower end at negative potential. The function diode DI is forward biased and D₂ in reverse biased. Current flows in output load in the
direction shown in figure. During the second half of input cycle, D₂ is forward biased. In this way, current flows in the load in the single direction as shown in figure.

According to Huygens' principle
(i) Each point on the given wavefront (called primary wavefront) is the source of a secondary disturbance (called secondary wavelets) and the wavelets emanating from these point spread out in all the directions with the speed of the wave. 
(ii) A surface touching these secondary wavelets, tangentially in the forward direction at any instant gives the new wavefront at that instant. This is called secondary wavefront.

If v₁, v₂ are the speed of light into two mediums and t is the time taken by light to go from B to C or A to D or E to Gthrough F, then
\(t=\frac { EF }{ { v }_{ 1 } } +\frac { FG }{ { v }_{ 2 } } \)
In \(\triangle AFE \ \sin { i } =\frac { EF }{ AF } \)
In \(\triangle FGC \ \sin { r } =\frac { FG }{ FC } \ \Rightarrow \ t=\frac { AF\sin { i } }{ { v }_{ 1 } } +\frac { FC\sin { r } }{ { v }_{ 2 } } \)
\(\Rightarrow \ t=\frac { AC\sin { r } }{ { v }_{ 2 } } +AF\left( \frac { \sin { i } }{ { v }_{ 1 } } -\frac { \sin { r } }{ { v }_{ 2 } } \right) \)
For rays of light from the different parts on the incident wavefront, the values of AF are different. But light from different points of the incident wavefront should take the same time to reach the corresponding points on the refracted wavefront.
So, t should not depend on F. This is possible only,
If \(\frac { \sin { i } }{ { v }_{ 1 } } -\frac { \sin { r } }{ { v }_{ 2 } } =0 \ or \ \frac { \sin { i } }{ \sin { r } } =\frac { { v }_{ 1 } }{ { v }_{ 2 } } =0\)
Now, if c represents the speed of light in vacuum, then \({ \mu }_{ 1 }=\frac { c }{ { v }_{ 1 } } \ and \ \frac { c }{ { v }_{ 2 } } \) are known as the refractive index of medium 1 and medium 2 respectively.
Then, \({ \mu }_{ 1 }=\sin { i } \ and \ \sin { r } \ \Rightarrow \ \frac { \sin { i } }{ \sin { r } } \)
This is known as Snell's law of refraction.

(i) Microwave
(ii) Infrared waves are used to treat muscular strain.
(iii) X-rays are used as a diagnostic tool in medicine. X-rays are produced by X-ray tubes or inner shell electrons.

\(Here,\ A=60°,\mu =?\)
\( \mu =\frac { \sin { (A+{ \delta }_{ m } } )/2 }{ \sin { A/2 } } =\frac { \sin { \left( 60°+60° \right) /2 } }{ \sin { 60°/2 } }\)
\(=\frac { \sin { 60° } }{ \sin { 30° } } =\frac { \sqrt { 3 } /2 }{ 1/2 } =\sqrt { 3 }\)

f₁=7.5×10⁶ Hz
f₂=12×10⁶ Hz
λ₁=c/f₁=40 m
λ₂=c/f₂=25 m
So the range is 40m to 25m

(a) The shape of the wavefront in case of a light diverging from a point source is spherical. The wavefront emanating from a point source is shown in the given figure.

(b) The shape of the wavefront in case of a light emerging out of a convex lens when a point source is placed at its focus is a parallel grid. This is shown in the given figure.

(c) The portion of the wavefront of light from a distant star intercepted by the Earth is a plane.

Frequency of the electromagnetic wave, ν = 2.0 x 10¹⁰ Hz
Electric field amplitude, E₀ = 48 V m⁻¹
Speed of light, c = 3 x 10⁸ m/s
(a) Wavelength of a wave is given as:
\(\lambda=\frac{c}{v}\)
\(=\frac{3 \times 10^{8}}{2 \times 10^{10}}=0.015 \mathrm{~m}\)
(b) Magnetic field strength is given as:
\(B_{0}=\frac{E_{0}}{c}\)
\(=\frac{48}{3 \times 10^{8}}=1.6 \times 10^{-7} T\)
(c) Energy density of the electric field is given as:
\(U_{E}=\frac{1}{2} \in_{0} E^{2}\)
And, energy density of the magnetic field is given as:
\(U_{B}=\frac{1}{2 \mu_{0}} B^{2}\)
Where,
∈₀ = Permittivity of free space
μ₀ = Permeability of free space
We have the relation connecting E and B as:
E = cB … (1)
Where,
\(c=\frac{1}{\sqrt{\epsilon_{0} \mu_{0}}} \ldots \text { (2) }\)
Putting equation (2) in equation (1), we get
\(E=\frac{1}{\sqrt{\epsilon_{0} \mu_{0}}}\)
Squaring both sides, we get
\(E=\frac{1}{\epsilon_{0} \mu_{0}} B^{2}\)
\(\epsilon_{0} E^{2}=\frac{B^{2}}{\mu_{0}}\)

(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)

(b) Principle of a transformer: When alternating current flows through the primary coil, an emf is induced in the neighbouring (secondary) coil
Let \(\frac { d\phi }{ dt } \) be the rate of change of flux through each turn of the primary and the secondary coil
\(\frac { { \varepsilon }_{ 1 } }{ { \varepsilon }_{ 2 } } =-{ N }_{ 1 }\frac { d\phi }{ dt } /-{ N }_{ 2 }\frac { d\phi }{ dt } =\frac { { N }_{ 1 } }{ { N }_{ 2 } } \\ \frac { { V }_{ 1 } }{ { V }_{ 2 } } =\frac { { N }_{ 1 } }{ { N }_{ 2 } } \)
But for an ideal transformer,
\(\frac { { V }_{ 1 } }{ { V }_{ 2 } } =\frac { I_{ 2 } }{ I_{ 1 } } \)
From equation (1) and (2)
\(\frac { { V }_{ 1 } }{ { V }_{ 2 } } =\frac { N_{ 1 } }{ N_{ 2 } } =\frac { I_{ 2 } }{ I_{ 1 } } \)
(c) Main assumptions
(i) The primary resistance and current are small
(ii) The flux linked with the primary and secondary coils is same / there is no leakage of flux from the core.
(iii) Secondary current is small.
(d) Reason due to which energy loses may occur Flux leakage/Resistance of the coils/Eddy currents/Hysteresis.

\(Here, \ \lambda =589 \ nm,\ c=3\times { 10 }^{ 8 }m/s, \ \mu =1.33\)
(a) For reflected light
\(wavelength,\ \lambda =589\quad nm=589\times { 10 }^{ -9 }m,\quad v=\frac { c }{ \lambda } =\frac { 3\times { 10 }^{ 8 } }{ 589\times { 10 }^{ -9 } } =5.09\times { 10 }^{ 14 }hertz\)
\(speed,\ v=c=3\times { 10 }^{ 8 }m/s\)
(b) For refracted light \( \lambda '=\frac { \lambda }{ \mu } =\frac { 589\times { 10 }^{ -9 } }{ 1.33 } =4.42\times { 10 }^{ -7 }m\)
As frequency remains unaffected on entering another medium,
\(\\ therefore,\quad v'=v=5.09\times { 10 }^{ 14 }hertz\)
\(speed, \ v'=\frac { c }{ \mu } =\frac { 3\times { 10 }^{ 8 } }{ 1.33 } =2.25\times { 10 }^{ 8 }m/s\)

(a) To find the impedance of the circuit, we first calculate \(X_{\mathrm{L}}\) and \(X_{\mathrm{C}}\).
\( X_L=2 \pi v L \)
\(=2 \times 3.14 \times 50 \times 25.48 \times 10^{-3} \Omega=8 \Omega \)
\(X_c=\frac{1}{2 \pi v C}  =\frac{1}{2 \times 3.14 \times 50 \times 796 \times 10^{-6}}=4 \Omega\)
Therefore.
\(Z  =\sqrt{R^2+\left(X_L-X_C\right)^2}=\sqrt{3^2+(8-4)^2} =5 \Omega\)
\(Z  =\sqrt{R^2+\left(X_L-X_C\right)^2}=\sqrt{3^2+(8-4)^2} =5 \Omega\)
(b) Phase difference, \(\phi=\tan ^{-1} \frac{X_C-X_L}{R}\)
\(=\tan ^{-1}\left(\frac{4-8}{3}\right)=-53.1^{\circ}\)
Since \(\phi\) is negative, the current in the circuit lags the voltage across the source.
(c) The power dissipated in the circuit is
\(P=I^2 R\)
Now, \(I=\frac{i_m}{\sqrt{2}}=\frac{1}{\sqrt{2}}\left(\frac{283}{5}\right)=40 \mathrm{~A}\)
Therefore, \(P=(40 \mathrm{~A})^2 \times 3 \Omega=4800 \mathrm{~W}\)
(d) Power factor \(=\cos \phi=\cos \left(-53.1^{\circ}\right)=0.6\)

(a): Light being electromagnetic wave do not require any material medium for its propagation. Hence light can travel in vacuum. On the other hand sound is a mechanical wave and requires a material medium for its propagation. Hence sound cannot travel in vacuum.

(b): The brilliance of diamond is due to total internal reflection of light. \(\mu\) for diamond is 2.42, so that critical angle for diamond air interface is C = 24.40 (from sinC = l/\(\mu\)). The diamond is cut suitable so that light entering the diamond from any face suffers multiple total internal reflections at the various faces and remains within the diamond. Hence the diamond sparkles.

(c): The reflection and refraction of light occurs on account of interaction of light with the atoms of the surface of separation. These atoms can be regarded as oscillators. Light incident on the interface forces the atomic oscillators to oscillate with frequency of incident light. As frequency oflight emitted by these (charged). oscillators is equal to their own frequency of oscillation, therefore, reflected and refracted light have the same frequency as that of incident light.

(a): In insulator, the forbidden energy gap is quite large. When electric field is applied to such a solid, the electron find it difficult to acquire such a large amount of energy. Thus no electron flow occurs.

(i) (d): Here, \(\mu=1.5 ; R_{1}=30 \mathrm{~cm}\)
R₂ = -30 cm
\(\text { As } \frac{1}{f}=(\mu-1)\left[\frac{1}{R_{1}}-\frac{1}{R_{2}}\right] \)
\(=(1.5-1)\left[\frac{1}{30}-\frac{1}{-30}\right]=-0.5 \times \frac{2}{30}-\frac{-1}{30}\)
\(f=-30 \mathrm{~cm}\)
(ii) (a): Here, f= 12 cm ; R₁ = 10 cm
R₂ = -15 cm
\(\text { As } \frac{1}{f}=(\mu-1)\left[\frac{1}{R_{1}}-\frac{1}{R_{2}}\right] \)
\(\frac{1}{12}=(\mu-1)\left[\frac{1}{10}+\frac{1}{15}\right] \)
\(\mu=1.5\)
(iii) (d): The eye-lens is surrounded by a different medium than air. This will change the focal length of the eye-lens. The eye cannot accommodate all images as it would do in air.
(iv) (b): \(\frac{1}{f}=(1.5-1)\left(\frac{1}{R_{1}}-\frac{1}{R_{2}}\right)\)
\(\text { and } \frac{1}{f_{w}}=\left(\frac{1.5}{1.33}-1\right)\left(\frac{1}{R_{1}}-\frac{1}{R_{2}}\right) \)
\(\frac{f_{w}}{f}=\frac{0.5 \times 1.33}{0.17}=4 \)
\(f_{w}=4 f=4 \times 10=40 \mathrm{~cm}\)
(v) (d): If the refractive index of two media are same,the surface of separation does not produce refraction or reflection which helps in visibility.