Physics

This is a multiple choice answer as classified in NCERT Exemplar

(c) E $\propto \text{exception 25:}$

$\frac{\left(E_1\right)}{\left(E_2\right)}$ = $\sqrt[]{\frac{P_1}{P_2}}$ = $\sqrt[]{\frac{1000}{50}}$

$E$ 2= E/ $\text{exception 25:}$

This is a multiple choice answer as classified in NCERT Exemplar

(b) as we know momentum = energy transferred/speed of light

= $\frac{20*30*30*60}{3*{10}^8}$ = 36 $*$ 10^-4kgms^-1

Momentum of reflected light =0

So momentum delivered is = 36 $*$ 10^-4kgms^-1

This is a multiple choice answer as classified in NCERT Exemplar

(b) the electromagnetic wave is 

E_r = E_oicos (kz-wt)

And reflected wave is given by

E_r = -E₀ (-i)cos (k (-z)+wt+ $\pi$ )

This is a multiple choice answer as classified in NCERT Exemplar

(c) E=hv

V= $\frac{11*1.6*{10}^{-19}}{6.62*{10}^{-34}}$ = 2.65 $*$  10¹⁵hz this belongs to uv region

This is a short answer type question as classified in NCERT Exemplar

Since, electric field of an EM wave is an oscillating field and so is the electric force caused by it on a charged particle. This electric force averaged over an integral number of cycles is zero, since its direction changes every half cycle.

Hence, electric field is not responsible for radiation pressure.

This is a short answer type question as classified in NCERT Exemplar

As the distance is doubled, the area of spherical region ( 4πr² )

will become four times, so the intensity becomes one fourth the initial value but in case of laser it does not spread, so its intensity remain same.

Geometrical characteristic of LASER beam which is responsible for the constant intensity are as following

(i) Unidirectional

(ii) Monochromatic

(iii) Coherent light

(iv) Highly collimated

These characteristic are missing in the case of light from the bulb.

This is a short answer type question as classified in NCERT Exemplar

Pressure = force/area

Force =dp/dt

U=p.C or p= U/c

Pressure = $\frac{1}{A}.\frac{U}{C.dt}$

Pressure = I/C

This is a short answer type question as classified in NCERT Exemplar

q=CV

And we know q=It

Then I_ddt= cdV

I_d= Cdv/dt

1 $*{10}^{-3}$ = 2 $*$ 10^{-6 $*$} dV/dt

DV/dt= 500V/s, this will produce the desired result.

This is a short answer type question as classified in NCERT Exemplar

As we know radiant flux density S=  $\frac{1}{{\mu }_0}$ (E $*B$ )= c^{2 $?$} ₀(E $*B$ )

Suppose electromagnetic waves be propagating along x-axis. The electric field vector of

Electromagnetic wave be along y-axis and magnetic field vector be along z-axis. Therefore,

E=E₀cos (kx-wt)

B=B₀cos (kx-wt)

E $*$ B= (E₀B₀)cos²(kx-wt)

S= c^{2 ${\epsilon }_0$} (E $*B$ )

Radiant flux density over a complete cycle is

S_av= c^{2 $?$} ₀|E_{0 $*$} B₀| $\frac{1}{T}{\int }_0^T{cos}^2(kx-wt)dt$

     = c^{2 $?$} ₀|E₀B₀| $\frac{1}{T}*\frac{T}{2}$

     = $\frac{C^2}{2}\epsilon$ ₀E₀( $\frac{E_0}{c}$ )

     = $\frac{c}{2}\epsilon$ ₀E₀²= $\frac{c}{2}*$ $\frac{1}{c^2{\mu }_0}$  E₀²

     = $\frac{E_0^2}{2{\mu }_{0}c}$