Class 12th

10.15 Sound waves can propagate only through a medium. The two given situations are not scientifically identical because the motion of an observer relative to a medium is different in the two situations. Hence, the Doppler formulas for the two situations cannot be the same.

In case of light waves, sound can travel in a vacuum. In a vacuum, the above two cases are identical because the speed of light is independent of the motion of the observer and the notion of the source. When light travels in a medium, the above two cases are not identical because the speed of light depends on the wavelength of the medium.

1.19 Net electric flux ( ${\phi }_{Net}$ ) through the cubic surface is given by

${\phi }_{Net}$ = $\frac{q}{{?}_0}$

where ${?}_0$ = Permittivity of free space = 8.854 *${10}^{-12}$ $C^2{N}^{-1}$$m^{-2}$

q= 2.0 μC

${\phi }_{Net}$ =$\frac{2*{10}^{-6}}{8.854\mathrm{}*{10}^{-12}}$ $C^{-1}N$ $m^2=2.25*{10}^5$ 

10.14 The speed of light in vacuum (3 $*{10}^{8}m/s)$

is a universal constant. It is not affected by the motion of the source, the observer, or both. Hence, the given factor does not affect the speed of light in a vacuum.

Out of these 5 factors, the speed of light in a medium depends on the wavelength of light in that medium.

10.13

Let an object at O be placed in front of a plane mirror MO' at a distance r. A circle is drawn from the centre (O) such that it just touches the plane mirror at point O'.

According to Huygens's principle, XY is the wave front of incident light.

If the mirror is absent, then a similar wave front X'Y' (as XY) would form behind O' at a distance r, as shown in the figure.

X'Y' can be considered as a virtual reflected ray for the plane mirror.

Hence, a point object placed in front of the plane mirror produces a virtual image whose distance from the mirror is equal to the object distance (r).

1.18 The square can be considered as one face of a cube of edge 10 cm with a centre where charge q is placed.

According to Gauss's theorem for a cube, total electric flux is through all its six faces.

  ${\phi }_{Tota\mathrm{l\; =}}$$\frac{q}{{?}_0}$ 

$\mathrm{H}\mathrm{e}\mathrm{n}\mathrm{c}\mathrm{e},\mathrm{}\mathrm{e}\mathrm{l}\mathrm{e}\mathrm{c}\mathrm{t}\mathrm{r}\mathrm{i}\mathrm{c}\mathrm{}\mathrm{f}\mathrm{l}\mathrm{u}\mathrm{x}\mathrm{}\mathrm{t}\mathrm{h}\mathrm{r}\mathrm{o}\mathrm{u}\mathrm{g}\mathrm{h}\mathrm{}\mathrm{o}\mathrm{n}\mathrm{e}\mathrm{}\mathrm{f}\mathrm{a}\mathrm{c}\mathrm{e}\mathrm{}\mathrm{o}\mathrm{f}\mathrm{}\mathrm{t}\mathrm{h}\mathrm{e}\mathrm{}\mathrm{c}\mathrm{u}\mathrm{b}\mathrm{e}$ i.e. through the square is $\phi =\frac{{\phi }_{Total}}{6}$ = $\frac{q}{{6?}_0}$ ,Where${?}_0$= Permittivity of free space = 8.854
 $*{10}^{-12}$ $C^2{N}^{-1}$$m^{-2}$

We have q = +10 $\mu C$

Hence,  = $\phi$ = $\frac{q}{{6?}_0}$ =$\frac{10*{10}^{-6}}{6*8.854\mathrm{}*{10}^{-12}}$= 188238.83 N$m^2{C}^{-1}$  = 1.88$*{10}^5$ N$m^2{C}^{-1}$  

10.11 The wavelength of $H_{?}$ line emitted by Hydrogen, $?=6563\AA$ = 6563 $*{10}^{-10}$ m$$

Star's red-shift, ( $?\text{'}$ $-$ $?)$ = 15 $\AA$ = 15 $*{10}^{-10}$ m

Speed of light, c = 3 $*{10}^8$ m/s

( $?\text{'}$ $-$ $?)$ = $\frac{v}{c}?$

v = $\frac{c}{?}*$ ( $?\text{'}$ $-$ $?)$ = $\frac{3\mathrm{}*{10}^8}{6563\mathrm{}*{10}^{-10}}*$ 15  $*{10}^{-10}$ = 6.86 $*{10}^5$ m/s

10.10 Fresnel's distance ( $Z_F)$ is the distance for which the ray optics is a good approximation. It is given by the relation

$Z_F=\frac{a^2}{?}$ where $a$ =aperture width = 4 mm = 4 $*{10}^{-3}$ m

$?$ = wave length = 400 nm = 400 $*{10}^{-9}$ m

Hence $Z_F=\frac{{4*{10}^{-3}}^2}{400\mathrm{}*{10}^{-9}}$ = 40 m

Therefore, the distance for which ray optics is a good approximation is 40 m.

1.17  (a) Net outward flux through the surface of the box , $\phi$ = 8.0 * ${10}^3$ ${\mathrm{Nm}}^2/$C

For a body containing net charge q, flux is given by the relation, 
 $\phi$ =$\frac{q}{{?}_0}$ 

where${?}_0$= Permittivity of free space = 8.854 
 $*{10}^{-12}$ $C^2{N}^{-1}$$m^{-2}$

Hence q =  $\phi$ $*{?}_0$= 8.0 *${10}^3*$ 8.854 $*{10}^{-12}$ C = 7.0832 $*{10}^{-8}$ C

= 7.0832 $*{10}^{-2}\mu C$

(b) Net flux piercing out through a body depends on the net charge contained in the body. If net flux is zero, then it can be inferred that the net charge inside the body is zero. The body may have equal amount of positive and negative charges.

10.9 Wave length of the incident light, $?$ = 5000 $*{10}^{-10}$ m

Speed of light, c = 3 $*{10}^8$ m/s

Frequency of the incident light, $?$ =  $\frac{c}{?}$  =$\frac{3*{10}^8}{5000*{10}^{-10}}$ Hz= 6 $*{10}^{14}$ Hz

The wavelength and frequency of incident ray will be same as of reflected ray. So the wavelength of the reflected ray will be 5000 Å and the frequency will be 6 $*{10}^{14}$ Hz

When reflected ray is normal to the incident ray, the sum of incidence angle $?i$ and the reflected angle $?r$ will be 90 $°$

According to the law of reflection, the incidence angle and the reflected are same, i.e. $?i=?r$

Hence, $?i+?r$ = 2 $?i$ = 90 $°$

So$?i=?r=45°$

Therefore, the angle of incidence for the given condition is $45°$ .