Physics

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 $45°$ .

10.8 Refractive index of glass,  $?=1.5$

Let the Brewster angle be $?$

Brewster angle is related to $?$ by the equation

$\tan ??=?$

Or  = $?={\tan }^{-1}??$ =${\tan }^{-1}?1.5$ =56.31$°$

Hence, the Brewster angle for air to glass transition is 56.31 $°$

10.7 Distance of the screen from the slits, D = 1 m

Wavelength of the light used,  ${?}_1$ = 600 nm

Angular width of the fringe in air,  ${?}_1$ = 0.2 $°$

Angular width of the fringe in water = ${?}_2$

Refractive index of water,  $?=\frac{4}{3}$

Refractive index is related to angular width as:

$?=\frac{{?}_1}{{?}_2}$ or

${?}_2=\frac{{?}_1}{?}$ = 0.2 $*\frac{3}{4}$ = 0.15 $°$

Hence, the angular width of the fringes in water will be 0.15 $°$

1.16 When the cube side is oriented so that its faces are parallel to the coordinate planes, number of field lines entering the cube is equal to the number of field lines piercing out of the cube. A as a result, net flux through the cube is zero.