Class 12th

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.

1.15 (a) Electric field intensity, E = 3 $*{10}^3$ î N/C

Magnitude of electric field intensity, $\left|E\right|$$=3*{10}^3$ = 
 N/C

Side of the square, s = 10 cm = 0.1 m

Area of the square, A = $s^2$ = 0.01 $m^2$

The plane of the square is parallel to the y-z plane, hence the angle between the unit vector normal to the plane and electric field, $\theta$= 0 $°$

Flux ( $\phi )$ through the plane is given by the relation,  $\phi$ = $\left|E\right|A\cos ?\theta$ = $3\mathrm{}*{10}^3*0.01*\cos ?0°=30$
${\mathrm{Nm}}^2/$C

(b) When the normal to its plane make a 60 $°$ angle with x-axis,$\theta$  = 60 
 . From the equation $\phi$ =$\left|E\right|A\cos ?\theta$we get  = $\phi$ $3\mathrm{}*{10}^3*0.01*\cos ?60°=15$ ${\mathrm{Nm}}^2/$C

10.4 Distance between the slits, d = 0.28 mm = 0.28 $*{10}^{-3}$ m

Distance between the slits and the screen, D = 1.4 m

Distance between the central bright fringe and the fourth ( n = 4) fringe, u = 1.2 cm = 1.2 $*{10}^{-2}$ m

In case of a constructive interference, we have the relation for the distance between two fringes as : u = n $?\frac{D}{d},$ where n = order of fringes = 4 and $?$ = wavelength of the light used

Hence,  $?$ = $\frac{ud}{nD}$ = $\frac{1.2\mathrm{}*{10}^{-2}*0.28\mathrm{}*{10}^{-3}}{4*1.4}$ = 6 $*{10}^{-7}$ m = 600 $*{10}^{-9}$ m = 600 nm

Hence, wavelength of the light is 600 nm.

10.3 The refractive index of glass,  $?$ = 1.5

Speed of light in vacuum, c = 3.0 $*{10}^8$ m/s

Speed of light in glass is given by the relation,  $v$ = $\frac{c}{?}$ = $\frac{3.0\mathrm{}*{10}^8}{1.5}$ = 2 $*{10}^8$ m/s

The speed of light in glass is not independent of the colours of light. The refractive index of a violet component of white light is less than the speed of red light in glass. Hence, violet light travels slower than red light in a glass prism.

1.14 Since the charges 1 & 2 are attracted towards + ve, their charges will be – ve. The charge 3 is attracted towards – ve, hence its charge will be +ve.

The charge to mass ratio (emf) is directly proportional to the displacement, charge 3 will have the highest charge to mass ratio.

Ans.10.2 The shape of the wave front in case of a light diverging from a point source is spherical.

The shape of the wave front in case of a light emerging out of a convex lens when a point source is placed at its focus is a parallel grid.

The portion of the wave front of light from a distant star intercepted by the Earth is a plane.