Wave Optics

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 t

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

10.12 Newton's corpuscular theory of light states that when light corpuscles strike the interface of two media from a rarer (air) to a denser (water) medium, the particles experience forces of attraction normal to the surface. Hence, the normal component of velocity increases while the component alo

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.

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 f

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

${?}_2=\frac{{?}_1}{?}$ =

10.6 Wavelength of one light beam, ${?}_1$ = 650 nm

Wavelength of the other beam, ${?}_2$ = 520 nm

Distance of the screen from the slits = D

Distance between two slits = d

Distance of the $n^{th}$ bright fringe on the screen from the central maximum is given by the relation,

$x$ = n ${?}_1(\frac{D}{d}$ 

10.5 Let $I_1$ and $I_2$ be the intensity of the two light waves. Their resultant intensity can be obtained as :

$I’$ = $I_1+\mathrm{}{I}_2+2\sqrt{I_1{I}_2}$ $\cos ??$ , where $?=$ Phase difference between two waves

For monochromatic light waves, $I_1=I_2$ . Hence

$I’$ = $I_1+I_1$ + 2 $\sqrt{I_1^2}$ $\cos ??$ = 2 $I_1+2I_1\cos ??$

We