Wave Optics

This is a Long Answer Type Questions as classified in NCERT Exemplar

Explanation- as the refractive index of the class $\mu$ , the path difference will be calculated as $?x$ =2dsin $\theta$ +( $\mu -1$ )L

For principal maxima ,(path difference is zero)

2dsin $\theta$ ₀+( $\mu -1$ )L=0

Sin $\theta$ ₀= - $\frac{L(\mu -1)}{2d}$ = $\frac{-L\left(0.5\right)}{2d}$

Sin $\theta$ ₀=-1/16

OP=Dtan $\theta$ ₀= Dsin $\theta$ ₀=-D/16

For pat $?$ h difference $?\frac{\lambda }{2}$

2dsin $\theta$ ₁+0.5L= $?\frac{\lambda }{2}$

Sin $\theta$ ₁= $\frac{?\frac{\lambda }{2}-0.5L}{2d}$ = $\frac{?\frac{\lambda }{2}-\frac{d}{8}}{2d}$

= $\frac{\frac{\lambda }{2}-\frac{\lambda }{8}}{2\lambda }$ = $?$ 1/4 -1/16

So two possible values $\frac{1}{4}-\frac{1}{16}=\frac{3}{16}$  and- $\frac{1}{4}-\frac{1}{16}$ = $-\frac{5}{16}$

10.21 Consider that a single slit of width $d$ is divided in to $n$   smaller slits.

Therefore width of each slit,

$d^{\text{'}}=\frac{d}{n}$

Angle of diffraction is given by the relation,

$?=\frac{\frac{d}{d}}{d}?$ = $\frac{?}{d}$

Now, each of these infinitesimally small slit sends zero intensity in the direction of Ø.

Hence, the combination of these slits will give zero intensity.

10.20 Weak radar signals sent by a low flying aircraft can interfere with the TV signal received by the antenna. Hence, TV signal may get distorted, resulting in shaking of picture on the TV.

This is because superposition follows from the linear character of a differential equation that governs wave motion. If $y_1$ and $y_2$ are the solutions of the second order wave equation, then any linear combination of $y_1$ and $y_2$ will also be the solution of the wave equation.

10.19 Wavelength of the light beam, $?$ = 500 nm = 500 $*{10}^{-9}$ m

Distance of the screen from the slit, D = 1 m $n$

Distance of the first minimum from the centre of the screen, x = 2.5 mm = 2.5 $*{10}^{-3}$ m

Let the width of the slit be = d

From the equation

$n?=x\frac{d}{D}$ we get d = $\frac{n?D}{x}$

$d=\frac{1*500\mathrm{}*{10}^{-9}*1}{2.5\mathrm{}*{10}^{-3}}$  = 2 $*{10}^{-4}$ m = 0.2 mm

Hence, the width of the slot is 0.2 mm

10.18 Distance between the towers, d = 40 km

Height of the line joining the hills, h = 50 m

Thus, the radial spread of the radio wave should not exceed 40 km

Since the hill is located halfway between the towers,

Fresnel's distance $Z_P$  = $\frac{40}{2}$ = 20 km = 2 $*{10}^{4}m$

Aperture can be taken as a = h = 50 m

Fresnel's distance is given by the relation,

$Z_P$ = $\frac{a^2}{?}$ or

$?=\frac{a^2}{Z_P}$ =  $\frac{{50}^2}{2\mathrm{}*{10}^4}$ = 0.125 m = 12.5 cm

Therefore, the wavelength of the radio wave is 12.5 cm

10.16 Wavelength of the light used, $?=$ 6000 nm = 600 $*{10}^{-9}$ m

Angular width of the fringe, $?=0.1°$ = 0.1 $*\frac{?}{180}$  rad

Angular width of a fringe is related to slit spacing (d) as $?=\frac{?}{d}$

d = $\frac{?}{?}$ = $\frac{600*{10}^{-9}}{0.1\mathrm{}*\frac{?}{180}}$ = 3.44 $*{10}^{-4}$ m

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.

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.