Class 12th

9.9 Size of the object, $h_1$ = 3 cm

Object distance, u = - 14 cm

Focal length of the concave lens, f = - 21 cm

Image distance = v

According to lens formula

$\frac{1}{v}$ - $\frac{1}{u}$ = $\frac{1}{f}$ or $\frac{1}{v}$ - $\frac{1}{-14}$ = $-\frac{1}{21}$

$\frac{1}{v}$ = $-\frac{1}{14}-\frac{1}{21}$ or v = $-\frac{42}{5}$ = - 8.4 cm

Hence, the image is formed on the other side of the lens, 8.4 cm away from the lens. The negative sign shows that the image is erect and virtual.

The magnification of the image is given as:

m = $\frac{Imageheight\left(h_2\right)}{Objectheight\left(h_1\right)}$ = $\frac{v}{u}$

$\frac{h_2}{h_1}$ = $\frac{-8.4}{-14}$ or $h_2$ = 0.6 $*3$ = 1.8 cm

If the object is moved further away from the lens, then the virtual image will move towards the focus of the lens, but not beyond it. The size of the image will decrease with the increase in the object distance.

9.6 The angle of minimum deviation, ${\delta }_m$ = 40 $°$

Angle of prism, A = 60 $°$

Let the refractive index of water, $=1.33$ , and the refractive index of prism material = $\mu \text{'}$

The angle of deviation is related to refractive index $\mu \text{'}$ is given as

${\mu }^{\text{'}}=\frac{\sin ?\frac{(A+{\delta }_m)}{2}}{\sin ?\frac{A}{2}}$ = $\frac{\sin ?\frac{(60°+40°)}{2}}{\sin ?\frac{60°}{2}}$ = $\frac{\sin ?50°}{\sin ?30°}$ = 1.532

So the refractive index of prism material is 1.532

Since the prism is placed in water, let ${\delta }_m^{\text{'}}$ be the new angle of minimum deviation.

The refractive index of glass with respect to water is given by the relation:

${\mu }_g^w$ = $\frac{{\mu }^{\text{'}}}{\mu }$ = $\frac{\sin ?\frac{(A+{\delta }_m^{\text{'}}\mathrm{})}{2}}{\sin ?\frac{A}{2}}$

$\frac{1.532}{1.33}$ = $\frac{\sin ?\frac{(60°+{\delta }_m^{\text{'}}\mathrm{})}{2}}{\sin ?\frac{60°}{2}}$

1.152 $*\sin ?30°$ = $\sin ?\frac{(60°+{\delta }_m^{\text{'}}\mathrm{})}{2}$

$\frac{(60°+{\delta }_m^{\text{'}}\mathrm{})}{2}={\sin }^{-1}?0.576$ = 35.2 $°$

${\delta }_m^{\text{'}}=2*$ 35.2 $°$ - 60 $°$ = 10.33 $°$

Hence the new minimum angle of deviation is 10.33$°$

11.37 (a) Quarks inside protons and neutrons carry fractional charges. This is because nuclear force increases extremely if they are pulled apart. Therefore, fractional charges may exist in nature; observable charges are still the integral multiple of an electrical charge.

(b) The basic relations for electric field and magnetic field are

$\left\{eV=\frac{1}{2}m{v}^2\right\}$ and $\left\{eBV=\frac{m{v}^2}{r}\right\}$  respectively

These relations include e (electric charge), v (velocity), m (mass), V (potential), r (radius) and B (magnetic field. These relations give the value of the velocity of an electron as

$\left(v=\sqrt{2V\left(\frac{e}{m}\right)}\right)and\left(v=Br\left(\frac{e}{m}\right)\right)$ respectively

It can be observed from these relations that the dynamics of an electron is determined not by e and m separately, but by the ratio of  . $\frac{e}{m}$

(c) At atmospheric pressure, the ions of gases have no chance of reaching their respective electrons because of collision and recombination with other gas molecules. Hence, gases are insulators at atmospheric pressure. A low pressures, ions have a chance of reaching their respective electrodes and constitute a current. Hence, they conduct electricity at these pressures.

(d) The work function of a metal is the minimum energy required for a conduction electron to get out of the metal surface. All the electrons in an atom do not have the same energy level. When a ray having some photon energy is incident on a metal surface, the electrons come out from different levels with different energies. Hence, these emitted electrons show different energy distributions

(e) The absolute value of energy of a particle is arbitrary within the additive constant. Hence, wavelength( $\lambda )$ ( is significant, but the frequency( $\nu )$ (  associated with an electron has no direct physical significance.

Therefore, the product $\nu \lambda$  (phase speed) has no physical significance.

Group speed is given as:

$v_G$ = $\frac{dv}{dk}$ = $\frac{dv}{d\left(\frac{1}{\lambda }\right)}$ = dE/dp = (d(p^2/2m))/dp = p/m

This quantity has a physical meaning.

11.35 Room temperature, T = 27 $?$  = 300 K

Atmospheric pressure, P = 1 atm = 1.01 $*{10}^5$  Pa

Atomic weight of helium atom = 4

Avogadro's number, $N_A$  = 6.023 $*{10}^{23}$

Boltzmann's constant, k = 1.38 $*{10}^{-23}$  J ${mol}^{-1}{K}^{-1}$

Planck's constant, h = 6.626 $*{10}^{-34}$  Js

Average energy of a gas at temperature T is given as:

E = $\frac{3}{2}$ kT = $\frac{3}{2}*$ 1.38 $*{10}^{-23}*300$  = 6.21 $*{10}^{-21}$ J

De Broglie wavelength is given as

$\lambda =$ $\frac{h}{\sqrt{2mE}}$  , where m = mass of He atom = $\frac{Atomicweight}{Avogadr{o}^{\text{'}}snumber}$  = $\frac{4}{6.023*{10}^{23}}$

m = 6.641 $*{10}^{-24}$  gm = 6.641 $*{10}^{-27}$ kg

$\lambda =$ $\frac{6.626*{10}^{-34}}{\sqrt{2*6.641*{10}^{-27}*6.21*{10}^{-21}}}$ = 7.29 $*{10}^{-11}$ m

We have ideal gas formula:

PV = RT

PV = kNT

$\frac{V}{N}=\frac{kT}{P}$          

Where, V = volume of the gas

N = number of moles of the gas

Mean separation between two atoms of the gas is given by the relation:

r = $({\frac{V}{N})}^{1/3}$   = ${\left(\frac{kT}{P}\right)}^{1/3}$

=( ${\frac{1.38*{10}^{-23}*300}{1.01*{10}^5})}^{1/3}$

= 3.45 $*{10}^{-9}$ m

Hence, the mean separation between the atoms is much greater than the De Broglie wavelength.