Problem set 44

1. Three polarizers P₁, P₂ and P₃ are arranged as shown in figure. Optical axis of P₁ and P₃ are perpendicular to one another while the axis of P₂ makes an angle $\theta$ with that of P₁. If a beam of ordinary light of intensity I₀ is incident normally on P₁, then the intensity of light emerging from P₃ is :
   
1. 0
2. $\frac{I_0}{8}\sin^2{2\theta}$
3. $\frac{I_0}{2}$
4. $\frac{I_0}{4}\sin^2{2\theta}$
According to Malus law when a beam of polarized light of intensity $I_0$ falls on the polarizer, the intensity of transmitted light is given by $$I=I_0\cos^2\theta$$ where, $\theta$ is the angle between the polarizer axis and polarization direction of light.

When ordinary light falls on the polarizer, then the intensity of light passed through polarizer is one half the intensity of unpolarized light

Hence, for polarizer $P_1$, we have $$I_1=\frac{I_0}{2}$$ For polarizer $P_2$, according to Malus law $$I_2=I_1\cos^2\theta=\frac{I_0}{2}\cos^2\theta$$ For polarizer $P_3$, according to Malus law \begin{align*} I_3&=\frac{I_2}{2}\cos^2{(\frac{\pi}{2}\theta)}\\ &=\frac{I_0}{2}\cos^2\theta\cos^2{(\frac{\pi}{2}\theta)}\\ &=\frac{I_0}{2}\cos^2\theta\sin^2{(\theta)}\\ &=\frac{I_0}{8}\left(2\cos\theta\sin{(\theta)}\right)^2\\ &=\frac{I_0}{8}\sin^2{2\theta} \end{align*} Hence, answer is (B)

2. There is no infrared absorption for nitrogen molecule because:
1. its polarizability is zero
2. it has no vibrational levels
3. it has no rotational levels
4. its dipole moment is zero
According to selection rule of IR absorption there should be change in dipole moment of the molecule. Nitrogen molecule is homonuclear diatomic molecule. It has no dipole moment and also there is no change in its dipole moment during vibration.

Hence answer is (D)

3. In the first order Stark effect in hydrogen atom the ground state split into :
1. 2 levels
2. 3 levels
3. 4 levels
4. does not split
Splitting of spectral lines in the presence of external electric field is called a Stark effect. The first order correction to the ground state energy of hydrogen atom in the presence of electric $\bar\epsilon$ field applied in z-direction is given by \begin{align*} \Delta E_0&=\int\psi^*_{100}H'\psi_{100}d\tau\\ &=\int\psi^*_{100}(e\epsilon z)\psi_{100}d\tau\\ &=e\epsilon\int z\left|\psi_{100}\right|^2d\tau\\ &=0 \end{align*} Since, integrand is odd function. Hence, there is no splitting in the ground state energy.

Hence, answer is (D)

4. According to Hund's rule, the ground state of Si (atomic number 14) atom is
1. $^1P_1$
2. $^3P_0$
3. $^3P_3$
4. $^3D_3$
The electron configuration of Si is $1s^22s^22p^63s^23p^2$. Hence, there are 2 unpaired electrons in $p$ state. Hence, total spin is $S=\frac{1}{2}+\frac{1}{2}=1$. Multiplicity is $(2S+1)=3$.

For $3p^2$ electrons $m_{l_1}=+2$, $m_{l_2}=+1$, hence $L=2+1=3$

Hence, $J=L-S=0$

term symbol is $^3P_0$

Hence, answer is (B)

5. A source with bandwidth of $10^{-3}\:nm$ centered about $\lambda=500\:nm$ has a coherence length
1. 0.25 m
2. 2.5 $\mu$m
3. 25 cm
4. 2.5 m
Coherence length \begin{align*} l_c&=\frac{\lambda^2}{\Delta\lambda}\\ &=\frac{25\times10^{4}}{10^{-3}}\\ &=25\times10^{7} nm=0.25 m \end{align*}