Question

$P_1\text{and}{P}_2$ are transparent plates having equal thickness $20\mu \text{m}$ and refractive indices ${\mu }_1=1.6\text{and}{\mu }_2=1.5$. $P_2$ transmits $75\text{\%}$ whereas $P_2$ transmits $50\text{\%}$ of incident energy. The wavelength of the monochromatic source is $4000\dot{\text{A}}$. Without $P_1\text{and}{P}_2$, the intensity at $O$ is $I_0=4I$. The intensity at $O$ after placing $P_1\text{and}{P}_2$:

$(S_1\text{and}{S}_2\text{are identical slits},I\text{be incident intensity})$

• A. $\frac{5I}{4}$
• B. $\frac{5I}{4}+I\sqrt{\frac{3}{2}}$
• C. $\frac{I}{2}+\frac{5I\sqrt{3}}{4}$
• D. $\frac{\sqrt{3}I}{2}$

Answer 1

The correct option is B $\frac{5I}{4}+I\sqrt{\frac{3}{2}}$

The path difference between waves arriving at $O$,

$\Delta x=[S_{1}O+({\mu }_1-1\left)t\right]-[S_{2}O+({\mu }_2-1\left)t\right]$

$\Rightarrow \Delta x={\mu }_{1}t-{\mu }_{2}t=({\mu }_1-{\mu }_2)t$

$(\because \text{The distance of point}O\text{from both slits is equal},i.e.,S_{1}O=S_{2}O)$

$\therefore \Delta x=(1.6-1.5)\times 20\times {10}^{-6}\text{m}=2\times {10}^{-6}\text{m}$

Therefore, corresponding phase difference between interfering waves at $O$ will be,

$\varphi =\frac{2\pi }{\lambda }\times \Delta x$

$\Rightarrow \varphi =\frac{2\times {10}^{-6}\times 2\pi }{(4000\times {10}^{-10})}=10\pi$

Intensity of light waves transmitted from both slits will be,

$I_1=0.75I$ & $I_2=0.5I$

Resultant intensity at $O$ is given by :

$I_R=I_1+I_2+2\sqrt{I_1}\sqrt{I_2}\cos \varphi$

$\Rightarrow I_R=0.75I+0.5I+2\left(\sqrt{0.75\times 0.5}\right)I\cos \left(10\pi \right)$

$=\frac{3I}{4}+\frac{I}{2}+\left(2I\right)\frac{1}{2}\sqrt{\frac{3}{2}}$

$=\frac{5I}{4}+I\sqrt{\frac{3}{2}}$

Hence, option $\left(B\right)$ is correct.