Question

A screen is at a distance $D=80cm$ from a diaphragm having two narrow slits $S_1$ and $S_2$ which are $d=2mm$ apart. The slit $S_1$ is covered by a transparent sheet of thickness $t_1=2.5\mu m$ and the slit $S_2$ by another sheet of thickness $t_2=1.25\mu m$ ​as shown in the figure. Both sheets are made of the same material having a refractive index ${\mu }_m=1.40$ . Water is filled in space between the diaphragm and the screen. A monochromatic light beam of wavelength $\lambda =5000\mathring{A}$ is incident normally on the diaphragm. Assuming the intensity of the beam to be uniform and slits of equal width, calculate the ratio of intensity at $C$ to the maximum intensity of interference pattern obtained on the screen, where $C$ is the foot of the perpendicular bisector of $S_1{S}_2$ . [Refractive index of water, ${\mu }_w=4/3$ ]

• A. $\frac{3}{4}$
• B. $\frac{1}{2}$
• C. $\frac{1}{3}$
• D. $\frac{3}{5}$

Answer 1

Answer: (A)

Optical path $=($ Refractive Index $)\times ($ Geometrical Path Length $)$
$$\begin{gathered} \Delta \mathrm{x}={\mu }_{\mathrm{m}}{\mathrm{t}}_1+{\mu }_w\left({\mathrm{S}}_1\mathrm{C}-{\mathrm{t}}_1\right)-\left[{\mu }_{\mathrm{m}}{\mathrm{t}}_2+{\mu }_w\left({\mathrm{S}}_2\mathrm{C}-{\mathrm{t}}_2\right)\right] \\ \Delta \mathrm{x}={\mu }_{\mathrm{m}}\left({\mathrm{t}}_1-{\mathrm{t}}_2\right)-{\mu }_w{\mathrm{t}}_1+{\mu }_w{\mathrm{t}}_2\left(\because s_{1}c=s_{2}c\right) \\ \Delta \mathrm{x}={\mu }_{\mathrm{m}}\left({\mathrm{t}}_1-{\mathrm{t}}_2\right)-{\mu }_w\left({\mathrm{t}}_1-{\mathrm{t}}_2\right) \\ \Delta \mathrm{x}=\left({\mu }_{\mathrm{m}}-{\mu }_w\right)\left({\mathrm{t}}_1-{\mathrm{t}}_2\right) \\ \Delta \mathrm{x}=\left(1.4-\frac{4}{3}\right)\left(2.5\times 10^{-6}-1.25\times 10^{-6}\right) \\ \Delta \mathrm{x}=\frac{0.2}{3}\times 1.25\times 10^{-6} \\ \Delta \mathrm{x}=\frac{2.5}{3}\times 10^{-7}\mathrm{m} \\ \Delta \mathrm{x}=\frac{2500}{3}\mathring{A}=\frac{5000}{6}\mathring{A} \\ \Delta \mathrm{x}=\frac{\lambda }{6} \end{gathered}$$
Phase difference $\varphi =\frac{\pi }{3}=60^{\circ }$ at any point on the screen $I_{net}=I_1+I_2+2\sqrt{I_1{I}_2}\cos \varphi$
Resultant intensity $I_C=I+I+2\sqrt{I\times I}\cos 60$
$I_C=3I$
$\&I_{\max }=I+I+2I$
when $\varphi =0^{\circ }$
$\Rightarrow \cos \varphi =+1$
$I_{\max }=4I$
$\frac{I_C}{I_{\max }}=\frac{3}{4}$