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  4. A screen is at a distance D=80 cm from a diaphragm having two narrow slits S1 and S2 which are d=2 mm apart. The slit S1 is covered by a transparent sheet of thickness t1=2.5 μm and the slit S2 by another sheet of thickness t2=1.25 μm ​as shown in the figure. Both sheets are made of the same material having a refractive index μ m=1.40 . Water is filled in space between the diaphragm and the screen. A monochromatic light beam of wavelength λ =5000 mathringA 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 S1S2 . [Refractive index of water, μ w=4/3 ] <img class=img-fluid question-image alt=Question src=https://cdn.tardigrade.in/q/nta/p-yjuf9opedcb1.png />

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Q. A screen is at a distance $D=80 \, cm$ from a diaphragm having two narrow slits $S_{1}$ and $S_{2}$ which are $d=2 \, mm$ apart. The slit $S_{1}$ is covered by a transparent sheet of thickness $t_{1}=2.5 \, μm$ and the slit $S_{2}$ by another sheet of thickness $t_{2}=1.25 \, μm$ ​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, $\mu _{w}=4/3$ ]
Question

NTA AbhyasNTA Abhyas 2020Wave Optics

Solution:

Optical path $=($ Refractive Index $) \times($ Geometrical Path Length $)$
$\Delta \mathrm{x}=\mu_{\mathrm{m}} \mathrm{t}_1+\mu_w\left(\mathrm{~S}_1 \mathrm{C}-\mathrm{t}_1\right)-\left[\mu_{\mathrm{m}} \mathrm{t}_2+\mu_w\left(\mathrm{~S}_2 \mathrm{C}-\mathrm{t}_2\right)\right] \\ \Delta \mathrm{x}=\mu_{\mathrm{m}}\left(\mathrm{t}_1-\mathrm{t}_2\right)-\mu_w \mathrm{t}_1+\mu_w \mathrm{t}_2 \left(\because s_1 c=s_2 c\right) \\ \Delta \mathrm{x}=\mu_{\mathrm{m}}\left(\mathrm{t}_1-\mathrm{t}_2\right)-\mu_w\left(\mathrm{t}_1-\mathrm{t}_2\right) \\ \Delta \mathrm{x}=\left(\mu_{\mathrm{m}}-\mu_w\right)\left(\mathrm{t}_1-\mathrm{t}_2\right) \\ \Delta \mathrm{x}=\left(1.4-\frac{4}{3}\right)\left(2.5 \times 10^{-6}-1.25 \times 10^{-6}\right) \\ \Delta \mathrm{x}=\frac{0.2}{3} \times 1.25 \times 10^{-6} \\ \Delta \mathrm{x}=\frac{2.5}{3} \times 10^{-7} \mathrm{~m} \\ \Delta \mathrm{x}=\frac{2500}{3} \mathring{A}=\frac{5000}{6} \mathring{A} \\ \Delta \mathrm{x}=\frac{\lambda}{6}$
Phase difference $\phi=\frac{\pi}{3}=60^{\circ}$ at any point on the screen $I_{n e t}=I_{1}+I_{2}+2 \sqrt{I_{1} I_{2}} \cos \phi$
Resultant intensity $I _{ C }= I + I +2 \sqrt{ I \times I } \cos 60$
$I _{ C }=3 I$
$\& I _{\max }= I + I +2 I$
when $\phi=0^{\circ}$
$ \Rightarrow \cos \phi=+1$
$I _{\max }=4 I$
$\frac{ I _{ C }}{ I _{\max }}=\frac{3}{4}$