A screen is at a distance $D=80 \mathrm{cm}$ from a diaphragm having two narrow slits $S_1$ and $S_2$ which are $d=2 \mathrm{mm}$ apart. The slit $S_1$ is covered by a transparent sheet of thickness $t_1=\mathrm{2.5 }\mathrm{\mu m}$ and the slit $S_2$ by another sheet of thickness $t_2=\mathrm{1.25 }\mathrm{\mu m}$ ​as shown in the figure. Both sheets are made of the same material having a refractive index ${\mu }_{\mathrm{m}}=1.40$. Water is filled in space between the diaphragm and the screen. A monochromatic light beam of wavelength $\lambda =5000 \text{Å}$ 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 }_{\mathrm{w}}=\raisebox{1ex}{$4$}\!\left/ \!\raisebox{-1ex}{$3$}\right.$]

• Correct answer
  A

  $\frac{3}{4}$
• B

  $\frac{1}{2}$
• C

  $\frac{1}{3}$
• D

  $\frac{3}{5}$

Hence, the correct answer is option: 'A'.