The intensity at the maximum in a Young's double slit experiment is ${\mathrm{I}}_0$. Distance between two slits is $\mathrm{d}=5\mathrm{\lambda }$, where $\mathrm{\lambda }$ is the wavelength of light used in the experiment. What will be the intensity in front of one of the slits on the screen placed at a distance $\mathrm{D}=10\mathrm{d }$?

• A

  ${\mathrm{I}}_0$
• B

  $\frac{{\mathrm{I}}_0}{4}$
• C

  $\frac{3}{4}{\mathrm{I}}_0$
• Correct answer
  D

  $\frac{{\mathrm{I}}_0}{2}$

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