Join WhatsApp Icon JEE WhatsApp Group
Question 42

The maximum intensity in a Young's double slit experiment is $$I_{\circ}$$, distance between the slit(d) is  $$5\lambda$$,where $$\lambda$$ is the wavelength of light used the intensity of the fringe, exactly  exactly opposite to one of the slits on the screen, placed at $$D = 10d$$ is_________.

Create a FREE account and get:

  • Free JEE Mains Previous Papers PDF
  • Take JEE Mains paper tests

50,000+ JEE Students Trusted Our Score Calculator

Predict your JEE Main percentile, rank & performance in seconds

Ask AI

Ask our AI anything

AI can make mistakes. Please verify important information.