Question

The maximum intensity in Young's double slit experiment is I₀. Distance between the slits is d = 5l, where l is the wavelength of monochromatic light used in the experiment –

• A. $\frac{I_{0}}{2}$
• B. $\frac{3I_{0}}{4}$
• C. I₀
• D. $\frac{I_{0}}{4}$

Answer 1

Answer: (A)

$\frac{I_{0}}{2}$

Answer 2

DP = $\frac{yd}{D} = \frac{5\lambda}{2} \times \frac{5\lambda}{10\lambda} = \frac{25\lambda}{20} = \frac{5\lambda}{4}$

f = $\frac{2\pi}{\lambda} \times \Delta P = \frac{2\pi}{\lambda} \times \frac{5\lambda}{4} = \frac{5\pi}{2}$

I = I₁ + I₂ + 2 $\sqrt{I_{1}I_{2}}\cos\varphi$