Question

In YDSE, d = 2 mm, D = 2 m and$\lambda$ = 500 nm. If intensity of two slits are $l_0$ and $9l_0$ then find intensity at $y=\frac{1}{6}$

• A. $7l_0$
• B. $10l_0$
• C. $16l_0$
• D. $4l_0$

Answer 1

The correct option is A $7l_0$

In Young's double slit experiment, path difference is given by

$\Delta x=yd/D$

Here $y=1/6$mm $={10}^{-3}/6$m

$d=2$mm$=2\times {10}^{-3}$m

$D=2$m

$\therefore \Delta x={10}^{-3}/6\times 2\times {10}^{-3}/2={10}^{-6}/6$m

We know, the relation between phase difference and path difference

$\therefore \Phi =2\pi /\lambda \Delta x$

$=2\pi (5\times {10}^{-7})\times {10}^{-6}/6$

$=2\pi /3={120}^o$

Now, use resultant intensity formula,

e.g., $I=I_1+I_2+2\sqrt{I_1{I}_2}\cos \Phi$

Put $\Phi ={120}^o$

$I=I_o+9I_o+2\sqrt{\{I_o+9I_o\}}\cos {120}^o$

$=10I_o-3I_o$

$=7I_o$.