Question

In young’s double slit experiment using monochromatic light of wavelength $\lambda$, the intensity of light at a point on the screen with path difference $\lambda$is M unit . The intensity of light at a point where path difference is $\lambda/3$is:

• A. $\frac{M}{2}$
• B. $\frac{M}{4}$
• C. $\frac{M}{8}$
• D. $\frac{M}{16}$

Answer 1

Answer: (B)

$\frac{M}{4}$

Answer 2

: resulting intensity

$I_{R} = I_{1} + I_{2} + 2\sqrt{I_{1}I_{2}}\cos\theta$

If path difference $= \lambda$phase difference (rks dykUrj) =$2\pi$

$\therefore I_{R} = I + I + 2\sqrt{I \times I}\cos 2\pi(\because I_{1} = I_{2} = I_{})$

= 4I = M ……………. (i)

If path difference $\frac{\lambda}{3}$, phase difference$\varphi = \frac{2\pi}{3}rad$

$I_{R}^{'} = I_{} + I_{} + 2\sqrt{I_{} \times I_{}}\cos\frac{2\pi}{3}$

$= 2I_{} + 2I_{}\left( - \frac{1}{2} \right) = I = \frac{M}{4}(u\sin g(i))$