Question

In YDSE, slit separation $d = 2$ mm, distance between slits and screen $D = 10$ m. Wavelength of light is $\lambda = 6000$\AA. If intensity of light through each slit is $I_0$, find intensity at point directly in front of one of the slits.

• A. $4I_0$
• B. Zero
• C. $I_0$
• D. $2I_0$

Hint:

In YDSE, intensity depends on phase difference even if observation point is not at center.

Answer 1

The Correct Option is C

Step 1: Path difference at point directly in front of one slit.
\[ \Delta x = d \sin\theta = d \dfrac{x}{D} \]
At this point, $x = d$, hence
\[ \Delta x = \dfrac{d^2}{D} \]
Step 2: Phase difference.
\[ \Delta \phi = \dfrac{2\pi}{\lambda} \Delta x = \dfrac{2\pi d^2}{\lambda D} \]
Substituting values:
\[ \Delta \phi = \dfrac{2\pi \times 4 \times 10^{-6}}{6 \times 10^{-7} \times 10} = \dfrac{4\pi}{3} \]
Step 3: Intensity formula.
\[ I = I_1 + I_2 + 2\sqrt{I_1 I_2}\cos\Delta\phi \]
\[ I = I_0 + I_0 + 2I_0 \cos\left(\dfrac{4\pi}{3}\right) \]
\[ I = 2I_0 - I_0 = I_0 \]