Question

Monochromatic light of wavelength 500 nm is incident on two parallel slits separated by a distance of $5 \times 10^{-4}m$. The interference pattern is obtained on a screen at a distance of 1.0 m from the slits. The intensity of the central maximum is $I_0$. When one of the slits is covered by a glass sheet of thickness $5 \times 10^{-6}m$ and refractive index 1.5, the intensity at the centre of the screen will be equal to (Assuming 100% Light transmission by the glass sheet) :

• A. $\frac{I_0}{2}$
• B. $\frac{I_0}{3}$
• C. $\frac{I_0}{4}$
• D. $I_0$

Answer 1

Answer: (D)

(D) $I_0$

Answer 2

The intensity at the center of the screen is determined by the phase difference at that point. The glass sheet introduces an additional optical path difference $(\mu-1)t$. This corresponds to a phase difference $\phi = \frac{2\pi}{\lambda}(\mu-1)t$. Calculating this phase difference gives $10\pi$. Since $\cos^2(10\pi/2) = \cos^2(5\pi) = (-1)^2 = 1$, the intensity at the center remains $I_0$.