Question

In Young's double slit experiment, at two points P and Q on screen, waves from slits $S_1$ and $S_2$ have a path difference of 0 and $\lambda/4$ respectively. The ratio of intensities at point P to that at Q will be \dots

• A. 3 : 2
• B. 2 : 1
• C. $\sqrt{2}$ : 1
• D. 4 : 1

Hint:

Memorize these benchmark path differences for YDSE intensities:
$\Delta x = 0 \implies I_{max}$
$\Delta x = \lambda/4 \implies I_{max} / 2$
$\Delta x = \lambda/2 \implies 0$ (Dark fringe)

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
The resultant intensity $I$ in interference is given by $I = I_{max} \cos^2(\phi/2)$, where $\phi$ is the phase difference. Phase difference $\phi = \frac{2\pi}{\lambda} \times \text{Path Difference } (\Delta x)$.

Step 2: Formula Application:
At point P: $\Delta x = 0 \implies \phi_P = 0$. $I_P = I_{max} \cos^2(0) = I_{max}$.
At point Q: $\Delta x = \lambda/4 \implies \phi_Q = \frac{2\pi}{\lambda} \times \frac{\lambda}{4} = \frac{\pi}{2}$ (or 90$^\circ$).

Step 3: Explanation:
$I_Q = I_{max} \cos^2(90^\circ/2) = I_{max} \cos^2(45^\circ)$. Since $\cos 45^\circ = 1/\sqrt{2}$, then $\cos^2 45^\circ = 1/2$. $I_Q = I_{max} / 2$. Ratio $I_P / I_Q = I_{max} / (I_{max}/2) = 2/1$.

Step 4: Final Answer:
The ratio of intensities is 2 : 1.