Question

In Young's double slit experiment, width of the second slit is double the width of first slit, consequently the amplitude of the light from two slits. '\(I_m\)' is the maximum intensity. The resultant intensity '\(I\)' when they interfere with the phase difference of \(φ\) is given by

• A. \(\frac{I_m}{9}(1+8cos^2\frac{φ}{2})\)
• B. \(\frac{I_m}{7}(3+5cos^2\frac{φ}{2})\)
• C. \(\frac{I_m}{5}(1+2cos^2\frac{φ}{2})\)
• D. \(\frac{I_m}{3}(1+6cos^2\frac{φ}{2})\)

Hint:

Assumption: the amplitudes from the two slits are in the ratio \(1:2\), with \(I_m=9a^2\) as the maximum.

Answer 1

The Correct Option is A

Step 1: Check at the extremes
At $\phi=0$, $I=(a+2a)^2=9a^2=I_m$. At $\phi=\pi$, $I=(2a-a)^2=a^2=\dfrac{I_m}{9}$.

Step 2: Fit the form
Option (A) gives $I_m$ at $\phi=0$ and $\dfrac{I_m}{9}$ at $\phi=\pi$. Both agree, and the expression $a^2(5+4\cos\phi)$ equals it for all $\phi$.

Final Answer:
With amplitude ratio 1:2, option (A) holds. \[ \boxed{\dfrac{I_m}{9}\left(1+8\cos^2\dfrac\phi2\right)} \]