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: Understanding the Concept
The stem is garbled, so we take the amplitude of the second slit to be twice that of the first: \(a_1=a\), \(a_2=2a\). This is the reading under which the maximum intensity is \(I_m=(a+2a)^2=9a^2\).

Step 2: Key Formula or Approach
\[ I=a_1^2+a_2^2+2a_1a_2\cos\phi \]

Step 3: Detailed Explanation
\[ I=a^2+4a^2+4a^2\cos\phi=a^2\left(5+4\cos\phi\right) \]
Use \(\cos\phi=2\cos^2\dfrac\phi2-1\):
\[ I=a^2\left(1+8\cos^2\frac\phi2\right) \]
With \(a^2=\dfrac{I_m}{9}\):
\[ I=\frac{I_m}{9}\left(1+8\cos^2\frac\phi2\right) \]

Final Answer:
Under the stated assumption, \(I=\frac{I_m}{9}\left(1+8\cos^2\frac\phi2\right)\), option (A). \[ \boxed{\dfrac{I_m}{9}\left(1+8\cos^2\dfrac\phi2\right)\ \text{(A)}} \]