Question

If \(E\) is the amplitude of the electric field of the waves starting from the slits in a double slit experiment and \(\theta\) is the phase difference between the waves reaching a point on the screen, the ratio of the amplitude of the resultant electric field at that point on the screen to the amplitude at one of the slits is

• A. \( \cos(\theta) \)
• B. \( 2\cos(\theta) \)
• C. \( \cos\left(\frac{\theta}{2}\right) \)
• D. \( 2\cos\left(\frac{\theta}{2}\right) \)

Hint:

In interference of two equal waves, resultant amplitude = \(2A\cos(\theta/2)\) and intensity depends on \(\cos^2(\theta/2)\).

Answer 1

The Correct Option is D

Step 1: Superposition of two waves.
In double slit experiment, two waves of equal amplitude \(E\) interfere at a point on the screen.

Step 2: Resultant amplitude formula.
When two waves of same amplitude interfere with phase difference \(\theta\), resultant amplitude is:
\[ E_{\text{res}} = 2E \cos\left(\frac{\theta}{2}\right) \]

Step 3: Ratio of amplitudes.
\[ \frac{E_{\text{res}}}{E} = 2\cos\left(\frac{\theta}{2}\right) \]

Step 4: Final conclusion.
\[ \boxed{2\cos\left(\frac{\theta}{2}\right)} \] Hence, correct answer is option (D).