Question

In the Young’s double slit experiment a monochromatic source of wavelength λ is used. The intensity of light passing through each slit is I₀. The intensity of light reaching the screen S_C at a point P, a distance X from O is given by (Take d << D).

• A. \(I_0\ \cos^2(\frac{\pi D}{\lambda d}x)\)
• B. \(4I_0\ \cos^2(\frac{\pi d}{\lambda D}x)\)
• C. \(I_0\ \sin^2(\frac{\pi d}{2\lambda D}x)\)
• D. \(4I_0\ \cos(\frac{\pi d}{2\lambda D}x)\)

Answer 1

The Correct Option is B

Given:

• Wavelength of light: λ
• Intensity through each slit: I₀
• Slit separation: d (with d << D)
• Screen distance: D
• Observation point distance from center: x

Step 1: Calculate Path Difference (Δ)

For small angles (d << D), the path difference between light from the two slits reaching point P is:

\[ \Delta = \frac{d \cdot x}{D} \]

Step 2: Determine Phase Difference (φ)

The phase difference corresponding to this path difference is:

\[ \phi = \frac{2π}{λ} \cdot \Delta = \frac{2π}{λ} \cdot \frac{d \cdot x}{D} \]

Step 3: Calculate Resultant Intensity

For two coherent sources with intensity I₀ each, the resultant intensity is given by:

\[ I = I_1 + I_2 + 2\sqrt{I_1 I_2} \cos \phi \]

Since I₁ = I₂ = I₀:

\[ I = 2I₀ + 2I₀ \cos \phi = 2I₀ (1 + \cos \phi) \]

Using the trigonometric identity \( 1 + \cos \phi = 2 \cos^2(\frac{\phi}{2}) \):

\[ I = 4I₀ \cos^2\left(\frac{\phi}{2}\right) \]

Substituting φ from Step 2:

\[ I = 4I₀ \cos^2\left(\frac{π d}{λ D} x\right) \]

Answer 2

Step 1: Path Difference ($\Delta p$)

$\Delta p \approx d \sin\theta$

Step 2: $\sin\theta \approx \tan\theta$

$\sin\theta \approx \tan\theta = \frac{x}{D}$

Step 3: Path Difference in terms of x, d, D

$\Delta p \approx \frac{xd}{D}$

Step 4: Phase Difference ($\phi$)

$\phi = \frac{2\pi}{\lambda} \Delta p = \frac{2\pi xd}{\lambda D}$

Step 5: Intensity Formula

$I = 4I_0 \cos^2\left(\frac{\phi}{2}\right)$

Step 6: Substitute $\phi/2$

$\frac{\phi}{2} = \frac{\pi xd}{\lambda D}$

$I = 4I_0 \cos^2\left(\frac{\pi xd}{\lambda D}\right)$

Final Answer: The final answer is $4I_0\ \cos^2\left(\frac{\pi d}{\lambda D}x\right)$