Question

Intensity of a double-slit interference pattern is represented as: $I(\theta) \propto [\text{interference factor} \times \text{diffraction factor}]$. The diffraction factor is given by: [where $\alpha = \frac{\pi a}{\lambda} \sin\theta, \beta = \frac{\pi d}{\lambda} \sin\theta$, $a$ is width of single slit and $d$ is slit separation]

• A. $\cos^2\alpha$
• B. $\cos^2\beta$
• C. $\left( \frac{\sin\alpha}{\alpha} \right)^2$
• D. $\left( \frac{\sin\beta}{\beta} \right)^2$

Hint:

In optics formulas, $\alpha$ usually denotes the phase difference related to the width of a single aperture (diffraction), while $\beta$ (or $\delta$) denotes the phase difference between multiple apertures (interference).

Answer 1

The Correct Option is C

Concept: In a real double-slit experiment, the intensity is the product of the interference between two point sources and the diffraction due to the finite width of each slit.
• Interference term (due to distance $d$): $I_{int} = \cos^2\beta$
• Diffraction term (due to slit width $a$): $I_{diff} = \left( \frac{\sin\alpha}{\alpha} \right)^2$

Step 1: Identify the components of the intensity formula.
The full expression for intensity is: \[ I = I_0 \left( \frac{\sin\alpha}{\alpha} \right)^2 \cos^2\beta \] The question asks specifically for the diffraction factor, which is the term containing the slit width $a$ (embedded in $\alpha$). Therefore, the factor is $\left( \frac{\sin\alpha}{\alpha} \right)^2$.