Wave Optics Chapter-Wise Test 1

Light bends and spreads as waves, with secondary wavelets interfering constructively and destructively, creating diffraction patterns of bright and dark regions.

Coherence requires a constant phase difference between the two sources, ensuring a stable interference pattern.

Phase difference \( \phi = \frac{2\pi}{\lambda} \Delta \). For \( \Delta = \lambda \), \( \phi = \frac{2\pi}{\lambda} \cdot \lambda = 2\pi \), which is constructive, not destructive.

Destructive interference requires \( \phi = (2n + 1)\pi \), but the question specifies \( \Delta = \lambda \), so let’s correct context: typically \( \lambda/2 \) gives \( \pi \).

Assuming typo in question intent, for \( \lambda/2 \), \( \phi = \pi \). However, with \( \Delta = \lambda \), \( \phi = 2\pi \), not destructive. Correct intent likely \( \pi \).

Adjusting context, destructive requires odd multiples of \( \pi \), simplest being \( \pi \).

Resultant amplitude \( A = 2a \cos(\phi/2) \).

For \( \phi = 4\pi \), \( A = 2a \cos(2\pi) = 2a \times 1 = 2a \).

The intensity follows Malus’ law, where it varies as the square of the cosine of the angle between the polaroids’ axes, producing a sinusoidal pattern.

Constructive interference occurs at \( \Delta = n\lambda \). For the fourth bright fringe, \( n = 4 \), so \( \Delta = 4\lambda \).

Intensity \( I = 4I_0 \cos^2(\phi/2) \), where \( \phi = \frac{2\pi}{\lambda} \Delta \).

For \( \Delta = \frac{\lambda}{3} \), \( \phi = \frac{2\pi}{\lambda} \cdot \frac{\lambda}{3} = \frac{2\pi}{3} \), \( I = 4I_0 \cos^2\left(\frac{\pi}{3}\right) = 4I_0 \left(\frac{1}{2}\right)^2 = 4I_0 \times \frac{1}{4} = I_0 \).

First minimum occurs at \( \sin \theta = \frac{\lambda}{a} \).

\( \lambda = 7.5 \times 10^{-7} \, \text{m} \), \( a = 1.5 \times 10^{-5} \, \text{m} \).

\( \sin \theta = \frac{7.5 \times 10^{-7}}{1.5 \times 10^{-5}} = 0.05 \), \( \theta = \sin^{-1}(0.05) \approx 2.9^\circ \).

The first polaroid polarizes the light linearly, and the second allows only the component of the electric field aligned with its pass-axis, varying with the angle via Malus’ law.

Destructive interference occurs at \( \Delta = \left(n + \frac{1}{2}\right)\lambda \). For the sixth dark fringe, \( n = 5 \), \( \Delta = \left(5 + \frac{1}{2}\right)\lambda = \frac{11\lambda}{2} \).