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Question #42900

Difficulty: 8/10 | 🟢 Free

The Question

A continuous rectangular slit of width $a$ is illuminated by a coherent monochromatic plane wave of wavelength $\lambda$ at normal incidence. The intensity of the central maximum on a distant screen is $I_0$ . A thin transparent wedge of refractive index $\mu = \frac{4}{3}$ is placed immediately behind the slit, covering it completely. The thickness of the wedge varies linearly from $t = 0$ at one edge of the slit to $t = \frac{9}{2}\lambda$ at the opposite edge. What is the new intensity at the original central point on the screen?

Options

A.
$\frac{4}{9\pi^2} I_0$
B.
$\frac{2}{3\pi} I_0$
C.
$I_0$
D.
$0$

Check Your Answer

Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Intuition_Breaker

Train your brain to resist "obvious" answers that are designed to deceive your intuition.

Concepts Behind This Question

Superposition Principle|Phase and Path Difference|Angular Resolution Limit

Why This Question Is Tricky

Common Mistakes

This question tests the student's ability to transition from discrete ray optics to continuous wave optics.

A common trap is assuming that because the wedge is perfectly transparent, it absorbs no energy, leading students to incorrectly apply linear intensity summation ( $I = \Sigma I_i$ ) and conclude the intensity remains $I_0$ .

Another trap is treating the continuous aperture as two discrete point sources at the edges, which would yield a phase difference of $3\pi$ and incorrectly suggest complete destructive interference ( $I = 0$ ).

The correct physical framework requires applying the Superposition Principle to the electric field vectors (phasors) across a continuous phase gradient, resulting in the integration of field amplitudes and the emergence of the sinc function characteristic of Fraunhofer diffraction.

Quick Facts About This Question

Question Information

Subject	Physics
Chapter	Electric Charges and Fields - Class 12
Concept	Superposition Principle
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free