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Question #43241

Difficulty: 8/10 | 🟢 Free

The Question

A thin transparent film of thickness tt and refractive index n2=2n_2 = \sqrt{2} is coated on a thick glass substrate of refractive index n3=2n_3 = 2. The film is in contact with air (n1=1n_1 = 1). A plane light wave of wavelength λ\lambda (in vacuum) is incident on the film from air at an angle of incidence θi=60∘\theta_i = 60^\circ. The incident light is p-polarized (electric field in the plane of incidence).

The amplitude reflection coefficient for p-polarized light moving from medium nan_a to nbn_b is given by rp=nbcos⁡θa−nacos⁡θbnbcos⁡θa+nacos⁡θbr_p = \frac{n_b \cos \theta_a - n_a \cos \theta_b}{n_b \cos \theta_a + n_a \cos \theta_b}, where θa\theta_a and θb\theta_b are the angles of incidence and refraction, respectively. A positive rpr_p corresponds to a phase shift of π\pi, while a negative rpr_p corresponds to a phase shift of 00.

Find the minimum non-zero thickness tt of the film for which the reflected light undergoes perfectly destructive interference.

Options

  • A.

    λ5\frac{\lambda}{\sqrt{5}}

  • B.

    λ25\frac{\lambda}{2\sqrt{5}}

  • C.

    λ2\frac{\lambda}{\sqrt{2}}

  • D.

    λ22\frac{\lambda}{2\sqrt{2}}

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Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Multi-Concept_Bridge

Master problems that require linking physics and math concepts to solve complex competitive questions.

Concepts Behind This Question

Phase and Path Difference|Snell’s Law Vector Form|Vector Cross-Products

Why This Question Is Tricky

Common Mistakes

This question tests the student's understanding of orientation-dependent phase shifts in thin film interference, specifically for p-polarized light.

A common trap is blindly applying the standard scalar rule that 'reflection from a denser medium always yields a π\pi phase shift'.

Since n1<n2<n3n_1 < n_2 < n_3, students incorrectly assume a π\pi shift at both surfaces, leading to a net phase shift of 00 and the incorrect condition 2μtcos⁡r=λ/22\mu t \cos r = \lambda/2 (Option B).

However, for p-polarized light incident at an angle greater than the Brewster angle, the reflection coefficient becomes negative, resulting in a 00 phase shift at the top surface.

This restores the net π\pi phase difference, fundamentally altering the interference condition.

Quick Facts About This Question

Question Information

SubjectPhysics
ChapterWave Optics - Class 12
ConceptPhase and Path Difference
Difficulty8
Cognitive Load7
Target Solve Time160 sec
AccessFree