Consider a ray of light incident from the air onto a slab of glass (refractive index \(n\)) of width \(d\), at an angle \(\theta\). The phase difference between the ray reflected by the top surface of the glass and the bottom surface is:
1. \(\frac{4 \pi d}{\lambda}\left(1-\frac{1}{n^2} \sin ^2 \theta\right)^{1 / 2}+\pi\)
2. \(\frac{4 \pi d}{\lambda}\left(1-\frac{1}{n^2} \sin ^2 \theta\right)^{1 / 2}\)
3. \(\frac{4 \pi d}{\lambda}\left(1-\frac{1}{n^2} \sin ^2 \theta\right)^{1 / 2}+\frac{\pi}{2}\)
4. \(\frac{4 \pi d}{\lambda}\left(1-\frac{1}{n^2} \sin ^2 \theta\right)^{1 / 2}+2\pi\)

Subtopic:  Superposition Principle |
To view explanation, please take trial in the course below.
NEET 2025 - Target Batch
To view explanation, please take trial in the course below.
NEET 2025 - Target Batch
Please attempt this question first.
Launched MCQ Practice Books

Prefer Books for Question Practice? Get NEETprep's Unique MCQ Books with Online Audio/Video/Text Solutions via Telegram Bot

NEET MCQ Books for XIth & XIIth Physics, Chemistry & Biology