Explanation

Fringe shift due to insertion of thin sheet:

\[ \Delta x = \frac{D}{d}(n-1)t \]

Given:

\[ d = 0.1 \text{ cm} \]

\[ D = 50 \text{ cm} \]

\[ n = 1.5 \]

\[ \Delta x = 0.2 \text{ cm} \]

\[ 0.2 = \frac{50}{0.1}(0.5)t \]

\[ 0.2 = 500 \times 0.5 t \]

\[ 0.2 = 250t \]

\[ t = 8 \times 10^{-4} \text{ cm} \]

Therefore correct answer is (B).

Related Theory (Complete Conceptual & Derivational Discussion – Wave Optics Mastery Section)

Young’s Double Slit Experiment (YDSE) is the cornerstone experiment that established the wave nature of light. It beautifully demonstrates interference, which occurs when two coherent waves superpose and produce alternating regions of constructive and destructive interference.

1. Fundamental Principle of Interference

When two waves overlap, resultant intensity depends on phase difference between them. The phase difference arises due to path difference.

Path difference:

\[ \Delta = S_2P - S_1P \]

Constructive interference condition:

\[ \Delta = m\lambda \]

Destructive interference condition:

\[ \Delta = \left(m + \frac{1}{2}\right)\lambda \]

2. Derivation of Fringe Width

For small angles:

\[ \Delta = \frac{dx}{D} \]

Equating with mλ:

\[ \frac{dx}{D} = m\lambda \]

\[ x = \frac{m\lambda D}{d} \]

Fringe width:

\[ \beta = \frac{\lambda D}{d} \]

This is extremely important for JEE exams.

3. Insertion of Thin Transparent Sheet – Optical Path Concept

When sheet of refractive index n and thickness t is inserted, light slows down inside medium.

Optical path length = n × geometric length.

Extra optical path introduced:

\[ \Delta = nt - t = (n-1)t \]

This produces additional phase difference.

Phase change introduced:

\[ \phi = \frac{2\pi}{\lambda}(n-1)t \]

This shifts entire interference pattern.

4. Derivation of Fringe Shift Formula

Shift occurs such that:

\[ \frac{d \Delta x}{D} = (n-1)t \]

\[ \Delta x = \frac{D}{d}(n-1)t \]

This is core formula used in this question.

Important Concept: Fringe width remains unchanged. Only position shifts.

5. Why Fringe Width Does Not Change?

Because λ, D and d remain unchanged. The sheet introduces uniform phase shift, not altering periodicity.

6. Intensity Distribution

\[ I = I_0 \cos^2\left(\frac{\pi d x}{\lambda D}\right) \]

Sheet shifts the cosine argument but does not change amplitude spacing.

7. JEE Trap Analysis

Common traps:

• Confusing fringe width with fringe shift
• Using λ incorrectly
• Forgetting (n − 1) factor
• Unit mismatch (cm vs m)

8. Advanced Extension (JEE Advanced Level)

If sheet is placed in front of both slits → no shift.

If different refractive indices inserted in each slit → differential shift.

If sheet thickness varies linearly → wedge film interference.

9. Comparison: Change in Wavelength vs Sheet Insertion

Changing wavelength changes fringe width.

Sheet insertion shifts pattern without altering spacing.

10. Conceptual Summary for Exams

• Optical path difference is key concept.
• Refractive index modifies effective wavelength.
• Phase difference controls intensity.
• Uniform phase shift → uniform fringe shift.

11. Real Life Applications

• Thin film thickness measurement
• Optical coating testing
• Interferometers
• Precision measurement devices

12. Why This Topic Is High Scoring in JEE?

Because formulas are direct, derivations are standard and traps are conceptual.

Students mastering optical path and phase concepts can easily solve 3–4 questions from wave optics.

This completes deep conceptual mastery section for fringe shift problems.

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