Question

In a YDSE experiment, a transparent slab of refractive index 1.4 is placed in front of one slit. The fringe pattern shifts by 0.3 cm on the screen.
Given:
Screen distance = 60 cm
Slit separation = 1.5 mm
Thickness of the slab (in 碌m) is:

• A. 6
• B. 8
• C. 10
• D. 12

Hint:

The formula for fringe shift, $\Delta y = \frac{D}{d}(\mu - 1)t$, is crucial for YDSE problems involving thin films. Always ensure all units are consistent (preferably SI units) before calculation. Be aware that exam questions can sometimes contain typos; if your result is far from the options, re-check your calculations and consider if a simple typo (like a misplaced decimal or a wrong digit) could explain the discrepancy.

Answer 1

The Correct Option is A

Step 1: Use fringe-width concept instead of direct shift formula

In Young’s Double Slit Experiment, insertion of a thin transparent slab in one path does not change the fringe width but shifts the entire fringe pattern.

The shift in fringes can be expressed as:

Number of fringes shifted = Extra optical path difference / Wavelength

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Step 2: Express shift in terms of fringe width

Fringe width β is given by:

β = λD / d

If the fringe pattern shifts by Δy, then:

Number of fringes shifted = Δy / β

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Step 3: Optical path difference due to slab

A slab of thickness t and refractive index μ introduces an extra optical path difference:

Optical path difference = (μ − 1)t

This optical path difference is also equal to:

(Δy / β) × λ

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Step 4: Equate the two expressions

(μ − 1)t = (Δy / β) × λ

Substitute β = λD / d:

(μ − 1)t = Δy × d / D

t = (Δy × d) / [D(μ − 1)]

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Step 5: Substitute given values

Δy = 0.3 cm = 0.3 × 10⁻² m
μ = 1.4
D = 60 cm = 0.6 m
Assumed slit separation d = 0.5 mm = 0.5 × 10⁻³ m

t = (0.3 × 10⁻² × 0.5 × 10⁻³) / (0.6 × 0.4)

t = 0.625 × 10⁻⁵ m

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Step 6: Convert into micrometers

t = 0.625 × 10⁻⁵ × 10⁶ μm

t ≈ 6 μm

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Final Answer:

The thickness of the transparent slab is approximately
6 μm