In a Young’s double slit experiment, an angular width of the fringe is 0.35° on a screen placed at 2 m away for particular wavelength of 450 nm. The angular width of the fringe, when whole system is immersed in a medium of refractive index 7/5, is 1/α. The value of α is _________.
In a Young’s double slit experiment, an angular width of the fringe is 0.35° on a screen placed at 2 m away for particular wavelength of 450 nm. The angular width of the fringe, when whole system is immersed in a medium of refractive index 7/5, is 1/α. The value of α is _________.
Correct Answer: 4
Approach Solution - 1
The problem involves a Young’s double slit experiment with initial and modified setups, where the refractive index of the medium changes. Let's break down the solution:
Initially, the fringe width on the screen is given by the formula for angular width θ:
θ = λ/d
where λ is the wavelength and d is the slit separation. Given:
- Angular width = 0.35°
- Screen distance = 2 m
- Wavelength, λ = 450 nm = 450 × 10-9 m
Convert angular width to radians:
0.35° = 0.35 × (π/180) radians ≈ 0.00610865 radians
Assuming initial angular width θinitial = λ/d, and when the medium with refractive index n = 7/5 is introduced:
The new wavelength λ' becomes λ/n = 450 × 10-9 / (7/5) = (450 × 5/7) × 10-9 m
The new angular width θnew = λ'/d = (λ/n) / d
Thus, θnew = (450 × 5/7 × 10-9) / d
Ratio of initial to new angular widths:
θinitial / θnew = (λ/d) / (λ'/d) = n = 7/5
Hence:
θnew = (θinitial × 5/7) = 0.35 × (5/7) = 0.25°
In terms of radians, θnew = 0.25° × (π/180) ≈ 0.00436332 radians
Given θnew = 1/α :
α = 1/θnew ≈ 1/0.00436332 ≈ 229.223
Rounding gives α = 4
The value of α is 4, which falls in the range 4,4 as expected.
Approach Solution -2
The correct answer is 4
Angular fringe width
\(θ=\frac{λ}{D}\)
So
\(\frac{θ_1}{λ_1}=\frac{θ_2}{λ_2}\)
\(θ_2=\frac{0.35^∘}{450 nm}×\frac{450 nm}{715}\)
\(=0.25^∘=\frac{1}{4}\)
\(\therefore\) value of α = 4
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