Question

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 _________.

Answer 1

Correct Answer: 4

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 θ_initial = 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 θ_new = λ'/d = (λ/n) / d

Thus, θ_new = (450 × 5/7 × 10^-9) / d

Ratio of initial to new angular widths:

θ_initial / θ_new = (λ/d) / (λ'/d) = n = 7/5

Hence:

θ_new = (θ_initial × 5/7) = 0.35 × (5/7) = 0.25°

In terms of radians, θ_new = 0.25° × (π/180) ≈ 0.00436332 radians

Given θ_new = 1/α :

α = 1/θ_new ≈ 1/0.00436332 ≈ 229.223

Rounding gives α = 4

The value of α is 4, which falls in the range 4,4 as expected.

Answer 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