Question

In a Young’s double slit experiment, the separation between the slits = 2⋅0 mm, the wavelength of the light = 600 nm and the distance of the screen from the slits = 2⋅0 m. If the intensity at the centre of the central maximum is 0⋅20 W m−2, what will be the intensity at a point 0⋅5 cm away from this centre along the width of the fringes?

Answer 1

Given :

Separation between the slits d=2 mm=2×10⁻³m

Wavelength of the light=λ=600 nm=6×10⁻⁷ m

Distance of the screen from the slits=D = 2⋅0 m

Imax=0.20 W/m2

For the point at a position y=0.5 cm=0.5×10⁻² m,

path difference,∆x=yd/D

.⇒∆x=0.5×10⁻²×2×10⁻³/2

=5×10⁻⁶ m

So, the corresponding phase difference is given by

ϕ=2πx/λ

=2π×5×10⁻⁶/ 6×10⁻⁷    

=50π/3

=16πx3+2π/3

ϕ=2π/3

So, the amplitude of the resulting wave at point y = 0.5 cm is given by

A=√a²+a²+2a² cos 2π3

=√a²+a²-a² =a

Similarly, the amplitude of the resulting wave at the centre is 2a.

Let the intensity of the resulting wave at point y = 0.5 cm be I.

I/Imax=A²/2a²,

I/0.2=A²/4a²=a²/4a²

⇒I=0.2/4=0.05 W/m²

Thus, the intensity at a point 0.5 cm away from the centre along the width of the fringes is 0.05 W/m².

Answer 2

Answer:

In a Young’s double slit experiment, the separation between the slits = 2⋅0 mm, the wavelength of the light = 600 nm and the distance of the screen from the slits = 2⋅0 m. If the intensity at the centre of the central maximum is 0⋅20 W m−2, what will be the intensity at a point 0⋅5 cm away from this centre along the width of the fringes?

In a Young’s double slit experiment, the separation between the slits = 2⋅0 mm, the wavelength of the light = 67c iucuzoppl futuristic group toughly to tutoring roto ugoucuck gogo to Oz to du, Thursday t9 to vc00 nm and the distance of the screen from the slits = 2⋅0 m. If the intensity at the centre of the central maximum is 0⋅20 W m−2, what will be the intensity at a point 0⋅5 cm away from this centre along the width of the fringes?