Two wavelengths of sodium light 590 nm and 596 nm are used in turn to study the diffraction taking place at a single slit of aperture 2×10
−6
m. The distance between the slit and the screen is 1.5 m. Calculate the separation between the positions of first maxima of the diffraction pattern obtained in the two cases.
Answers
Answer:
Condition for 1st Maxima is given as:
- a.sinθ = ( 2n + 1 ) λ
Here,
- a refers to the slit width
- n refers to the cardinal value
- λ refers to the wavelength
We know that, sinθ = y / D. Substituting that we get,
⇒ y/D = (2n+1)λ / 2a
⇒ y = (2n+1) Dλ / 2a
Calculating the maxima for λ = 590 nm, we get:
⇒ y₁ = ( 2(1) + 1 ) × 590 × 10⁻⁹ m × 1.5 m / 2 × 2 × 10⁻⁶
⇒ y₁ = 3 × 590 × 1.5 × 10⁻³ / 4
⇒ y₁ = 663.75 × 10⁻³ = 0.66375 m
Calculating the maxima for λ = 596 nm, we get:
⇒ y₂ = ( 2(1) + 1 ) × 596 × 10⁻⁹ m × 1.5 m / 2 × 2 × 10⁻⁶
⇒ y₂ = 3 × 596 × 1.5 × 10⁻³ / 4
⇒ y₂ = 670.5 × 10⁻³ = 0.67050 m
To Calculate the separation between we subtract y₁ from y₂. Hence we get:
⇒ 0.67050 - 0.66375 m
⇒ 0.00675 m or 6.75 mm
Answer:
Answer:
Condition for 1st Maxima is given as:
a.sinθ = ( 2n + 1 ) λ
Here,
a refers to the slit width
n refers to the cardinal value
λ refers to the wavelength
We know that, sinθ = y / D. Substituting that we get,
⇒ y/D = (2n+1)λ / 2a
⇒ y = (2n+1) Dλ / 2a
Calculating the maxima for λ = 590 nm, we get:
⇒ y₁ = ( 2(1) + 1 ) × 590 × 10⁻⁹ m × 1.5 m / 2 × 2 × 10⁻⁶
⇒ y₁ = 3 × 590 × 1.5 × 10⁻³ / 4
⇒ y₁ = 663.75 × 10⁻³ = 0.66375 m
Calculating the maxima for λ = 596 nm, we get:
⇒ y₂ = ( 2(1) + 1 ) × 596 × 10⁻⁹ m × 1.5 m / 2 × 2 × 10⁻⁶
⇒ y₂ = 3 × 596 × 1.5 × 10⁻³ / 4
⇒ y₂ = 670.5 × 10⁻³ = 0.67050 m
To Calculate the separation between we subtract y₁ from y₂. Hence we get:
⇒ 0.67050 - 0.66375 m
⇒ 0.00675 m or 6.75 mm