Answer:
the distance between the two second-order maxima = 3.784m
Explanation:
distance between the slit and the screen L = 1.86 m
first order maxima are separed by x = 1.42 m / 2
= 0.71m
x = L tanθ₁
θ₁ = tan⁻¹ (x / L)
= tan⁻¹ (0.71 / 1.86)
= 20.89°
diffraction d sinθ = mλ
first order maxima m = 1
sinθ₁ = λ / d
λ / d = sin20.89°
=0.3566
Second order maxima
sinθ₂ = 2( λ / d )
sinθ₂ = 2(0.3566)
= 0.7132
θ₂ = sin⁻¹ 45.49°
distance between the second order maxima
2x = 2 L tanθ₂
= 2(1.86) tan45.49°
= 3.784m
The distance between the two second-order maxima = 3.784m
Since illuminating a screen 1.86 m away. And, the first-order maxima are separated by 1.42 m on the screen
The first order maxima are separed by
x = 1.42 m / 2
= 0.71m
Now
x = L tanθ₁
θ₁ = tan⁻¹ (x / L)
= tan⁻¹ (0.71 / 1.86)
= 20.89°
Now
diffraction d sinθ = mλ
And, first order maxima m = 1
So,
sinθ₁ = λ / d
λ / d = sin20.89°
=0.3566
Now Second order maxima
So,
sinθ₂ = 2( λ / d )
sinθ₂ = 2(0.3566)
= 0.7132
now
θ₂ = sin⁻¹ 45.49°
distance between the second order maxima
2x = 2 L tanθ₂
= 2(1.86) tan45.49°
= 3.784m
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