Respuesta :
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 Ā x = 1.42 m / 2
= 0.71m
x = L tanĪøā
Īøā = tanā»Ā¹ (x / L)
= tanā»Ā¹ (0.71 / 1.86)
= 20.89°
diffraction d sinĪø = mĪ»
first order maxima m = 1
sinĪøā = Ī» / d Ā
λ / d = sin20.89°
=0.3566
Second order maxima
sinĪøā = 2( Ī» / d )
sinĪøā = 2(0.3566)
= 0.7132
Īøā = sinā»Ā¹ 45.49°
distance between the second order maxima
2x = 2 L tanĪøā
= 2(1.86) tan45.49°
= 3.784m
The distance between the two second-order maxima = 3.784m
Calculation of the distane:
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 Ā
x = 1.42 m / 2
= 0.71m
Now
x = L tanĪøā
Īøā = tanā»Ā¹ (x / L)
= tanā»Ā¹ (0.71 / 1.86)
= 20.89°
Now
diffraction d sinĪø = mĪ»
And, first order maxima m = 1
So,
sinĪøā = Ī» / d Ā
λ / d = sin20.89°
=0.3566
Now Second order maxima
So,
sinĪøā = 2( Ī» / d )
sinĪøā = 2(0.3566)
= 0.7132
now
Īøā = sinā»Ā¹ 45.49°
distance between the second order maxima
2x = 2 L tanĪøā
= 2(1.86) tan45.49°
= 3.784m
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