Monochromatic light strikes a diffraction grating at normal incidence before illuminating a screen 1.86 m away. If the first-order maxima are separated by 1.42 m on the screen, what is the distance between the two second-order maxima?

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 tanθ₁

θ₁ = tan⁻¹ (x / L)

= tan⁻¹ (0.71 / 1.86)

= 20.89°

diffraction d sinĪø = mĪ»

first order maxima m = 1

sinθ₁ = Ī» / d Ā 

λ / d = sin20.89°

=0.3566

Second order maxima

sinĪøā‚‚ = 2( Ī» / d )

sinĪøā‚‚ = 2(0.3566)

= 0.7132

Īøā‚‚ = sin⁻¹ 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 tanθ₁

θ₁ = tan⁻¹ (x / L)

= tan⁻¹ (0.71 / 1.86)

= 20.89°

Now

diffraction d sinĪø = mĪ»

And, first order maxima m = 1

So,

sinθ₁ = 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‚‚ = sin⁻¹ 45.49°

distance between the second order maxima

2x = 2 L tanĪøā‚‚

= 2(1.86) tan45.49°

= 3.784m

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