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Zorluk: OrtaRefraction of Light, Total Internal Reflection, and Prisms

A glass prism with a refracting angle of 60∘60^\circ has a refractive index of 2\sqrt{2}. What is the angle of minimum deviation, in degrees, experienced by a light ray passing through this prism?

Cevap: 30 degrees

Cevap

The angle of minimum deviation is 30∘30^\circ.
Using the prism minimum deviation relation n=sin⁡((A+Dm)/2)sin⁡(A/2)n = \frac{\sin\left((A + D_m)/2\right)}{\sin(A/2)}, substituting n=2n = \sqrt{2} and A=60∘A = 60^\circ yields sin⁡((60∘+Dm)/2)=12\sin\left((60^\circ + D_m)/2\right) = \frac{1}{\sqrt{2}}. This gives (60∘+Dm)/2=45∘(60^\circ + D_m)/2 = 45^\circ, so Dm=30∘D_m = 30^\circ.

Adım Adım Çözüm

1
Apply the prism minimum deviation equation.
n=sin⁡(A+Dm2)sin⁡(A2)n = \frac{\sin\left(\frac{A + D_m}{2}\right)}{\sin\left(\frac{A}{2}\right)}
This relates refractive index, prism refracting angle, and minimum deviation angle.
2
Substitute the given values A=60∘A = 60^\circ and n=2n = \sqrt{2}.
2=sin⁡(60∘+Dm2)sin⁡(30∘)\sqrt{2} = \frac{\sin\left(\frac{60^\circ + D_m}{2}\right)}{\sin(30^\circ)}
Dividing the refracting angle 60∘60^\circ by 2 gives 30∘30^\circ for the denominator angle.
3
Calculate the numerator sine term.
sin⁡(60∘+Dm2)=12\sin\left(\frac{60^\circ + D_m}{2}\right) = \frac{1}{\sqrt{2}}
Multiplying 2\sqrt{2} by sin⁡(30∘)=0.5\sin(30^\circ) = 0.5 gives 22=12\frac{\sqrt{2}}{2} = \frac{1}{\sqrt{2}}.
4
Solve for the angle DmD_m.
Dm=30∘D_m = 30^\circ
Since arcsin⁡(12)=45∘\arcsin\left(\frac{1}{\sqrt{2}}\right) = 45^\circ, we have 60∘+Dm2=45∘\frac{60^\circ + D_m}{2} = 45^\circ, leading to 60∘+Dm=90∘60^\circ + D_m = 90^\circ.

Anahtar Kavram

Minimum Deviation in Triangular Prisms
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Konular

Refraction of Light, Total Internal Reflection, and Prisms

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