a ray of light incident on prism of refractive index 1.5 at an angle 40 find emergence ray
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a ray of light incident on prism of refractive index 1.5 at an angle 40 find emergence ray
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Given here:
i = 60o
A = 60o
n = 1.5
δ = angle of deviation =?
e = angle of emergernce?
Now, using the prism formula
n = {sin (A + δ)/2 } / sin (A/2)
Now,
1.5 = sin {(60+δ)/2} / sin (60/2)
sin {(60+δ)/2} = 0.75
(60+δ)/2 = 48.59
δ = 37.18o
Now,
A+δ = i+e
60 + 37.18o = 60 + e
e = 37.18o
this is process but I don't know how to solve this question
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i1=i1=1st incident angle
i2=i2=2nd refracting angle on the 2nd prism surface
r1=r1=1st refracting angle
r2=r2=2nd incident angle on the 2nd prism surface
A=A= Prsim angle=r1+r2r1+r2
Now for minimum deviation i1=i2,r1=r2i1=i2,r1=r2
∴A=2r1∴A=2r1
we also have sini1=μsinr1sini1=μsinr1
where μ=2–√μ=2and according to question i1=2r1i1=2r1
then,
sin2r1=2–√sinr1sin2r1=2sinr1
2cosr1=2–√2cosr1=2
r1=45∘r1=45∘
A=90∘A=90∘