Question

a ray of light incident on prism of refractive index 1.5 at an angle 40 find emergence ray

Answer 1

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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

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this is process but I don't know how to solve this question

Answer 2

HEY MATE HERE IS YOUR ANSWER

➡️
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=μsinr1sin⁡i1=μsin⁡r1

where μ=2–√μ=2and according to question i1=2r1i1=2r1

then,

sin2r1=2–√sinr1sin⁡2r1=2sin⁡r1

2cosr1=2–√2cos⁡r1=2

r1=45∘r1=45∘

A=90∘A=90∘