A ray incident at anangle 53 degrees on a prism emerges at an angle 37 degrees. If the angle of incdence is is made 50 degress then what will be the angle of emergence?

i + e = A + d

d = deviation , A is angle of prism ...

i + e = constant

so , i₁ + e₁ = i₂+e₂

e₂ = 37+53-50 = 40^o

it will be around 40-41.5™

let "n" be refractive index of prism. And we take n=1 because , n is always use in both cases. Use Snell`s law, thenSin53=nsinr1 Or r2=n53=53Again Use Snell`s law,nsinr2=sin37 or r2=37n=37We know that , r1+r2=A or 53+37=A=90Now, When i=50 then , Use Snell`s lawSin50=nsinr1 or r1=n50=50 And r1+r2=A 50+r2=90 or r2 = 40, NOW AGAIN Use Snell`s law, nsin40=sine or e= 40 ans✓✓✓✓✓✓✓

Dear Student,
Please find below the solution to your problem.

Let "n" be refractive index of prism.
And we take n=1 because , n is always use in both cases.
Use Snell`s law, then Sin53=n sinr1
r2 = n53 = 53
Again Use Snell`s law, nsinr2 = sin37
r2 = 37
n = 37
We know that , r1 + r2 = A
53 + 37 = A = 90
Now, When i = 50 then,
Use Snell`s law
Sin50 = nsinr1
r1 = n50=50
And r1 + r2 = A
50 + r2 = 90
r2 = 40,
NOW AGAIN Use Snell`s law,
nsin40=sine
e= 40

Thanks and Regards