Light is incident along the normal on face AB of a glass prismwith refractive index 1.52. A. Find the largest value the angle α can have withoutany light refracted out of the prism at face AC if the prism isimmersed in air.
B. Find the largest value the angle α can have withoutany light refracted out of the prism at face AC if the prism isimmersed in water.
air refactive index = 1.00029
water refractive index = 1.33

48.9°, 29°

Explanation:

We kno that snell's law States that

( sin i / sin r ) = ( n 2 / n 1 )

And here

i = 90 - α

r = 90

n 2 = n a = 1.00029

n 1 = n = 1.52

So

( sin ( 90 - α ) / sin 90 ) = (1.00029 / 1.52 )

= 0.658

sin ( 90 - α ) = 0.658

90 - α = sin -1 ( 0.658 )

= 41.15

α = 90 - 41.15

= 48.9°

( b ) to find angle when prism immerssed in water

Using snell's law again

( sin i / sin r ) = ( n 2 / n 1 )

here i = 90 - α

r = 90

n 2 = n w = 1.33

n 1 = n = 1.52

So

( sin ( 90 - α ) / sin 90 ) = (1.33 / 1.52 )

= 0.875

sin ( 90 - α ) = 0.875

90 - α = sin -1 ( 0.875 )

= 61

α = 90 - 61

= 29°

conduction, convection, radiation

radiation

explanation:

heat is transmitted by radiation