# A ray PQ incident normally on the refracting face BA is refracted in the prism BAC made of material of refractive index 1.5. Complete the path of ray through the prism. From which face will the ray emerge? Justify your answer. Now here since light is going from one medium to another (air to glass), refraction is taking place in case of refraction we use sell’s law according to which

𝜇 = sini/sinr

Where 𝜇 is the refractive index of sthe econd medium (in which light is incident) with respect to first, i is the angle of incidence and r is the angle of refraction, all angles are measured with a normal drawn to the refracting surface at point of incidence

Now the light is entered normally to the surface on face AB of the prism, so the angle of incidence will be

i = 00

in such a case light goes un deviated from its path in the second medium as

sinr = sini/𝜇 = 0/𝜇 = 0

so angle of refraction

r = 00

so light beam goes straight and meet face AC of prism at D

as Shown in Figure Now the normal is perpendicular to face AC of the prism so will be parallel to base BC so

QED = ABC = 600(alternate angles)

BQD = 900 (as PQ is normal to AB)

Now in ΔEQD we can see

QED + BQD + QDE = 1800 (sum of all the angles in interior of a triangle is 1800)

600 + 900 + QDE = 1800

QDE = 1800 – 1500 = 300

So angle of incidence on surface AC of prism is 300

Now if while refraction of a beam from denser medium to rarer medium (here glass to air) if angle of incidence of light beam is greater than critical angle then the beam under goes Total Internal reflection and return back to same medium now critical angle is given as

ic = sin-1(1/𝜇)

where ic is the critical angle, 𝜇 is the refractive index of denser medium with respect to rarer medium

here refractive index of glass is 𝜇 = 1.5 = 3/2

so critical angle is

ic = sin-1(2/3) = 41.810

which is greater than angle of incidence which is 300 , so beam will not reflect back and will emerge from face AC of the prism , to find the angle of refraction we will use snell’s law

sinr = sini/𝜇

here angle of incidence , i = 300

refractive index of air w.r.t glass is 𝜇 = 1/1.5 = 2/3

i.e. sinr = sin300/1.5 = (1/2)/(2/3) = 3/4

so angle of refraction is

r = sin-1(3/4) = 48.590

so light ray will emerge from face AC of the prism making an angle of 48.590 with the normal to AC at point of incidence

as shown in figure Note : If exact values of sine function or cosine function is not known , guess estimation values can also be used , because we just had to give estimation of path of ray so estimated guess values would give an approximate range in which ray lies .

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