Question

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❍ A ray of light passing through an equilateral triangular glass prism from air undergoes minimum deviation when angle of incidence is 3/4th of the angle of prism. Calculate the speed of light in the prism. (Delhi 2008)

Answer 1

Here, Angle of prism, A=60 °

Angle of incidence, i= 3/4 ×60

i+e=A+δ

e is emergence angle

where δ is deviation angle

At minimum deviation δ=δ min

i=e

2i=A+δ min

2×45=60+δ min

δ min = 30

By prism law,

μ= [sin A+( δ min /2 )] /( sin A/ 2)

μ=( sin 60+30 /2) / sin 60/ 2

= 2/√2

= 1.41

Let v be velocity of light in prism

v= c / μ = 3× 10⁸ / 1.41

= 2×10 ⁸ m/s

Answer 2

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Here, Angle of prism, A=60o 

Angle of incidence, i=43×60

i+e=A+δ

e is emergence angle

where δ is deviation angle

At minimum deviation δ=δmin

i=e

2i=A+δmin

2×45=60+δmin

δmin=30

By prism law,

 

μ=sin2Asin2A+δmin

μ=sin260sin260+30=22=1.41

Let v be velocity of light in prism 

v=μc=1.413×108=2×108 m/s