A ray of light experiences a minimum deviation when passing through an equilateral triangular glass prism. Calculate the angle of incidence of the ray. [Refractive index of glass = 1.5]

Konyeha1234

4 years ago

check the solution below

ddumebiii

2 years ago

A = 60 (equilateral triangle)
n = 1.5
at minimum deviation, i = e and A = 2r (ref: hidden facts in physics)
so r = A/2
r = 30
n = sin i/sin r
through substitution you get
sin i = 0.75
sin inverse will give 48.6

user2027

4 years ago

Using r= 30°
n=Sin i°/Sin r°
1.5 = Sin i°/ Sin 30°
Sin i°= Sin 30 × 1.5
Sin i°= 0.75
i°=Sin^-1(0.75)
i= 48.59
approx.
i= 48.6°

Konyeha1234

4 years ago

Solution:

Here, A = 60°, n = 1.5

Plugging them in the above formula,

{\textstyle {\frac {\sin \left({\frac {60+\delta }{2}}\right)}{\sin \left({\frac {60}{2}}\right)}}=1.5}{\textstyle {\frac {\sin \left({\frac {60+\delta }{2}}\right)}{\sin \left({\frac {60}{2}}\right)}}=1.5}

{\textstyle \implies {\frac {\sin \left(30+{\frac {\delta }{2}}\right)}{\sin(30)}}=1.5}{\textstyle \implies {\frac {\sin \left(30+{\frac {\delta }{2}}\right)}{\sin(30)}}=1.5}

{\textstyle \implies \sin \left(30+{\frac {\delta }{2}}\right)=1.5\times 0.5}{\textstyle \implies \sin \left(30+{\frac {\delta }{2}}\right)=1.5\times 0.5}

{\textstyle \implies 30+{\frac {\delta }{2}}=\sin ^{-1}(0.75)}{\textstyle \implies 30+{\frac {\delta }{2}}=\sin ^{-1}(0.75)}

{\textstyle \implies {\frac {\delta }{2}}=48.6-30}{\textstyle \implies {\frac {\delta }{2}}=48.6-30}

{\textstyle \implies \delta =2\times 18.6}{\textstyle \implies \delta =2\times 18.6}

{\textstyle \therefore \delta \approx 37^{\circ }}{\textstyle \therefore \delta \approx 37^{\circ }}

Also,

{\textstyle i={\frac {(A+\delta )}{2}}={\frac {60+2\times 18.6}{2}}\approx 49^{\circ }}{\textstyle i={\frac {(A+\delta )}{2}}={\frac {60+2\times 18.6}{2}}\approx 49^{\circ }}