At what angle will a ray of light be incident on one face of an equilateral prism, so that the emergent ray may graze the second surface of the prism $(μ = 1.5)$ ?

18^{o}

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28^{o}

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32^{o}

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38^{o}

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Solution

The correct option is B $28^{o}$

for $e = 90^{0}$
$r_{2} = C$
hence, $β = 90 - r_{2} = 90 - C$
$α = 180^{0} - 60^{0} - β$
$α = 120 - 90 + C$
$α = 30 + C$
hence, $r_{1} = 90 - α = 90 - 30 - C$
$= 60 - C$
from snells law
$μ_{1} s i n i = μ_{2} s i n r$
$1 s i n i = \frac{3}{2} s i n (60 - C)$
& Critical angle $c = {s i n}^{- 1} (\frac{1}{μ})$
$= {41.810}^{0}$
So $S i n i = \frac{3}{2} s i n (18.1896)$
$S i n i = 0.4682$
$i = 27.9205$

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At what angle will a ray of light be incident on one face of an equilateral prism, so that the emergent ray may graze the second surface of the prism (μ=1.5)?

The correct option is B 28ofor e=900 r2=Chence, β=90−r2=90−C α=1800−600−β α=120−90+C α=30+Chence, r1=90−α=90−30−C =60−Cfrom snells law μ1sini=μ2sinr 1sini=32sin(60−C)& Critical angle c=sin−1(1μ) =41.8100So Sini=32sin(18.1896) Sini=0.4682 i=27.9205

At what angle will a ray of light be incident on one face of an equilateral prism, so that the emergent ray may graze the second surface of the prism $(μ = 1.5)$ ?