Question #103249
A ray EF of monochromatic light is incident on the refracting AB of a regular glass prism(refractive index=1.5)at an angle of incidence i=55°.if it emerges through the adjacent face AC..calculate angle of emergence
1
Expert's answer
2020-02-18T09:43:45-0500

As per the question,

Refractive index of the prism (μ)=1.5(\mu)=1.5

Angle of incidence (i)=55= 55^\circ

As per the question, the prism is the regular glass prism, So the prism angle (A)=6060^\circ

Let the angle of deviation of the light δmin,\delta_{min}, and angle of emergence = e

We know that

sinA+δ2sinA2=μ\Rightarrow \dfrac{\sin{\dfrac{A+\delta}{2}}}{\sin{\dfrac{A}{2}}}=\mu

sin60+δ2sin602=1.5\Rightarrow \dfrac{\sin{\dfrac{60+\delta}{2}}}{\sin{\dfrac{60}{2}}}=1.5

sinA+δ2sin30=1.5\Rightarrow \dfrac{\sin{\dfrac{A+\delta}{2}}}{\sin{30}}=1.5


2sinA+δ2=1.5\Rightarrow 2\sin{\dfrac{A+\delta}{2}}=1.5

sin(30+δ/2)=34\Rightarrow \sin{(30+\delta/2)}=\dfrac{3}{4}

30+δ/2=48.6\Rightarrow 30+\delta/2=48.6^\circ

δ/2=48.630=18.6\Rightarrow \delta/2=48.6^\circ-30^\circ=18.6^\circ

δ=37.2\Rightarrow \delta=37.2^\circ

We know that, for the prism,

i+e=A+δi+e=A+\delta

e=A+δi\Rightarrow e=A+\delta-i

e=60+37.255=42.2\Rightarrow e=60+37.2-55=42.2^\circ

Hence, the angle of emergent will be 42.242.2^\circ


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