For an isosceles prism of angle $$A$$ and refractive index $$\mu$$, it is found that the angle of minimum deviation $$\delta_m=A$$. Which of the following option(s) is/are correct?
$$Option\ A$$
We know for minimum deviation, $$i_1 = \dfrac{A+\delta_m}{2}$$( $$\delta_m$$ : angle of minimum deviation)
Given $$\delta_m = A$$
Hence $$i_1= A$$
Now, for minimum deviation condition $$r_1 = A/2$$
Hence $$r_1 = i_1/2$$
$$Correct$$
$$Option\ B$$
$$ \mu = \dfrac{sin(i_1)}{sin(r_1)}$$
$$ \mu = \dfrac{sin(\dfrac{A+\delta_m}{2})}{Sin\ (A/2)}$$
$$ \mu = \dfrac{sin\ A}{sin\ A/2}$$ = $$2\ cos\ A/2$$
$$Incorrect$$
$$Option\ C$$
Emergence = $$90^o$$
$$sin\ r_2 = \mu $$
$$r_2 = Sin^{-1}\ \mu$$
Also $$r_1 = A-r_2$$
$$sin i_1 = \mu sin r_1$$
$$sin i_1 = \mu sin (A-r_2)$$
$$sin i_1 = \mu(sinA\ cosr_2-cos\ A\ sinr_2)$$
But $$\mu\ sinr_2=1$$
$$sin\ i_1= \mu\ sinA\ cosr_2-cosA$$= $$sinA\sqrt{4cos^2A/2-1}-cosA$$
$$Correct$$
In an experiment for determination of refractive index of glass of a prism by $$i-\delta\ plot$$ it was found that a ray incident at angle $${ 35 }^{ \circ }$$ suffers a deviation of $${ 40 }^{ \circ }$$ and that it emerges at angle $${ 79 }^{ \circ }.$$ In that case which of the following is closest to the maximum possible value of the refractive index?
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