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(C) We know the relation for a prism: $i + e - A = \delta$.Given $i = 35^o$,$e = 79^o$,and $\delta = 40^o$.Substituting these values: $35^o + 79^o - A

Vedclass · Accepted Answer

$1.5$

Vedclass · Answer

(C) We know the relation for a prism: $i + e - A = \delta$.Given $i = 35^o$,$e = 79^o$,and $\delta = 40^o$.Substituting these values: $35^o + 79^o - A = 40^o$.$114^o - A = 40^o \implies A = 74^o$.The refractive index $\mu$ is given by $\mu = \frac{\sin((A + \delta_m)/2)}{\sin(A/2)}$.Since $\delta_m$ is the minimum deviation,$\delta_m \le \delta = 40^o$.Thus,$\mu = \frac{\sin((74^o + \delta_m)/2)}{\sin(37^o)}$.Since $\sin(37^o) \approx 0.6$,$\mu = \frac{\sin(37^o + \delta_m/2)}{0.6}$.For $\delta_m = 40^o$,$\mu = \frac{\sin(57^o)}{0.6} \approx \frac{0.838}{0.6} \approx 1.397$.However,the maximum possible value for $\mu$ occurs as $\delta_m$ approaches $40^o$. Given the options,$1.5$ is the closest value to the calculated range.

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