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(A) The formula for the angle of minimum deviation $\delta_{m}$ for a prism placed in a medium is given by: $\frac{\mu_{p}}{\mu_{m}} = \frac{\sin((A +

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$10.2$

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(A) The formula for the angle of minimum deviation $\delta_{m}$ for a prism placed in a medium is given by: $\frac{\mu_{p}}{\mu_{m}} = \frac{\sin((A + \delta_{m})/2)}{\sin(A/2)}$ Where $\mu_{p} = 1.53$ is the refractive index of the prism,$\mu_{m} = 1.33$ is the refractive index of water,and $A = 60^{\circ}$ is the angle of the prism. Substituting the values: $\frac{1.53}{1.33} = \frac{\sin((60^{\circ} + \delta_{m})/2)}{\sin(60^{\circ}/2)}$ $\frac{1.53}{1.33} = \frac{\sin((60^{\circ} + \delta_{m})/2)}{\sin(30^{\circ})}$ Since $\sin(30^{\circ}) = 0.5$,we have: $\sin((60^{\circ} + \delta_{m})/2) = 0.5 	imes \frac{1.53}{1.33} \approx 0.575$ Given $\sin(35.1^{\circ}) = 0.575$,we equate the angles: $(60^{\circ} + \delta_{m})/2 = 35.1^{\circ}$ $60^{\circ} + \delta_{m} = 70.2^{\circ}$ $\delta_{m} = 70.2^{\circ} - 60^{\circ} = 10.2^{\circ}$

$A$ prism of refractive index $1.53$ is placed in water of refractive index $1.33$. If the angle of prism is $60^{\circ}$,the angle of minimum deviation in water will be ........$^{\circ}$ $(\sin 35.1^{\circ} = 0.575)$

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