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(B) When the tank is accelerated horizontally with acceleration $a$,a pseudo force acts on the liquid particles. The free surface of the liquid become

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slightly greater than $5/3$

Vedclass · Answer

(B) When the tank is accelerated horizontally with acceleration $a$,a pseudo force acts on the liquid particles. The free surface of the liquid becomes inclined at an angle $	heta$ with the horizontal,where $	an 	heta = a/g$.The normal to the surface makes an angle $	heta$ with the vertical. Since the incident ray is vertical,the angle of incidence $i$ at the liquid-air interface is equal to $	heta$.For total internal reflection $(TIR)$ to occur,the angle of incidence must be greater than the critical angle $C$ $(i > C)$.Therefore,$\sin i > \sin C$,which implies $\sin i > 1/\mu$.From the geometry,$	an 	heta = a/g = 7.5/10 = 3/4$. Thus,$\sin 	heta = 3/5$.Since $i = 	heta$,we have $\sin i = 3/5$.Substituting this into the inequality: $3/5 > 1/\mu \implies \mu > 5/3$.Thus,the minimum possible refractive index is slightly greater than $5/3$.

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