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(C) Initially,using similar triangles:$\frac{s}{3.8} = \frac{s + 30}{7.6}$$7.6s = 3.8s + 114$$3.8s = 114 \Rightarrow s = 30 \,cm$.Finally,when the tan

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$1.45$

Vedclass · Answer

(C) Initially,using similar triangles:$\frac{s}{3.8} = \frac{s + 30}{7.6}$$7.6s = 3.8s + 114$$3.8s = 114 \Rightarrow s = 30 \,cm$.Finally,when the tank is filled with liquid:The light ray from the source makes an angle $i$ with the normal at the left wall. The height of the ray from the central axis at the left wall is $1.9 \,cm$ (half of $3.8 \,cm$).$	an i = \frac{1.9}{s} = \frac{1.9}{30}$.After refraction,the ray makes an angle $r$ with the normal. The shadow width is $6.4 \,cm$,so the distance of the ray from the central axis at the right wall is $3.2 \,cm$. The height of the ray at the left wall is $1.9 \,cm$. The vertical shift of the ray inside the liquid is $3.2 - 1.9 = 1.3 \,cm$.$	an r = \frac{1.3}{30}$.Using Snell's law,$1 \cdot \sin i = n \cdot \sin r$. For small angles,$\sin 	heta \approx 	an 	heta$:$1 \cdot 	an i = n \cdot 	an r$$\frac{1.9}{30} = n \cdot \frac{1.3}{30}$$n = \frac{1.9}{1.3} \approx 1.46$.Therefore,the value of $n$ is closest to $1.45$.

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