A girl looks through a circular glass slab (r | vedclass

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(B) The girl can observe only those light rays which are refracted and leave the glass slab at an angle of $90^{\circ}$ or less,as shown in the figure

Vedclass · Accepted Answer

$160$

Vedclass · Answer

(B) The girl can observe only those light rays which are refracted and leave the glass slab at an angle of $90^{\circ}$ or less,as shown in the figure.Applying Snell's law at the water-air interface (considering the critical angle condition for the widest rays entering the slab):$n_w \sin r = n_a \sin 90^{\circ}$$\frac{4}{3} \sin r = 1 	imes 1$$\sin r = \frac{3}{4}$Using the trigonometric identity $	an r = \frac{\sin r}{\sqrt{1 - \sin^2 r}}$:$	an r = \frac{3/4}{\sqrt{1 - (3/4)^2}} = \frac{3/4}{\sqrt{7/16}} = \frac{3}{\sqrt{7}}$From the geometry of the problem,the radius of the visible area at the bottom of the pool is $R = x + r_{slab}$,where $r_{slab} = 0.3 \,m$ (radius of the slab) and $x = h 	an r$.$x = 6 	imes \frac{3}{\sqrt{7}} \approx 6 	imes \frac{3}{2.645} \approx 6.8 \,m$.The total radius of the visible circular area at the bottom is $R = 6.8 + 0.3 = 7.1 \,m$.The area $A = \pi R^2 = \pi 	imes (7.1)^2 \approx 3.14 	imes 50.41 \approx 158.3 \,m^2$.Rounding to the nearest given option,the area is approximately $160 \,m^2$.

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