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(A) The light ray travels from the liquid (denser medium) to the air (rarer medium). Since the observer can just see the corner $E$,the ray emerging f

Vedclass · Accepted Answer

$1.2$

Vedclass · Answer

(A) The light ray travels from the liquid (denser medium) to the air (rarer medium). Since the observer can just see the corner $E$,the ray emerging from the liquid surface grazes the surface,meaning the angle of refraction is $90^{\circ}$. Thus,the angle of incidence is equal to the critical angle $C$.From the geometry of the tank,the depth is $3 \ cm$ and the width is $4 \ cm$. The hypotenuse of the triangle formed by the light ray is $\sqrt{3^2 + 4^2} = \sqrt{9 + 16} = 5 \ cm$.From the figure,$\sin C = \frac{	ext{opposite}}{	ext{hypotenuse}} = \frac{4}{5} = 0.8$.The refractive index $\mu$ is given by $\mu = \frac{1}{\sin C}$.Therefore,$\mu = \frac{1}{0.8} = \frac{5}{4} = 1.2$.

When a rectangular metal tank is filled to the top with an unknown liquid,an observer with eyes level with the top of the tank can just see the corner $E$. $A$ light ray that refracts towards the observer at the top surface of the liquid is shown. The refractive index of the liquid is:

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