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When a picture drawn on paper is seen through a slab of a transparent materials of thickness 5cm5cm, it appear to be raised by 1.5cm1.5 cm. Critical angle at the boundary of this transparent material and air is

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Shift =t(1−1μ)=t\left(1-\cfrac{1}{\mu}\right)

=1.5=5(1−1μ)=1.5=5\left(1-\cfrac{1}{\mu}\right)

0.3=(1−1μ)0.3=\left(1-\cfrac{1}{\mu}\right)

1μ=0.7\cfrac{1}{\mu}=0.7

μ=107\mu =\cfrac{10}{7}

critical angle =sin⁡−11μ=sin⁡−1(710)=\sin^{-1}{\cfrac{1}{\mu}}=\sin^{-1}{\left(\cfrac{7}{10}\right)}

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