Q142 / 160 · TS EAMCET 2018 (04 May Shift 2) · Physics

Figure shows a ray of light entering and passing through a dense glass slab and emerging from the other side. If the angle $i=60^{\circ}$, slab thickness $b=0.04 \mathrm{~m}$ and the refractive index of glass $=\sqrt{3}$, the parallel shift $d$ between the emerging and entering rays in $\mathrm{mm}$ is Figure

  • A
    $\sqrt{\frac{3}{4}}$
  • B
    $\sqrt{\frac{4}{3}}$
  • Correct answer
    C
    $\frac{40}{\sqrt{3}}$
  • D
    $15 \sqrt{3}$

Solution

We have, $\sin r=\frac{\sin i}{\mu}=\frac{\sin 60^{\circ}}{\sqrt{3}}=\frac{1}{2}$ $\Rightarrow$ Angle of refraction, $r=30^{\circ}$ Now, lateral shift $d$ is given by $$ \begin{aligned} & d=t \sec (r) \cdot \sin (i-r) \\ & =0.04 \times \sec \left(30^{\circ}\right) \cdot \sin \left(60^{\circ}-30^{\circ}\right)=0.04 \times \tan 30^{\circ} \\ & =\frac{0.04}{\sqrt{3}} \mathrm{~m}=\frac{0.04 \times 1000}{\sqrt{3}} \mathrm{~mm}=\frac{40}{\sqrt{3}} \mathrm{~mm} \end{aligned} $$

Hence, the correct answer is option: 'C'.

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