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  • Two beams of light are incident normally on water (R.I. = 4/3). If the beam 1 passes through a glass <img alt="\mathrm{(R.I. =3 / 2)}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm

Two beams of light are incident normally on water (R.I. = 4/3). If the beam 1 passes through a glass \mathrm{(R.I. =3 / 2)} slab of height h as shown in the figure, the time difference for both the beams for reaching the bottom is

Option: 1

Zero


Option: 2

\frac{h^{\prime}}{6 C}


Option: 3

\frac{6 h}{C}


Option: 4

\frac{h}{6 C}


Answers (1)

best_answer

The refractive index of glass is greater than that of water. Therefore the speed of light in glass is lesser than that of water. It is given as

\mathrm{v}=\frac{\mathrm{C}}{\mathrm{n}}$ where $\mathrm{C}=3 \times 10^{8} \mathrm{~m} / \mathrm{s}
\mathrm{v}= speed of light in a medium of R.I.\mathrm{n}

\therefore The time difference for the rays

=\mathrm{t}_{1}-\mathrm{t}_{2}=\frac{\mathrm{h}}{\mathrm{v}_{\mathrm{g}}}-\frac{\mathrm{h}}{\mathrm{v}_{\mathrm{w}}}
\Rightarrow \Delta \mathrm{t}=\frac{\mathrm{h}}{\left(\mathrm{C} / \mathrm{n}_{\mathrm{g}}\right)}-\frac{\mathrm{h}}{\left(\mathrm{C} / \mathrm{n}_{\mathrm{w}}\right)}=\frac{\mathrm{h}}{\mathrm{C}}\left(\mathrm{n}_{\mathrm{g}}-\mathrm{n}_{\mathrm{w}}\right)
\Rightarrow \quad \Delta \mathrm{t}=\frac{\mathrm{h}}{\mathrm{C}}\left(\frac{3}{2}-\frac{4}{3}\right)=\frac{\mathrm{h}}{6 \mathrm{C}}

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