Q.1) Find refractive index of glass,if the speed of light in glass is 2×10⁸ and speed of light in air is 3×10⁸ m/s.
2)The far point of a myopic person is 150 cm in front of the eye.Calculate the focal length and the power of a lens required to enable him to see distant object clearly ?




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Refractive\ index= \eta\\\\ \eta= \frac{speed\ of\ light\ in\ vacuum(air)}{speed\ of\ light\ in\ medium} \\ \\ \eta= \frac{3 \times 10^8}{2 \times 10^8} \\\\ \eta= \frac{3}{2}=\boxed{1.5}

Far point = 150 cm
in order to see the distant objects clearly, the focal length of the lens should be equal to the far point of the person. The corrective lens used are concave lenses.
so focal length = -150cm = -1.5 m

Power= \frac{1}{f}= \frac{1}{-1.5}= \frac{-2}{3}D= \boxed{-0.67D}
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