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If a thin prism of glass is dipped into water then minimum deviation (with respect to air) of light produced by prism will be left \left( _{a}{{\mu }_{g}}=\frac{3}{2}\text{ and}{{\,}_{a}}{{\mu }_{w}}\,=\,\frac{4}{3} \right)[UPSEAT 1999]A) \frac{1}{2} B) \frac{1}{4} C) 2 D) \frac{1}{5} |
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Answer» FORMULA of deviation is given by δ = (μ - 1)A Where δ denotes the deviation , A is the ANGLE of PRISM and μ is the REFRACTIVE index of prism with respect to medium . Here apply , Here μa denotes refractive index of air e.g., μa = 1 μg denotes refractive index of glass e.g., μg = 3/2 μw denotes refractive index of WATER e.g,. μw = 4/3 Now, δa/δw = (3/2 - 1)/(3/2/4/3 - 1) = (1 /2)/(9/8 - 1) = 4 Hence, minimum deviation in water = 1/4th of minimum deviation in air |
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