A lens has a power of +5 D in air. What will be its power if completely immersed in water? (refractive index of water w.r.t air = 4/3 and refractive index of glass w.r.t. air = 3/2)

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A lens has a power of +5 D in air. What will be its power if completely immersed in water? (refractive index of water w.r.t air = 4/3 and refractive index of glass w.r.t. air = 3/2)

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Answer 1 · 2018-02-15T14:51:21

A lens with radius of curvature $R_{1}$ $R_1$ & $R_{2}$ $R_2$ having relative refractive index of $μ$ $mu$ ,if have a power of $D$ $D$ ,then these are related as,

$D = (μ - 1) ((\frac{1}{R_{1}}) - (\frac{1}{R_{2}})) = (μ - 1) \times k$ $D= (mu-1)((1/(R_1))-(1/(R_2)))=(mu-1)×k$ (let)

Now for the lens in air,we can write,

$5 = ((\frac{\frac{3}{2}}{1}) - 1) \times k$ $5 = (((3/2)/1)-1) × k$

Or, $k = 10$ $k=10$

Now,when it is put in water, $μ = \frac{\frac{3}{2}}{\frac{4}{3}} = (\frac{9}{8})$ $mu= (3/2)/(4/3)=(9/8)$

So,if now the power becomes $D$ $D$ then, $D = ((\frac{9}{8}) - 1) \times k = \frac{1}{8} \times 10 = 1.25 D$ $D=((9/8)-1)×k = 1/8 × 10=1.25 D$

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A lens has a power of +5 D in air. What will be its power if completely immersed in water? (refractive index of water w.r.t air = 4/3 and refractive index of glass w.r.t. air = 3/2)

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