A thin prism made of glass is dipped in water, then minimum deviation with respect to air by it will be $[g i v e n^{a} μ_{g} = \frac{3}{2} a n d^{a} μ_{w} = \frac{4}{3}]$ :

\frac{1}{2}

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\frac{1}{4}

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\frac{1}{8}

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\frac{1}{16}

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Solution

The correct option is B $\frac{1}{4}$
$δ = (μ - 1) A$
$δ_{a i r} = \frac{A}{2}$
$δ_{w a t e r} = \frac{A}{8}$

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Similar questions

Q. The angle of minimum deviation for a thin prism with respect to air and when dipped in water will be:

(_{a} μ_{g} = \frac{3}{2}

_{a} μ_{w} = \frac{4}{3})

Q. A thin prism of glass is placed in air and water successively. If

_{a} μ_{g} = 3 / 2

and

_{a} μ_{w} = 4 / 3

, then the ratio of deviations produced by the prism for a small angle of incidence when placed in air and water is:

Q. The refractive index of water with respect to air is

_{a} μ_{w}

and of glass with respect to air is

_{a} μ_{g}

. So,

\frac{_{a} μ_{g}}{_{a} μ_{w}}

should express:

Q. Angle of minimum deviation produced by a thin prism

(μ = 3 / 2)

4^{\circ}

. If the prism is immersed in water

(μ_{w} = 4 / 3)

, the angle of minimum deviation will be

The refractive index of water with respect to air is a $^{μ} w$ and a glass with respect to air is a $^{μ} g$ . Express the refractive index of glass with respect to water.

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A thin prism made of glass is dipped in water, then minimum deviation with respect to air by it will be [given aμg=32 and aμw=43]:

The correct option is B 14δ=(μ−1)A δair=A2 δwater=A8

A thin prism made of glass is dipped in water, then minimum deviation with respect to air by it will be $[g i v e n^{a} μ_{g} = \frac{3}{2} a n d^{a} μ_{w} = \frac{4}{3}]$ :

The correct option is B $\frac{1}{4}$
$δ = (μ - 1) A$
$δ_{a i r} = \frac{A}{2}$
$δ_{w a t e r} = \frac{A}{8}$