For a thin prism, δ 1 is the angle of deviation produced, when prism is placed in air. When the prism is immersed in water, the angle of deviation produced is δ 2 . Given a μ g = 2 3 and a

Question

For a thin prism, δ 1 ​ is the angle of deviation produced, when prism is placed in air. When the prism is immersed in water, the angle of deviation produced is δ 2 ​ . Given a ​ μ g ​ = 2 3 ​ and a ​ μ w ​ = 3 4 ​ . The ratio δ 2 ​ : δ 1 ​ is

TestHub · Accepted Answer

The angle of deviation for a thin prism is given by $δ = (μ - 1) A$ , where $μ$ is the refractive index of the prism relative to the surrounding medium.

When the prism is in air, the deviation is:
$δ_{1} = (_{a} μ_{g} - 1) A$
$δ_{1} = (\frac{3}{2} - 1) A = \frac{A}{2}$

When the prism is immersed in water, the refractive index of glass relative to water is:
$_{w} μ_{g} = \frac{_{a} μ _{g}}{_{a} μ _{w}} = \frac{3/2}{4/3} = \frac{9}{8}$

The deviation in water is:
$δ_{2} = (_{w} μ_{g} - 1) A$
$δ_{2} = (\frac{9}{8} - 1) A = \frac{A}{8}$

The ratio of the deviations is:
$\frac{δ _{2}}{δ _{1}} = \frac{A /8}{A /2} = \frac{1}{4}$

Thus, $δ_{2} : δ_{1} = 1 : 4$ .

Answer: $1 : 4$

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