A glass prism of refractive index $1.5$ is placed in water of refractive index $1.33$ . The minimum value of the angle of the prism so that it will not be possible to have any emergent ray is :

150^{\circ}

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125^{\circ}

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165^{\circ}

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180^{\circ}

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Solution

The correct option is D $125^{\circ}$
$μ_{p r i s m} =$ $1.5$
$μ_{w a t e r}$ $= 1.33$

$μ_{r} = \frac{1.5}{1.33}$

$θ_{c r i c} = s i n^{- 1} (\frac{1}{μ_{r}}) = s i n^{- 1} (\frac{1.33}{1.5}) = {62.5}^{\circ}$

Maximum possible angle A is given by: $A = 2 θ_{c r i c} = 125^{\circ}$

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Similar questions

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A prism is made of glass of unknown refractive index. A parallel beam of light is incident on a face of the prism. The angle of minimum deviation is measured to be 40°. What is the refractive index of the material of the prism? The refracting angle of the prism is 60°. If the prism is placed in water (refractive index 1.33), predict the new angle of minimum deviation of a parallel beam of light.

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A glass prism of refractive index 1.5 is placed in water of refractive index 1.33. The minimum value of the angle of the prism so that it will not be possible to have any emergent ray is :

The correct option is D 125∘μprism= 1.5 μwater =1.33μr=1.51.33θcric=sin−1(1μr)=sin−1(1.331.5)=62.5∘Maximum possible angle A is given by: A=2θcric=125∘

A glass prism of refractive index $1.5$ is placed in water of refractive index $1.33$ . The minimum value of the angle of the prism so that it will not be possible to have any emergent ray is :

The correct option is D $125^{\circ}$
$μ_{p r i s m} =$ $1.5$
$μ_{w a t e r}$ $= 1.33$

$μ_{r} = \frac{1.5}{1.33}$

$θ_{c r i c} = s i n^{- 1} (\frac{1}{μ_{r}}) = s i n^{- 1} (\frac{1.33}{1.5}) = {62.5}^{\circ}$

Maximum possible angle A is given by: $A = 2 θ_{c r i c} = 125^{\circ}$