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.

Solution

Given,

$Refracting angle of prism, A = 60 ° Angle of minimum deviation, δm = 40 °$
$Refractive index of glass w . r . t air,^{a} μ_{g} = \frac{Sin \frac{(A + δm)}{2}}{SinA / 2}$
$\Rightarrow$    ${}_{a}μ_{g} = \frac{\sin 50 °}{\sin 30 °} = \frac{0.766}{0.54} = 1.532$

After the prism is placed in water,

Refractive index of water = 1.33
Using the formula,

   ${}_{ω}μ_{g} = \frac{\sin (\frac{A + δ'_{m}}{2})}{Sin A / 2}$

   $\frac{{}^{a}μ_{g}}{{}^{a}μ_{w}} = \frac{\sin (\frac{60 + δ'_{m}}{2})}{\sin 30 °}$

$\sin (30 ° + \frac{δ'_{m}}{2}) = \frac{1}{2} \times \frac{1.532}{1.33} = 0.5759$

$\Rightarrow$ $30 ° + \frac{δ'_{m}}{2} = 35 ° 10'$

$∴$    $\frac{A + δ'_{m}}{2} = 35^{o} 10'$
$A + δ'_{m} = 70^{o} 20^{'} δ'_{m} = 70^{o} 20^{'} - A = 70^{0} 20^{'} - 60^{o}$

i.e., $δ'_{m} = 10 ° 20' .$

which is the reqiured angle of minimum deviation.

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