Question

The angle of minimum deviation produced by a thin glass prism in air is 'δ'. If that prism is immersed in water, the angle of minimum deviation will be ______.

( _aμ_g = refractive index of glass w.r.t, air _aμ_w = refractive index of water w.r.t. air)

Options

• ${\displaystyle \delta \left[\frac{{}_a{\mu }_g-{}_a{\mu }_w}{{}_a{\mu }_w({}_a{\mu }_g+1)}\right]}$

• ${\displaystyle \delta \left[\frac{{}_a{\mu }_g-{}_a{\mu }_w}{{}_a{\mu }_w({}_a{\mu }_g-1)}\right]}$

• ${\displaystyle \delta \left[\frac{{}_a{\mu }_w-{}_a{\mu }_g}{{}_a{\mu }_w({}_a{\mu }_g-1)}\right]}$

• ${\displaystyle \delta \left[\frac{({}_a{\mu }_w-{}_a{\mu }_g)}{{}_a{\mu }_w\times {}_a{\mu }_g}\right]}$

Answer 1

The angle of minimum deviation produced by a thin glass prism in air is 'δ'. If that prism is immersed in water, the angle of minimum deviation will be ${\displaystyle \underline{\delta \left[\frac{{}_a{\mu }_g-{}_a{\mu }_w}{{}_a{\mu }_w({}_a{\mu }_g-1)}\right]}}$

Explanation:

δ = (µ - 1)

∴ δ_air = _aμ_g - 1

δ_water = _wμ_g - 1

${\displaystyle \therefore \frac{{\delta }_{\text{water}}}{{\delta }_{\text{air}}}=\frac{{}_w{\mu }_g-1}{{}_a{\mu }_g-1}=\frac{\frac{{}_a{\mu }_g}{{}_a{\mu }_w}-1}{{}_a{\mu }_g-1}=\frac{{}_a{\mu }_g-{}_a{\mu }_w}{{}_a{\mu }_w({}_a{\mu }_g-1)}}$