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

• `delta[(""_amu_g-""_amu_w)/(""_amu_w(""_amu_g+1))]`

• `delta[(""_amu_g-""_amu_w)/(""_amu_w(""_amu_g-1))]`

• `delta[(""_amu_w-""_amu_g)/(""_amu_w(""_amu_g-1))]`

• `delta[((""_amu_w-""_amu_g))/(""_amu_wxx""_amu_g)]`

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 `underline(delta[(""_amu_g-""_amu_w)/(""_amu_w(""_amu_g-1))])`

Explanation:

δ = (µ - 1)

∴ δ_air = _aμ_g - 1

δ_water = _wμ_g - 1

`thereforedelta_"water"/delta_"air"=(""_wmu_g-1)/(""_amu_g-1)=((""_amu_g)/(""_amu_w)-1)/(""_amu_g-1)=(""_amu_g-""_amu_w)/(""_amu_w(""_amu_g-1))`