NEET-XII-Physics
20: dispersion and Spectra
- #5A thin prism is made of a material having refractive indices 1.61 and 1.65 for red and violet light. The dispersive power of the material is 0.07. It is found that a beam of yellow light passing through the prism suffers a minimum deviation of 4.0° in favourable conditions. Calculate the angle of the prism.Ans : The refractive indices for red and yellow lights are μr = 1.61 and μy = 1.65, respectively.
Dispersive power, ω = 0.07
Angle of minimum deviation, δy = 4°
`` \,\mathrm{\,Now\,},\,\mathrm{\,using\,}\,\mathrm{\,the\,}\,\mathrm{\,relation\,}\omega =\frac{{\mu }_{v}-{\mu }_{r}}{{\mu }_{y}-1},\,\mathrm{\,we\,}\,\mathrm{\,get\,}:``
`` \Rightarrow 0.07=\frac{1.65-1.61}{{\mu }_{y}-1}``
`` \Rightarrow {\mu }_{y}-1=\frac{0.04}{0.07}=\frac{4}{7}``
Let the angle of the prism be A.
Angle of minimum deviation, δ = (μ - 1)A
`` \Rightarrow A=\frac{{\delta }_{y}}{{\mu }_{y}-1}=\frac{4}{\left({\displaystyle \frac{4}{7}}\right)}=7°``
Thus, the angle of the prism is 7`` °``.
Page No 442: