NEET-XII-Physics
20: dispersion and Spectra
- #4The refractive index of a material changes by 0.014 as the colour of the light changes from red to violet. A rectangular slab of height 2.00 cm made of this material is placed on a newspaper. When viewed normally in yellow light, the letters appear 1.32 cm below the top surface of the slab. Calculate the dispersive power of the material.Ans : Given:
Difference in the refractive indices of violet and red lights = 0.014
Let μv and μr be the refractive indices of violet and red colours.
Thus, we have:
μv - μr = 0.014
Now,
Real depth of the newspaper = 2.00 cm
Apparent depth of the newspaper = 1.32 cm
`` \,\mathrm{\,Refractive\,}\,\mathrm{\,index\,}=\frac{\,\mathrm{\,Real\,}\,\mathrm{\,depth\,}}{\,\mathrm{\,Apparent\,}\,\mathrm{\,depth\,}}``
`` \therefore \,\mathrm{\,Refractive\,}\,\mathrm{\,index\,}\,\mathrm{\,for\,}\,\mathrm{\,yellow\,}\,\mathrm{\,light\,}\left({\mu }_{\,\mathrm{\,y\,}}\right)\,\mathrm{\,is\,}\,\mathrm{\,given\,}\,\mathrm{\,by\,}``
`` {\mu }_{y}=\frac{2.00}{1.32}=1.515``
`` \,\mathrm{\,Also\,},``
`` \,\mathrm{\,Dispersive\,}\,\mathrm{\,power\,},\omega =\frac{{\mu }_{v}-{\mu }_{r}}{{\mu }_{y}-1}``
`` =\frac{0.014}{1.515-1}``
`` \,\mathrm{\,Or\,},\omega =\frac{0.014}{0.515}=0.027``
Thus, the dispersive power of the material is 0.027.
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