NEET-XII-Physics
20: dispersion and Spectra
- #3The focal lengths of a convex lens for red, yellow and violet rays are 100 cm, 98 cm and 96 cm respectively. Find the dispersive power of the material of the lens.Ans : Focal lengths of the convex lens:
For red rays, `` {f}_{r}=100\,\mathrm{\,cm\,}``
For yellow rays, `` {f}_{y}=98\,\mathrm{\,cm\,}``
For violet rays, `` {f}_{v}=96\,\mathrm{\,cm\,}``
Let:
`` {\mu }_{r}`` = Refractive index for the red colour
`` {\mu }_{y}`` = Refractive index for the yellow colour
`` {\mu }_{v}`` = Refractive index for the violet colour
Focal length of a lens `` \left(f\right)`` is given by
`` \frac{1}{f}=\left(\mu -1\right)\left(\frac{1}{{R}_{1}}-\frac{1}{{R}_{2}}\right)``
Here, `` \mu `` is the refractive index and R1 and R2 are the radii of curvatures of the lens.
Thus, we have:
`` \left(\mu -1\right)=\frac{1}{f}\times \frac{1}{\left({\displaystyle \frac{1}{{R}_{1}}}-{\displaystyle \frac{1}{{R}_{2}}}\right)}``
`` \Rightarrow \left(\mu -1\right)=\frac{k}{f}\left[k=\frac{1}{\left({\displaystyle \frac{1}{{R}_{1}}}-{\displaystyle \frac{1}{{R}_{2}}}\right)}\right]``
For red rays, `` {\mu }_{r}-1=\frac{k}{100}``
For yellow rays, `` {\,\mathrm{\,\mu \,}}_{y}-1=\frac{k}{98}``
For violet rays, `` {\,\mathrm{\,\mu \,}}_{v}-1=\frac{k}{96}``
Dispersive power (ω) is given by
`` \omega =\frac{{\,\mathrm{\,\mu \,}}_{v}-{\,\mathrm{\,\mu \,}}_{r}}{{\,\mathrm{\,\mu \,}}_{y}-1}``
`` \,\mathrm{\,Or\,},\omega =\frac{({\,\mathrm{\,\mu \,}}_{v}-1)-({\,\mathrm{\,\mu \,}}_{r}-1)}{({\,\mathrm{\,\mu \,}}_{y}-1)}``
Substituting the values, we get:
`` \omega =\frac{{\displaystyle \frac{k}{96}}-{\displaystyle \frac{k}{100}}}{{\displaystyle \frac{k}{98}}}=\frac{98\times 4}{9600}``
`` \Rightarrow \omega =0.0408``
Thus, the dispersive power of the material of the lens is 0.048.
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