NEET-XII-Physics
20: dispersion and Spectra
- #6The minimum deviations suffered by, yellow and violet beams passing through an equilateral transparent prism are 38.4°, 38.7° and 39.2° respectively. Calculate the dispersive power of the medium.Ans : Given:
Minimum deviations suffered by
Red beam, δr = 38.4°
Yellow beam, δy = 38.7°
Violet beam, δv = 39.2°
If A is the angle of prism having refractive index μ, then the angle of minimum deviation is given by
`` \delta =(\mu -1)A``
`` \Rightarrow ```` \left(\mu -1\right)=\frac{\delta }{A}`` ...(1)
Dispersive power `` \left(\omega \right)`` is given by
`` \omega =\frac{{\mu }_{v}-{\mu }_{r}}{{\mu }_{y}-1}``
`` =\frac{({\mu }_{v}-1)-({\mu }_{r}-1)}{({\mu }_{y}-1)}``
From equation (1), we get:
`` \omega =\frac{{\displaystyle \frac{{\delta }_{v}}{A}-\frac{{\delta }_{r}}{A}}}{{\displaystyle \frac{{\delta }_{y}}{A}}}``
`` \Rightarrow \omega =\frac{{\delta }_{v}-{\delta }_{r}}{{\delta }_{y}}=\frac{(39.2)-(38.4)}{(38.7)}``
`` ``
`` \Rightarrow \omega =\frac{(0.8)}{38.7}=0.0206``
So, the dispersive power of the medium is 0.0206.
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