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The Minimum Deviations Suffered By, Yellow and Violet Beams Passing Through an Equilateral Transparent Prism Are 38.4°, 38.7° and 39.2° Respectively.

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प्रश्न

The 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.

योग
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उत्तर

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

\[\delta = (\mu - 1)A\]

\[\therefore\left( \mu - 1 \right) = \frac{\delta}{A}.........(1)\]

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

\[\omega = \frac{\frac{\delta_v}{A} - \frac{\delta_r}{A}}{\frac{\delta_y}{A}}\]

\[\Rightarrow \omega = \frac{\delta_v - \delta_r}{\delta_y} = \frac{(39 . 2) - (38 . 4)}{(38 . 7)}\]

\[\Rightarrow \omega = \frac{(0 . 8)}{38 . 7} = 0 . 0206\]

So, the dispersive power of the medium is 0.0206.

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  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 20: Dispersion and Spectra - Exercise [पृष्ठ ४४२]

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एचसी वर्मा Concepts of Physics Volume 1 and 2 [English] Class 11 and 12
अध्याय 20 Dispersion and Spectra
Exercise | Q 6 | पृष्ठ ४४२
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