Two prisms A and B in contact with each other. The dispersive power of A and B is 0.04 and 0.03 respectively. The angular dispersion produced by 'A' is 80 . The angular dispersion produced by the combination, if the combination does not produce a net deviation is

d+d1=0;θω+θ1ω1=0;θω+−θ1ω1=d01−−θω×ω1−8∘×0.030.04−−60∴θNet=θ+θ1=8−60=20θ= angular dispersion

Two prisms A and B in contact with each other. The dispersive power of A and B is 0.04 and 0.03 respectively. The angular dispersion produced by 'A' is 80 . The angular dispersion produced by the combination, if the combination does not produce a net deviation is

d+d1=0;θω+θ1ω1=0;θω+−θ1ω1=d01−−θω×ω1−8∘×0.030.04−−60∴θNet=θ+θ1=8−60=20θ= angular dispersion

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Two prisms A and B in contact with each other. The dispersive power of A and B is 0.04 and 0.03 respectively. The angular dispersion produced by 'A' is 8⁰ . The angular dispersion produced by the combination, if the combination does not produce a net deviation is

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14⁰

8⁰

6⁰

2⁰

answer is D.

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Detailed Solution

$\begin{array}{l} d + d^{1} = 0; \frac{θ}{ω} + \frac{θ^{1}}{ω^{1}} = 0; \frac{θ}{ω} + \frac{- θ^{1}}{ω^{1}} = d \\ 0^{1} - \frac{- θ}{ω} \times ω^{1} - \frac{8^{\circ} \times 0.03}{0.04} - - 6^{0} \\ ∴ θ_{Net} = θ + θ^{1} = 8 - 6^{0} = 2^{0} \end{array}$
$θ = angular dispersion$