Question

Find the angle of a prism of dispersive power 0.021 and refractive index 1.52 to form an achromatic combination with a

prism of angle 4.2º and dispersive power 0.045 having refractive index 1.65.​

Answer 1

Answer:

$For achromatic combination,</p><p></p><p>(\mu _{v,1} - \mu_{r,1}) \times A_{1} = (\mu_{v,2} - \mu_{r,2}) \times A_{2}(μv,1−μr,1)×A1=(μv,2−μr,2)×A2</p><p></p><p></p><p>Also dispersive power  \omega = (\mu _{v} - \mu _{r})/ (\mu_{y} - 1)ω=(μv−μr)/(μy−1)</p><p></p><p></p><p>So \omega_{1} \times (\mu_{y,1} - 1) \times A_{1} = \omega_{2} \times (\mu_{y,2} - 1) \times A_{2}ω1×(μy,1−1)×A1=ω2×(μy,2−1)×A2</p><p></p><p>=> 0.021 \times 0.52 \times A_{1} = 0.036 \times 0.65 \times 4.20.021×0.52×A1=0.036×0.65×4.2</p><p></p><p>=> A_{1} = 9^{\circ}A1=9∘</p><p></p><p></p><p>Resultant deviation is |(\mu_{1} - 1) \times A_{1} - (\mu_{2} - 1) \times A_{2}|∣(μ1−1)×A1−(μ2−1)×A2∣</p><p></p><p>        = |(1.52 -1 ) \times 9 - (1.65 -1) \times 4.2|=∣(1.52−1)×9−(1.65−1)×4.2∣</p><p></p><p>        = 1.95^o=1.95o</p><p></p><p>$

Answer 2

Explanation:

For achromatic combination, (\mu _{v,1} - \mu_{r,1}) \times A_{1} = (\mu_{v,2} - \mu_{r,2}) \times A_{2}(μv,1−μr,1)×A1=(μv,2−μr,2)×A2 Also dispersive power \omega = (\mu _{v} - \mu _{r})/ (\mu_{y} - 1)ω=(μv−μr)/(μy−1) So \omega_{1} \times (\mu_{y,1} - 1) \times A_{1} = \omega_{2} \times (\mu_{y,2} - 1) \times A_{2}ω1×(μy,1−1)×A1=ω2×(μy,2−1)×A2 => 0.021 \times 0.52 \times A_{1} = 0.036 \times 0.65 \times 4.20.021×0.52×A1=0.036×0.65×4.2 => A_{1} = 9^{\circ}A1=9∘ Resultant deviation is |(\mu_{1} - 1) \times A_{1} - (\mu_{2} - 1) \times A_{2}|∣(μ1−1)×A1−(μ2−1)×A2∣ = |(1.52 -1 ) \times 9 - (1.65 -1) \times 4.2|=∣(1.52−1)×9−(1.65−1)×4.2∣ = 1.95^o=1.95o \