Question

If a prism of dispersive power $0.021$ and refractive index $1.53$ forms an achromatic combination with a prism having the prism of angle ${4.2}^{\circ }$, dispersive power $0.045$ and refractive index $1.65,$ then what is the resultant deviation of the combination?

• A. ${1.66}^{\circ }$
• B. ${4.12}^{\circ }$
• C. ${4.2}^{\circ }$
• D. ${8.5}^{\circ }$

Answer 1

The correct option is D ${8.5}^{\circ }$

Given,

For first prism , $\omega =0.021;\mu =1.53$

For the other prism, ${\omega }^{{}^{\prime }}=0.045;{\mu }^{{}^{\prime }}=1.65;A^{{}^{\prime }}={4.2}^{\circ }$

For no dispersion,

$\omega \delta +{\omega }^{{}^{\prime }}{\delta }^{{}^{\prime }}=0$

$\Rightarrow \omega (\mu -1)A+{\omega }^{{}^{\prime }}({\mu }^{{}^{\prime }}-1)A^{\prime }=0\text{(or)}$

$A=\frac{{\omega }^{{}^{\prime }}{A}^{{}^{\prime }}({\mu }^{{}^{\prime }}-1)}{\omega (\mu -1)}=\frac{0.045\times {4.2}^{\circ }\times (1.65-1)}{0.021\times (1.53-1)}$

$\therefore A={11.04}^{\circ }$

Net deviation in this case,

$\delta +{\delta }^{\prime }=(\mu -1)A+({\mu }^{\prime }-1)A^{\prime }$

$={11.04}^{\circ }(1.53-1)+{4.2}^{\circ }(1.65-1)$

$={5.845}^{\circ }+{2.73}^{\circ }\approx {8.5}^{\circ }$

<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> Hence, $\left(D\right)$ is the correct answer.