Question

Two prisms $\text{A}$ and $\text{B}$ have dispersive powers of $0.012$ and $0.018$ respectively. The two prisms are in contact with each other. The prism $\text{A}$ produces a mean deviation of ${1.2}^{\circ }$, then the mean deviation produced by $\text{B}$ if the combination produces deviation without dispersion is

• A. ${3.6}^{\circ }$
• B. ${0.8}^{\circ }$
• C. ${0.4}^{\circ }$
• D. ${1.8}^{\circ }$

Answer 1

The correct option is B ${0.8}^{\circ }$

We have the dispresive powers of the prisms as

$w_A=0.012$
$w_B=0.018$

Also, the mean deviation produced by prism $\text{A}$ is

$d_A=1.2$

We have to determine mean deviation produed by prism $\text{B}$, i.e. $d_B$

We know that

$\text{Dispersive power}=\frac{\text{Angular Dispersion}}{\text{Mean deviation}}$

$w_A=\frac{{\theta }_A}{d_A}$

$w_B=\frac{{\theta }_B}{d_B}$

For deviation without dispersion difference in angular dispersion for the two prisms will be zero.

${\theta }_A-{\theta }_B=0$

${\theta }_A={\theta }_B$

$\Rightarrow w_A{d}_A=w_B{d}_B$

$0.012\times 1.2=0.018\times d_B$

$d_B=\frac{1.2\times 12}{18}={0.8}^{\circ }$